ARM Assembly Language Programming

Peter Knaggs

April 7, 2016

Warning: Work in progress

As a work in progress this book will contain errors

Preface

Broadly speaking, you can divide the history of computers into four periods: the mainframe, the
mini, the microprocessor, and the modern post-microprocessor. The mainframe era was charac-
terized by computers that required large buildings and teams of technicians and operators to keep
them going. More often than not, both academics and students had little direct contact with the
mainframe—you handed a deck of punched cards to an operator and waited for the output to ap-
pear hours later. During the mainfame era, academics concentrated on languages and compilers,
algorithms, and operating systems.

The minicomputer era put computers in the hands of students and academics, because university
departments could now buy their own minis. As minicomputers were not as complex as main-
frames and because students could get direct hands-on experience, many departments of computer
science and electronic engineering taught students how to program in the native language of the
In those days, the mid 1970s, assembly language programming
computer—assembly language.
was used to teach both the control of I/O devices, and the writing of programs (i.e., assembly
language was taught rather like high level languages). The explosion of computer software had
not taken place, and if you wanted software you had to write it yourself.

The late 1970s saw the introduction of the microprocessor. For the first time, each student was
able to access a real computer. Unfortunately, microprocessors appeared before the introduction
of low-cost memory (both primary and secondary). Students had to program microprocessors
in assembly language because the only storage mechanism was often a ROM with just enough
capacity to hold a simple single-pass assembler.

The advent of the low-cost microprocessor system (usually on a single board) ensured that virtually
every student took a course on assembly language. Even today, most courses in computer science
include a module on computer architecture and organization, and teaching students to write
programs in assembly language forces them to understand the computer’s architecture. However,
some computer scientists who had been educated during the mainframe era were unhappy with
the microprocessor, because they felt that the 8-bit microprocessor was a retrograde step—its
architecture was far more primitive than the mainframes they had studied in the 1960s.

The 1990s is the post-microprocessor era. Today’s personal computers have more power and
storage capacity than many of yesterday’s mainframes, and they have a range of powerful software
tools that were undreamed of in the 1970s. Moreover, the computer science curriculum of the
1990s has exploded. In 1970 a student could be expected to be familiar with all field of computer
science. Today, a student can be expected only to browse through the highlights.

The availability of high-performance hardware and the drive to include more and more new ma-
terial in the curriculum, has put pressure on academics to justify what they teach. In particular,
many are questioning the need for courses on assembly language.

If you regard computer science as being primarily concerned with the use of the computer, you
can argue that assembly language is an irrelevance. Does the surgeon study metallurgy in order
to understand how a scalpel operates? Does the pilot study thermodynamics to understand how
a jet engine operates? Does the news reader study electronics to understand how the camera

i

ii

PREFACE

operates? The answer to all these questions is “no”. So why should we inflict assembly language
and computer architecture on the student?

First, education is not the same as training. The student of computer science is not simply being
trained to use a number of computer packages. A university course leading to a degree should
also cover the history and the theoretical basis for the subject. Without a knowledge of computer
architecture, the computer scientist cannot understand how computers have developed and what
they are capable of.

Is assembly language today the same as assembly language yesterday?

Two factors have influenced the way in which we teach assembly language—one is the way in which
microprocessors have changed, and the other is the use to which assembly language teaching is
put. Over the years microprocessors have become more and more complex, with the result that
the architecture and assembly language of a modern state-of-the-art microprocessor is radically
different to that of an 8-bit machine of the late 1970s. When we first taught assembly language in
the 1970s and early 1980s, we did it to demonstrate how computers operated and to give students
hands-on experience of a computer. Since all students either have their own computer or have ac-
cess to a computer lab, this role of the single-board computer is now obsolete. Moreover, assembly
language programming once attempted to ape high-level language programming— students were
taught algorithms such as sorting and searching in assembly language, as if assembly language
were no more than the (desperately) poor person’s C.

The argument for teaching assembly language programming today can be divided into two com-
ponents: the underpinning of computer architecture and the underpinning of computer software.

Assembly language teaches how a computer works at the machine (i.e., register) level. It is there-
fore necessary to teach assembly language to all those who might later be involved in computer
architecture—either by specifying computers for a particular application, or by designing new
architectures. Moreover, the von Neumann machine’s sequential nature teaches students the limi-
tation of conventional architectures and, indirectly, leads them on to unconventional architectures
(parallel processors, Harvard architectures, data flow computers, and even neural networks).

It is probably in the realm of software that you can most easily build a case for the teaching of
assembly language. During a student’s career, he or she will encounter a lot of abstract concepts in
subjects ranging from programming languages, to operating systems, to real-time programming,
to AI. The foundation of many of these concepts lies in assembly language programming and
computer architecture. You might even say that assembly language provides bottom-up support
for the top-down methodology we teach in high-level languages. Consider some of the following
examples (taken from the teaching of Advanced RISC Machines Ltd (ARM) assembly language).

Data types

Students come across data types in high-level languages and the effects of strong and weak
data typing. Teaching an assembly language that can operate on bit, byte, word and long
word operands helps students understand data types. Moreover, the ability to perform any
type of assembly language operation on any type of data structure demonstrates the need
for strong typing.

Addressing modes

A vital component of assembly language teaching is addressing modes (literal, direct, and
indirect). The student learns how pointers function and how pointers are manipulated. This
aspect is particularly important if the student is to become a C programmer. Because an
assembly language is unencumbered by data types, the students’ view of pointers is much
simplified by an assembly language. The ARM has complex addressing modes that support
direct and indirect addressing, generated jump tables and handling of unknown memory
offsets.

PREFACE

The stack and subroutines

iii

How procedures are called, and parameters passed and returned from procedures. By using
an assembly language you can readily teach the passing of parameters by value and by
reference. The use of local variables and re-entrant programming can also be taught. This
supports the teaching of task switching kernels in both operating systems and real-time
programming.

Recursion

The recursive calling of subroutines often causes a student problems. You can use an assem-
bly language, together with a suitable system with a tracing facility, to demonstrate how
recursion operates. The student can actually observe how the stack grows as procedures are
called.

Run-time support for high-level languages

A high-performance processor like the ARM provides facilities that support run-time check-
ing in high-level languages. For example, the programming techniques document lists a
series of programs that interface with ’C’ and provide run-time checking for errors such as
an attempt to divide a number by zero.

Protected-mode operation

Members of the ARM family operate in either a priviledge mode or a user mode. The
operating system operates in the priviledge mode and all user (applications) programs run in
the user mode. This mechanism can be used to construct secure or protected environments in
which the effects of an error in one application can be prevented from harming the operating
system (or other applications).

Input-output

Many high-level languages make it difficult to access I/O ports and devices directly. By
using an assembly language we can teach students how to write device drivers and how to
control interfaces. Most real interfaces are still programmed at the machine level by accessing
registers within them.

All these topics can, of course, be taught in the appropriate courses (e.g., high-level languages,
operating systems). However, by teaching them in an assembly language course, they pave the
way for future studies, and also show the student exactly what is happening within the machine.

Conclusion

A strong case can be made for the continued teaching of assembly language within the computer
science curriculum. However, an assembly language cannot be taught just as if it were another
general-purpose programming language as it was once taught ten years ago. Perhaps more than
any other component of the computer science curriculum, teaching an assembly language supports
a wide range of topics at the heart of computer science. An assembly language should not be used
just to illustrate algorithms, but to demonstrate what is actually happening inside the computer.

iv

Acknowledgements

PREFACE

As usual there are many people without whom this book could not exist. The first of these
is Stephen Welsh, without whom I may never have started this project and for reading and
commenting on many revisions of the text. Andrew Main for finding me the time to work on it.
Andrew Watson for his support, comments and ideas.

The students of the Computing and Software Engineering Management courses for unwittingly
debugging the examples, exercises, and the development of the main text.

Contents

Preface

Contents

List of Programs

1 Introduction

Instruction Code Mnemonics

1.6.1 Additional Features of Assemblers
1.6.2 Choosing an Assembler

1.1 The Meaning of Instructions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.1.1 Binary Instructions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 A Computer Program . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3 The Binary Programming Problem . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.4 Using Octal or Hexadecimal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.5
1.6 The Assembler Program . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.7 Disadvantages of Assembly Language . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.8 High-Level Languages
1.8.1 Advantages of High-Level Languages . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . .
1.8.2 Disadvantages of High-Level Languages
1.9 Which Level Should You Use? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.9.1 Applications for Machine Language . . . . . . . . . . . . . . . . . . . . . . .
1.9.2 Applications for Assembly Language . . . . . . . . . . . . . . . . . . . . . .
1.9.3 Applications for High-Level Language . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.9.4 Other Considerations
1.10 Why Learn Assembler? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2 Assemblers
2.1 Fields

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.1.1 Delimiters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.1.2 Labels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Operation Codes (Mnemonics)
. . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3 Directives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . .
2.3.1 The DEFINE CONSTANT (Data) Directive
2.3.2 The EQUATE Directive . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3.3 The AREA Directive . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3.4 Housekeeping Directives . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3.5 When to Use Labels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.4 Operands and Addresses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.4.1 Decimal Numbers
2.4.2 Other Number Systems
. . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.4.3 Names . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.4.4 Character Codes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

v

i

v

xi

1
1
1
1
2
2
3
4
4
5
5
6
6
7
8
8
8
8
8
8

11
11
11
12
14
14
14
15
16
16
17
17
17
17
18
18

vi

CONTENTS

2.4.5 Arithmetic and Logical Expressions
2.4.6 General Recommendations

. . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . .
2.5 Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.6 Types of Assemblers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.7 Errors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.8 Loaders

3 ARM Architecture
3.1 Processor modes
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2 Registers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2.1 The stack pointer, SP or R13 . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2.2 The Link Register, LR or R14 . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2.3 The program counter, PC or R15 . . . . . . . . . . . . . . . . . . . . . . . .
3.2.4 Current Processor Status Registers: CPSR . . . . . . . . . . . . . . . . . . .
3.3 Flags . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.4 Exceptions
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.5 Register Transfer Language . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.5.1 Memory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . .
3.5.2 Arithmetic and Logic Unit
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Instruction Fetch . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.6.1
3.6.2
Instruction Decode . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.6.3 Operand Fetch . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.6.4 Execute . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.6.5 Operand Store . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.6.6

3.6 Control Unit

4 Instruction Set

4.3 Flow Control

4.1 Data Movement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.1.1
Set Flags Variant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.1.2 Conditional Execution: (cid:104)cc(cid:105) . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2 Arithmetic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2.1 Addition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2.2
Subtraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2.3 Multiplication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2.4 Division . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.3.1 Comparisons
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.3.2 Branching . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Jumping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.3.3
4.4 Memory Access . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.4.1 Load and Store Register . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.4.2 Load and Store Register Byte . . . . . . . . . . . . . . . . . . . . . . . . . .
4.4.3 Load and Store Multiple registers . . . . . . . . . . . . . . . . . . . . . . . .
4.5 Logical and Bit Manipulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.6 System Control / Privileged . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.6.1
Software Interrupt . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.6.2
Semaphores . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Status Register Access . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.6.3
4.6.4 Coprocessor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.6.5 Privileged Memory Access . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.6.6 Undefined Instructions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

5 Addressing Modes

18
18
19
19
20
21

23
23
25
26
27
27
28
28
29
30
31
31
33
34
35
36
36
36
36

39
41
41
42
43
43
43
43
44
44
44
45
45
46
46
46
46
47
48
48
48
48
49
49
50

51

CONTENTS

5.1 Data Processing Operands: (cid:104)op1 (cid:105) . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.1.1 Unmodified Value
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.1.2 Logical Shift Left . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.1.3 Logical Shift Right . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.1.4 Arithmetic Shift Right . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.1.5 Rotate Right . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.1.6 Rotate Right Extended . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2 Memory Access Operands: (cid:104)op2 (cid:105) . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2.1 Offset Addressing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2.2 Pre-Index Addressing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2.3 Post-Index Addressing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

6 Programs

6.1 Example Programs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.1.1 Program Listing Format . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.1.2 Guidelines for Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.2 Trying the examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.3 Trying the examples from the command line . . . . . . . . . . . . . . . . . . . . . .
Setting up TextPad . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.4 Program Initialization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.5 Special Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.6 Problems

6.3.1

7 Data Movement

7.1 Program Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.1.1
16-Bit Data Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.1.2 One’s Complement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
32-Bit Addition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.1.3
7.1.4
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Shift Left One Bit
7.1.5 Byte Disassembly . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.1.6 Find Larger of Two Numbers . . . . . . . . . . . . . . . . . . . . . . . . . .
64-Bit Adition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.1.7
7.1.8 Table of Factorials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.2 Problems
64-Bit Data Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.2.1
32-Bit Subtraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.2.2
7.2.3
Shift Right Three Bits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.2.4 Halfword Assembly . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.2.5 Find Smallest of Three Numbers . . . . . . . . . . . . . . . . . . . . . . . .
Sum of Squares . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.2.6
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Shift Left n bits
7.2.7

8 Logic

8.1 Program Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.1.1 Find Larger of Two Numbers . . . . . . . . . . . . . . . . . . . . . . . . . .
8.1.2
64-Bit Adition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.1.3 Table of Factorials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.2 Problems
64-Bit Data Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.2.1
32-Bit Subtraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.2.2
8.2.3
Shift Right Three Bits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.2.4 Halfword Assembly . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.2.5 Find Smallest of Three Numbers . . . . . . . . . . . . . . . . . . . . . . . .
Sum of Squares . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.2.6

vii

51
51
51
52
52
53
54
54
55
56
57

59
59
59
59
60
61
62
63
63
63

65
65
65
66
67
68
69
69
71
72
73
73
73
73
73
74
74
74

75
75
75
76
77
78
78
78
79
79
79
79

viii

CONTENTS

8.2.7

Shift Left n bits

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

80

9 Program Loops

9.1 Program Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Sum of Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.1.1
9.1.2
. . . . . . . . . . . . . . . . . . . . . . . . . . . . .
64-bit Sum of Numbers
9.1.3 Number of negative elements . . . . . . . . . . . . . . . . . . . . . . . . . .
9.1.4 Find Maximum Value . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . .
9.1.5 Normalise A Binary Number
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.2.1 Checksum of data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.2.2 Number of Zero, Positive, and Negative numbers . . . . . . . . . . . . . . .
9.2.3 Find Minimum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.2.4 Count 1 Bits
. . . . . . . . . . . . . . . . . . . . . . . . .
9.2.5 Find element with most 1 bits

9.2 Problems

81
82
82
83
84
85
86
87
87
88
88
88
88

10 Strings

10.1 Handling data in ASCII . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.2 A string of characters
10.2.1 Fixed Length Strings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.2.2 Terminated Strings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.2.3 Counted Strings
10.3 International Characters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.4 Program Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.4.1 Length of a String of Characters . . . . . . . . . . . . . . . . . . . . . . . .
10.4.2 Find First Non-Blank Character
. . . . . . . . . . . . . . . . . . . . . . . .
10.4.3 Replace Leading Zeros with Blanks . . . . . . . . . . . . . . . . . . . . . . .
10.4.4 Add Even Parity to ASCII Chatacters . . . . . . . . . . . . . . . . . . . . .
10.4.5 Pattern Match . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

91
91
92
93
93
93
94
94
94
95
96
97
98
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100
10.5.1 Length of a Teletypewriter Message
. . . . . . . . . . . . . . . . . . . . . . 100
10.5.2 Find Last Non-Blank Character . . . . . . . . . . . . . . . . . . . . . . . . . 100
10.5.3 Truncate Decimal String to Integer Form . . . . . . . . . . . . . . . . . . . 100
10.5.4 Check Even Parity and ASCII Characters . . . . . . . . . . . . . . . . . . . 101
10.5.5 String Comparison . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101

10.5 Problems

11 Code Conversion

103
11.1 Program Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103
11.1.1 Hexadecimal to ASCII . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103
11.1.2 Decimal to Seven-Segment . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104
11.1.3 ASCII to Decimal
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105
11.1.4 Binary-Coded Decimal to Binary . . . . . . . . . . . . . . . . . . . . . . . . 106
11.1.5 Binary Number to ASCII String . . . . . . . . . . . . . . . . . . . . . . . . 107
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
11.2.1 ASCII to Hexadecimal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
11.2.2 Seven-Segment to Decimal . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108
11.2.3 Decimal to ASCII
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108
11.2.4 Binary to Binary-Coded-Decimal . . . . . . . . . . . . . . . . . . . . . . . . 108
11.2.5 Packed Binary-Coded-Decimal to Binary String . . . . . . . . . . . . . . . . 109
11.2.6 ASCII string to Binary number . . . . . . . . . . . . . . . . . . . . . . . . . 109

11.2 Problems

12 Arithmetic

111
12.1 Program Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111
12.1.2 64-Bit Addition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

CONTENTS

ix

12.2 Problems

12.1.3 Decimal Addition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112
12.1.4 Multiplication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113
12.1.5 32-Bit Binary Divide . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115
12.2.1 Multiple precision Binary subtraction . . . . . . . . . . . . . . . . . . . . . 115
12.2.2 Decimal Subtraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115
12.2.3 32-Bit by 32-Bit Multiply . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

13 Tables and Lists

117
13.1 Program Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117
13.1.1 Add Entry to List
13.1.2 Check an Ordered List . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118
13.1.3 Remove an Element from a Queue . . . . . . . . . . . . . . . . . . . . . . . 119
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120
13.1.4 Sort a List
. . . . . . . . . . . . . . . . . . . . . . . . . 121
13.1.5 Using an Ordered Jump Table
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121
13.2.1 Remove Entry from List . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121
13.2.2 Add Entry to Ordered List
. . . . . . . . . . . . . . . . . . . . . . . . . . . 121
13.2.3 Add Element to Queue . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121
13.2.4 4-Byte Sort . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122
13.2.5 Using a Jump Table with a Key . . . . . . . . . . . . . . . . . . . . . . . . 122

13.2 Problems

14 Subroutines

123
14.1 Types of Subroutines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123
14.2 Subroutine Documentation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124
14.3 Parameter Passing Techniques . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124
14.3.1 Passing Parameters In Registers
. . . . . . . . . . . . . . . . . . . . . . . . 125
14.3.2 Passing Parameters In A Parameter Block . . . . . . . . . . . . . . . . . . . 125
14.3.3 Passing Parameters On The Stack . . . . . . . . . . . . . . . . . . . . . . . 126
14.4 Types Of Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126
14.5 Program Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133
14.6 Problems
14.6.1 ASCII Hex to Binary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133
14.6.2 ASCII Hex String to Binary Word . . . . . . . . . . . . . . . . . . . . . . . 133
14.6.3 Test for Alphabetic Character . . . . . . . . . . . . . . . . . . . . . . . . . . 133
14.6.4 Scan to Next Non-alphabetic . . . . . . . . . . . . . . . . . . . . . . . . . . 134
14.6.5 Check Even Parity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134
14.6.6 Check the Checksum of a String . . . . . . . . . . . . . . . . . . . . . . . . 134
14.6.7 Compare Two Counted Strings . . . . . . . . . . . . . . . . . . . . . . . . . 135

A ARM Instruction Definitions

137
A.1 ADC: Add with Carry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137
A.2 ADD: Add . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138
A.3 AND: Bitwise AND . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139
A.4 B, BL: Branch, Branch and Link . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139
A.5 BIC: Bit Clear . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140
A.6 CMN: Compare Negative
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140
A.7 CMP: Compare . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141
A.8 EOR: Exclusive OR . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141
A.9 LDM: Load Multiple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141
A.10 LDR: Load Register . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142
A.11 LDRB: Load Register Byte . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143
A.12 MLA: Multiply Accumulate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143
A.13 MOV: Move . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144

x

CONTENTS

A.14 MRS: Move to Register from Status . . . . . . . . . . . . . . . . . . . . . . . . . . . 144
A.15 MSR: Move to Status from Register . . . . . . . . . . . . . . . . . . . . . . . . . . . 145
A.16 MUL: Multiply . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146
A.17 MVN: Move Negative
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147
A.18 ORR: Bitwise OR . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147
A.19 RSB: Reverse Subtract
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147
A.20 RSC: Reverse Subtract with Carry . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148
A.21 SBC: Subtract with Carry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149
A.22 STM: Store Multiple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149
A.23 STR: Store Register . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150
A.24 STRB: Store Register Byte . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151
A.25 SUB: Subtract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151
A.26 SWI: Software Interrupt
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152
A.27 SWP: Swap . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153
A.28 SWPB: Swap Byte . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153
A.29 TEQ: Test Equivalence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154
A.30 TST: Test

B ARM Instruction Summary

155

List of Programs

move16.s
invert.s

7.1
7.2
7.3a add.s
7.3b add2.s
7.4
7.5
7.6
7.7
7.8

16bit data transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Find the one’s compliment (inverse) of a number . . . . . . . . . . . .
Add two numbers
. . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Add two numbers and store the result . . . . . . . . . . . . . . . . . .
Shift Left one bit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
shiftleft.s
Disassemble a byte into its high and low order nibbles . . . . . . . . .
nibble.s
Find the larger of two numbers . . . . . . . . . . . . . . . . . . . . . .
bigger.s
64 bit addition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
add64.s
factorial.s Lookup the factorial from a table by using the address of the memory
location . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

8.1
8.2
8.3

Find the larger of two numbers . . . . . . . . . . . . . . . . . . . . . .
bigger.s
64 bit addition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
add64.s
factorial.s Lookup the factorial from a table by using the address of the memory
location . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

bigsum.s

9.1a sum1.s
9.1b sum2.s
9.2
9.3a cntneg1.s
9.3b cntneg2.s
9.4
largest.s
9.5a normal1.s
9.5b normal2.s

. . . . . . . . . . . . . . . . . . . . . .
Add a table of 32 bit numbers
. . . . . . . . . . . . . . . . . . . . . .
Add a table of 32 bit numbers
. . . . . . . . . .
Add a table of 32-bit numbers into a 64-bit number
. .
Count the number of negative values in a table of 32-bit numbers
. .
Count the number of negative values in a table of 32-bit numbers
Find largest unsigned 32 bit value in a table
. . . . . . . . . . . . . .
Normalize a binary number . . . . . . . . . . . . . . . . . . . . . . . .
Normalise a binary number . . . . . . . . . . . . . . . . . . . . . . . .

65
66
67
67
68
69
70
71

72

75
76

77

82
83
83
84
84
85
86
87

Find the length of a Carage Return terminated string . . . . . . . . .
10.1a strlencr.s
10.1b strlen.s
Find the length of a null terminated string . . . . . . . . . . . . . . .
10.2 skipblanks.s Find first non-blank . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Supress leading zeros in a string . . . . . . . . . . . . . . . . . . . . .
10.3 padzeros.s
Set the parity bit on a series of characters store the amended string in
10.4 setparity.s
Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Compare two counted strings for equality . . . . . . . . . . . . . . . .
Compare null terminated strings for equality assume that we have no
knowledge of the data structure so we must assess the individual strings 98

10.5a cstrcmp.s
10.5b strcmp.s

94
95
95
96

97
98

11.1a nibtohex.s
Convert a single hex digit to its ASCII equivalent
11.1b wordtohex.s Convert a 32 bit hexadecimal number to an ASCII string and output

. . . . . . . . . . . 103

to the terminal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104
Convert a decimal number to seven segment binary . . . . . . . . . . 104
11.2 nibtoseg.s
11.3 dectonib.s
. . . . . . . . . . . . 105
Convert an ASCII numeric character to decimal
11.4a ubcdtohalf.s Convert an unpacked BCD number to binary . . . . . . . . . . . . . . 106
11.4b ubcdtohalf2.s Convert an unpacked BCD number to binary using MUL . . . . . . . 106
Store a 16bit binary number as an ASCII string of ’0’s and ’1’s . . . . 107
11.5 halftobin.s

xi

xii

12.2 add64.s
12.3 addbcd.s
12.4a mul16.s
12.4b mul32.s
12.5 divide.s

13.1a insert.s

13.1b insert2.s

13.2 search.s
13.3 head.s
13.4 sort.s

14.1a init1.s
14.1b init2.s
14.1c init3.s
14.1d init3a.s
14.1e byreg.s

14.1f bystack.s

LIST OF PROGRAMS

64 Bit Addition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111
Add two packed BCD numbers to give a packed BCD result
. . . . . 112
16 bit binary multiplication . . . . . . . . . . . . . . . . . . . . . . . . 113
Multiply two 32 bit number to give a 64 bit result (corrupts R0 and R1)113
Divide a 32 bit binary no by a 16 bit binary no store the quotient and
remainder there is no ’DIV’ instruction in ARM! . . . . . . . . . . . . 114

Examine a table for a match - store a new entry at the end if no match
found . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117
Examine a table for a match - store a new entry if no match found
extends insert.s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118
Examine an ordered table for a match . . . . . . . . . . . . . . . . . . 118
Remove the first element of a queue . . . . . . . . . . . . . . . . . . . 119
Sort a list of values – simple bubble sort . . . . . . . . . . . . . . . . . 120

Initiate a simple stack . . . . . . . . . . . . . . . . . . . . . . . . . . . 127
Initiate a simple stack . . . . . . . . . . . . . . . . . . . . . . . . . . . 127
Initiate a simple stack . . . . . . . . . . . . . . . . . . . . . . . . . . . 128
Initiate a simple stack . . . . . . . . . . . . . . . . . . . . . . . . . . . 128
A simple subroutine example program passes a variable to the routine
in a register . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129
A more complex subroutine example program passes variables to the
routine using the stack . . . . . . . . . . . . . . . . . . . . . . . . . . 129
A 64 bit addition subroutine . . . . . . . . . . . . . . . . . . . . . . . 131
. . . . . . . . . . . . . 132

14.1g add64.s
14.1h factorial.s A subroutine to find the factorial of a number

1 Introduction

A computer program is ultimately a series of numbers and therefore has very little meaning to a
human being. In this chapter we will discuss the levels of human-like language in which a computer
program may be expressed. We will also discuss the reasons for and uses of assembly language.

1.1 The Meaning of Instructions

The instruction set of a microprocessor is the set of binary inputs that produce defined actions
during an instruction cycle. An instruction set is to a microprocessor what a function table is to a
logic device such as a gate, adder, or shift register. Of course, the actions that the microprocessor
performs in response to its instruction inputs are far more complex than the actions that logic
devices perform in response to their inputs.

1.1.1 Binary Instructions

An instruction is a binary digit pattern — it must be available at the data inputs to the micropro-
cessor at the proper time in order to be interpreted as an instruction. For example, when the ARM
receives the binary pattern 111000000100 as the input during an instruction fetch operation, the
pattern means subtract. Similary the microinstruction 111000001000 means add. Thus the 32
bit pattern 11100000010011101100000000001111 means:

“Subtract R15 from R14 and put the answer in R12.”

The microprocessor (like any other computer) only recognises binary patterns as instructions or
data; it does not recognise characters or octal, decimal, or hexadecimal numbers.

1.2 A Computer Program

A program is a series of instructions that causes a computer to perform a particular task.

Actually, a computer program includes more than instructions, it also contains the data and the
memory addresses that the microprocessor needs to accomplish the tasks defined by the instruc-
tions. Clearly, if the microprocessor is to perform an addition, it must have two numbers to add
and a place to put the result. The computer program must determine the sources of the data and
the destination of the result as well as the operation to be performed.

All microprocessors execute instructions sequentially unless an instruction changes the order of
execution or halts the processor. That is, the processor gets its next instruction from the next
higher memory address unless the current instruction specifically directs it to do otherwise.

1

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CHAPTER 1.

INTRODUCTION

Ultimately, every program is a set of binary numbers. For example, this is a snippet of an ARM
program that adds the contents of memory locations 809416 and 809816 and places the result in
memory location 809C16:

11100101100111110001000000010000
11100101100111110001000000001000
11100000100000010101000000000000
11100101100011110101000000001000

This is a machine language, or object, program. If this program were entered into the memory of
an ARM-based microcomputer, the microcomputer would be able to execute it directly.

1.3 The Binary Programming Problem

There are many difficulties associated with creating programs as object, or binary machine lan-
guage, programs. These are some of the problems:

• The programs are difficult to understand or debug.

(Binary numbers all look the same,

particularly after you have looked at them for a few hours.)

• The programs do not describe the task which you want the computer to perform in anything

resembling a human-readable format.

• The programs are long and tiresome to write.

• The programmer often makes careless errors that are very difficult to locate and correct.

For example, the following version of the addition object program contains a single bit error. Try
to find it:

11100101100111110001000000010000
11100101100111110001000000001000
11100000100000010101000000000000
11100110100011110101000000001000

Although the computer handles binary numbers with ease, people do not. People find binary
programs long, tiresome, confusing, and meaningless. Eventually, a programmer may start re-
membering some of the binary codes, but such effort should be spent more productively.

1.4 Using Octal or Hexadecimal

We can improve the situation somewhat by writing instructions using octal or hexadecimal num-
bers, rather than binary. We will use hexadecimal numbers because they are shorter, and because
they are the standard for the microprocessor industry. Table 1.1 defines the hexadecimal digits
and their binary equivalents. The ARM program to add two numbers now becomes:

E59F1010
E59f0008
E0815000
E58F5008

At the very least, the hexadecimal version is shorter to write and not quite so tiring to examine.

Errors are somewhat easier to find in a sequence of hexadecimal digits. The erroneous version of
the addition program, in hexadecimal form, becomes:

1.5.

INSTRUCTION CODE MNEMONICS

3

Hexadecimal
Digit
0
1
2
3
4
5
6
7
8
9
A
B
C
D
E
F

Binary
Equivalent
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111

Decimal
Equivalent
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15

Table 1.1: Hexadecimal Conversion Table

E59F1010
E59f0008
E0815000
E68F5008

The mistake is far more obvious.

The hexadecimal version of the program is still difficult to read or understand; for example, it
does not distinguish operations from data or addresses, nor does the program listing provide any
suggestion as to what the program does. What does 3038 or 31C0 mean? Memorising a card full
of codes is hardly an appetising proposition. Furthermore, the codes will be entirely different for
a different microprocessor and the program will require a large amount of documentation.

1.5

Instruction Code Mnemonics

An obvious programming improvement is to assign a name to each instruction code. The instruc-
tion code name is called a “mnemonic” or memory jogger.

In fact, all microprocessor manufacturers provide a set of mnemonics for the microprocessor in-
struction set (they cannot remember hexadecimal codes either). You do not have to abide by the
manufacturer’s mnemonics; there is nothing sacred about them. However, they are standard for
a given microprocessor, and therefore understood by all users. These are the instruction codes
that you will find in manuals, cards, books, articles, and programs. The problem with selecting
instruction mnemonics is that not all instructions have “obvious” names. Some instructions do
(for example, ADD, AND, ORR), others have obvious contractions (such as SUB for subtraction, EOR
for exclusive-OR), while still others have neither. The result is such mnemonics as BIC, STMIA,
and even MRS. Most manufacturers come up with some reasonable names and some hopeless ones.
However, users who devise their own mnemonics rarely do much better.

Along with the instruction mnemonics, the manufacturer will usually assign names to the CPU
registers. As with the instruction names, some register names are obvious (such as A for Accumu-
lator) while others may have only historical significance. Again, we will use the manufacturer’s
suggestions simply to promote standardisation.

If we use standard ARM instruction and register mnemonics, as defined by Advanced RISC Ma-
chines, our ARM addition program becomes:

4

CHAPTER 1.

INTRODUCTION

LDR
LDR
ADD
STR

R1, num1
R0, num2
R5, R1, R0
R5, num3

The program is still far from obvious, but at least some parts are comprehensible. ADD is a
considerable improvement over E59F. The LDR mnemonic does suggest loading data into a register
or memory location. We now see that some parts of the program are operations and others are
addresses. Such a program is an assembly language program.

1.6 The Assembler Program

How do we get the assembly language program into the computer? We have to translate it, either
into hexadecimal or into binary numbers. You can translate an assembly language program by
hand, instruction by instruction. This is called hand assembly.

The following table illustrates the hand assembly of the addition program:

Instruction Mnemonic Register/Memory Location Hexadecimal Equivalent
R1, num1
R0, num2
R5, R1, R0
R5, num3

E59F1010
E59F0008
E0815000
E58F5008

LDR
LDR
ADD
STR

Hand assembly is a rote task which is uninteresting, repetitive, and subject to numerous minor
errors. Picking the wrong line, transposing digits, omitting instructions, and misreading the codes
are only a few of the mistakes that you may make. Most microprocessors complicate the task even
further by having instructions with different lengths. Some instructions are one word long while
others may be two or three. Some instructions require data in the second and third words; others
require memory addresses, register numbers, or who knows what?

Assembly is a rote task that we can assign to the microcomputer. The microcomputer never
makes any mistakes when translating codes; it always knows how many words and what format
each instruction requires. The program that does this job is an “assembler.” The assembler
program translates a user program, or “source” program written with mnemonics, into a machine
language program, or “object” program, which the microcomputer can execute. The assembler’s
input is a source program and its output is an object program.

Assemblers have their own rules that you must learn. These include the use of certain markers
(such as spaces, commas, semicolons, or colons) in appropriate places, correct spelling, the proper
control of information, and perhaps even the correct placement of names and numbers. These
rules are usually simple and can be learned quickly.

1.6.1 Additional Features of Assemblers

Early assemblers did little more than translate the mnemonic names of instructions and registers
into their binary equivalents. However, most assemblers now provide such additional features as:

• Allowing the user to assign names to memory locations, input and output devices, and even

sequences of instructions

• Converting data or addresses from various number systems (for example, decimal or hex-
adecimal) to binary and converting characters into their ASCII or EBCDIC binary codes

• Performing some arithmetic as part of the assembly process

1.7. DISADVANTAGES OF ASSEMBLY LANGUAGE

5

• Telling the loader program where in memory parts of the program or data should be placed

• Allowing the user to assign areas of memory as temporary data storage and to place fixed

data in areas of program memory

• Providing the information required to include standard programs from program libraries, or

programs written at some other time, in the current program

• Allowing the user to control the format of the program listing and the input and output

devices employed

1.6.2 Choosing an Assembler

All of these features, of course, involve additional cost and memory. Microcomputers generally
have much simpler assemblers than do larger computers, but the tendency is always for the size of
assemblers to increase. You will often have a choice of assemblers. The important criterion is not
how many off-beat features the assembler has, but rather how convenient it is to use in normal
practice.

1.7 Disadvantages of Assembly Language

The assembler does not solve all the problems of programming. One problem is the tremendous gap
between the microcomputer instruction set and the tasks which the microcomputer is to perform.
Computer instructions tend to do things like add the contents of two registers, shift the contents
of the Accumulator one bit, or place a new value in the Program Counter. On the other hand, a
user generally wants a microcomputer to do something like print a number, look for and react to
a particular command from a teletypewriter, or activate a relay at the proper time. An assembly
language programmer must translate such tasks into a sequence of simple computer instructions.
The translation can be a difficult, time-consuming job.

Furthermore, if you are programming in assembly language, you must have detailed knowledge of
the particular microcomputer that you are using. You must know what registers and instructions
the microcomputers has, precisely how the instructions affect the various registers, what addressing
methods the computer uses, and a mass of other information. None of this information is relevant
to the task which the microcomputer must ultimately perform.

In addition, assembly language programs are not portable. Each microcomputer has its own
assembly language which reflects its own architecture. An assembly language program written for
the ARM will not run on a 486, Pentium, or Z8000 microprocessor. For example, the addition
program written for the Z8000 would be:

LD
ADD
LD

R0,%6000
R0,%6002
%6004,R0

The lack of portability not only means that you will not be able to use your assembly language
program on a different microcomputer, but also that you will not be able to use any programs that
were not specifically written for the microcomputer you are using. This is a particular drawback
for new microcomputers, since few assembly language programs exist for them. The result, too
frequently, is that you are on your own. If you need a program to perform a particular task, you
are not likely to find it in the small program libraries that most manufacturers provide. Nor are
you likely to find it in an archive, journal article, or someone’s old program File. You will probably
have to write it yourself.

6

CHAPTER 1.

INTRODUCTION

1.8 High-Level Languages

The solution to many of the difficulties associated with assembly language programs is to use,
insted, high-level or procedure-oriented langauges. Such languages allow you to describe tasks in
forms that are problem-oriented rather than computer-oriented. Each statement in a high-level
language performs a recognisable function; it will generally correspond to many assembly language
instruction. A program called a compiler translates the high-level language source program into
object code or machine language instructions.

Many different hgih-level languages exist for different types of tasks.
If, for exampe, you can
express what you want the computer to do in algebraic notation, you can write your FORTRAN
(For mula Translation Language), the oldest of the high-level languages. Now, if you want to add
two numbers, you just tell the computer:

sum = num1 + num2;

That is a lot simpler (and shorter) than either the equivalent machine language program or the
equivalent assembly language program. Other high-level languages include COBOL (for business
applications), BASIC (a cut down version of FORTRAN designed to prototype ideas before codeing
them in full), C (a systems-programming language), C++ and JAVA (object-orientated general
development languages).

1.8.1 Advantages of High-Level Languages

Clearly, high-level languages make program easier and faster to write. A common estimate is
that a programmer can write a program about ten times as fast in a high-level langauge as in
assembly language. That is just writing the program; it does not include problem definition,
program design, debugging testing or documentation, all of which become simpler and faster. The
high-level language program is, for instance, partly self-documenting. Even if you do not know
FORTRAN, you could probably tell what the statement illustrated above does.

Machine Independence

High-level languages solve many other problems associated with assembly language programming.
The high-level language has its own syntax (usually defined by an international standard). The
language does not mention the instruction set, registers, or other features of a particular computer.
The compiler takes care of all such details. Programmers can concentrate on their own tasks; they
do not need a detailed understanding of the underlying CPU architecture — for that matter, they
do not need to know anything about the computer the are programming.

Portability

Programs written in a high-level language are portable — at least, in theory. They will run on
any computer that has a standard compiler for that language.

At the same time, all previous programs written in a high-level language for prior computers and
available to you when programming a new computer. This can mean thousands of programs in
the case of a common language like C.

1.8. HIGH-LEVEL LANGUAGES

7

1.8.2 Disadvantages of High-Level Languages

If all the good things we have said about high-level languages are true — if you can write programs
faster and make them portable besides — why bother with assebly languages? Who wants to
worry about registers, instruction codes, mnemonics, and all that garbage! As usual, there are
disadvantages that balance the advantages.

Syntax

One obvious problem is that, as with assembly language, you have to learn the “rules” or syntax
of any high-level language you want to use. A high-level langauge has a fairly complicated set of
rules. You will find that it takes a lot of time just to get a program that is syntactically correct
(and even then it probably will not do what you want). A high-level computer language is like
If you have talent, you will get used to the rules and be able to turn out
a foreign language.
programs that the compiler will accept. Still, learning the rules and trying to get the program
accepted by the compiler does not contribute directly to doing your job.

Cost of Compilers

Another obvious problem is that you need a compiler to translate program written in a high-level
language into machine language. Compilers are expensive and use a large amount of memory.
While most assemblers occupy only a few KBytes of memory, compilers would occupy far larger
amounts of memory. A compiler could easily require over four times as much memory as an
assembler. So the amount of overhead involved in using the compiler is rather large.

Adapting Tasks to a Language

Furthermore, only some compilers will make the implementation of your task simpler. Each
language has its own target proglem area, for example, FORTRAN is well-suited to problems
that can be expressed as algebraic formulas.
If however, your problem is controlling a display
terminal, editing a string of characters, or monitoring an alarm system, your problem cannot
be easily expressed. In fact, formulating the solution in FORTRAN may be more awkward and
more difficult than formulating it in assembly language. The answer is, of course, to use a more
suitable high-level language. Languages specifically designed for tasks such as those mentioned
above do exist — they are called system implementation languages. However, these languages are
less widely used.

Inefficiency

High-level languages do not produce very efficient machine language program. The basic reason
for this is that compilation is an automatic process which is riddled with compromises to allow for
many ranges of possibilities. The compiler works much like a computerised language translator —
sometimes the words are right but the sentence structures are awkward. A simpler compiler connot
know when a variable is no longer being used and can be discarded, when a register should be
used rather than a memory location, or when variables have simple relationships. The experienced
programmer can take advantage of shortcuts to shorten execution time or reduce memory usage.
A few compiler (known as optimizing cmpilers) can also do this, but such compilers are much
larger than regular compilers.

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CHAPTER 1.

INTRODUCTION

1.9 Which Level Should You Use?

Which language level you use depends on your particulr application. Let us briefly note some of
the factors which may favor particular levels:

1.9.1 Applications for Machine Language

Virtually no one programs in machine language because it wastes human time and is difficult to
document. An assembler costs very little and greatly reduces programming time.

1.9.2 Applications for Assembly Language

• Limited data processing

• Short to moderate-sized programs

• High-volume applications

• Application where memory cost is a factor

• Real-Time control applications

• Applications involving more input/output

or control than computation

1.9.3 Applications for High-Level Language

• Long programs

• Compatibility with similar applications

using larger computers

• Low-volume applications

• Applications involing more computation

than input/output or control

• Programs which are expected
to undergo many changes

• Applications where the amout of memory

required is already very large

• Availability of a specific program in a high-level language which can be used

in the application.

1.9.4 Other Considerations

Many other factors are also important, such as the availability of a large computer for use in
development, experience with particular languages, and compatibility with other applications.

If hardware will ultimately be the largest cost in your application, or if speed is critical, you should
favor assembly language. But be prepared to spend much extra time in software development in
exchange for lower memory costs and higher execution speeds. If software will be the largest cost
in your application, you should favor a high-level language. But be prepared to spend the extra
money required for the supporting hardware and software.

Of course, no one except some theorists will object if you use both assembly and high-level lan-
guages. You can write the program originally in a high-level language and then patch some sections
in assembly language. However, most users prefer not to do this because it can create havoc in
debugging, testing, and documentation.

1.10 Why Learn Assembler?

Given the advance of high-level languages, why do you need to learn assembly language program-
ming? The reasons are:

1.10. WHY LEARN ASSEMBLER?

9

1. Most industrial microcomputer users program in assembly language.

2. Many microcomputer users will continue to program in assembly language since they need

the detailed control that it provides.

3. No suitable high-level language has yet become widely available or standardised.

4. Many application require the efficiency of assembly language.

5. An understanding of assembly language can help in evaluating high-level languages.

6. Almost all microcomputer programmers ultimately find that they need some knowledge of
assembly language, most often to debug programs, write I/O routines, speed up or shorten
critical sections of programs written in high-level languages, utilize or modify operating
system functions, and undertand other people’s programs.

The rest of these notes will deal exclusively with assembler and assembly language programming.

10

CHAPTER 1.

INTRODUCTION

2 Assemblers

This chapter discusses the functions performed by assemblers, beginning with features common
to most assemblers and proceeding through more elaborate capabilities such as macros and con-
ditional assembly. You may wish to skim this chapter for the present and return to it when you
feel more comfortable with the material.

As we mentioned, today’s assemblers do much more than translate assembly language mnemonics
into binary codes. But we will describe how an assembler handles the translation of mnemonics
before describing additional assembler features. Finally we will explain how assemblers are used.

2.1 Fields

Assembly language instructions (or “statements”) are divided into a number of “fields”.

The operation code field is the only field which can never he empty; it always contains either an
instruction mnemonic or a directive to the assembler, sometimes called a “pseudo-instruction,”
“pseudo-operation,” or “pseudo-op.”

The operand or address field may contain an address or data, or it may be blank.

The comment and label fields are optional. A programmer will assign a label to a statement or
add a comment as a personal convenience: namely, to make the program easier to read and use.

Of course, the assembler must have some way of telling where one field ends and another begins.
Assemblers often require that each field start in a specific column. This is a “fixed format.”
However, fixed formats are inconvenient when the input medium is paper tape; fixed formats are
also a nuisance to programmers. The alternative is a “free format” where the fields may appear
anywhere on the line.

2.1.1 Delimiters

If the assembler cannot use the position on the line to tell the fields apart, it must use something
else. Most assemblers use a special symbol or “delimiter” at the beginning or end of each field.

Label
Field

VALUE1
VALUE2
RESULT
START

NEXT:

Operation Code
or Mnemonic
Field

Operand or
Address
Field

Comment Field

DCW
DCW
DCW
MOV
ADD
STR
?

0x201E
0x0774
1
R0, VALUE1
R0, R0, VALUE2
RESULT, R0
?

;FIRST VALUE
;SECOND VALUE
;16-BIT STORAGE FOR ADDITION RESULT
;GET FIRST VALUE
;ADD SECOND VALUE TO FIRST VALUE
;STORE RESULT OF ADDITION
;NEXT INSTRUCTION

11

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CHAPTER 2. ASSEMBLERS

label (cid:104)whitespace(cid:105) instruction (cid:104)whitespace(cid:105) ; comment

whitespace Between label and operation code, between operation code and ad-

comma
asterisk
semicolon

dress, and before an entry in the comment field
Between operands in the address field
Before an entire line of comment
Marks the start of a comment on a line that contains preceding code

Table 2.1: Standard ARM Assembler Delimiters

The most common delimiter is the space character. Commas, periods, semicolons, colons, slashes,
question marks, and other characters that would not otherwise be used in assembly language
programs also may serve as delimiters. The general form of layout for the ARM assembler is:

You will have to exercise a little care with delimiters. Some assemblers are fussy about extra spaces
or the appearance of delimiters in comments or labels. A well-written assembler will handle these
minor problems, but many assemblers are not well-written. Our recommendation is simple: avoid
potential problems if you can. The following rules will help:

• Do not use extra spaces,

in particular, do not put spaces after commas that separate

operands, even though the ARM assembler allows you to do this.

• Do not use delimiter characters in names or labels.

• Include standard delimiters even if your assembler does not require them. Then it will be

more likely that your programs are in correct form for another assembler.

2.1.2 Labels

The label field is the first field in an assembly language instruction; it may be blank. If a label
is present, the assembler defines the label as equivalent to the address into which the first byte
of the object code generated for that instruction will be loaded. You may subsequently use the
label as an address or as data in another instruction’s address field. The assembler will replace
the label with the assigned value when creating an object program.

The ARM assembler requires labels to start at the first character of a line. However, some other
assemblers also allow you to have the label start anywhere along a line, in which case you must
use a colon (:) as the delimiter to terminate the label field. Colon delimiters are not used by the
ARM assembler.

Labels are most frequently used in Branch or SWI instructions. These instructions place a new
value in the program counter and so alter the normal sequential execution of instructions. B 15016
means “place the value 15016 in the program counter.” The next instruction to be executed will
be the one in memory location 15016. The instruction B START means “place the value assigned
to the label START in the program counter.” The next instruction to be executed will be the on at
the address corresponding to the label START. Figure 2.1 contains an example.

Why use a label? Here are some reasons:

• A label makes a program location easier to find and remember.

• The label can easily be moved, if required, to change or correct a program. The assembler
will automatically change all instructions that use the label when the program is reassembled.

• The assembler can relocate the whole program by adding a constant (a “relocation constant”)
to each address in which a label was used. Thus we can move the program to allow for the
insertion of other programs or simply to rearrange memory.

2.1. FIELDS

13

Assembly language Program

START

MOV
.
.
.
BAL

R0, VALUE1

(Main Program)

START

When the machine language version of this program is executed, the instruction
B START causes the address of the instruction labeled START to be placed in the
program counter That instruction will then be executed.

Figure 2.1: Assigning and Using a Label

• The program is easier to use as a library program; that is, it is easier for someone else to

take your program and add it to some totally different program.

• You do not have to figure out memory addresses. Figuring out memory addresses is partic-

ularly difficult with microprocessors which have instructions that vary in length.

You should assign a label to any instruction that you might want to refer to later.

The next question is how to choose a label. The assembler often places some restrictions on the
number of characters (usually 5 or 6), the leading character (often must be a letter), and the
trailing characters (often must be letters, numbers, or one of a few special characters). Beyond
these restrictions, the choice is up to you.

Our own preference is to use labels that suggest their purpose, i.e., mnemonic labels. Typical
examples are ADDW in a routine that adds one word into a sum, SRCHETX in a routine that searches
for the ASCII character ETX, or NKEYS for a location in data memory that contains the number of
key entries. Meaningful labels are easier to remember and contribute to program documentation.
Some programmers use a standard format for labels, such as starting with L0000. These labels are
self-sequencing (you can skip a few numbers to permit insertions), but they do not help document
the program.

Some label selection rules will keep you out of trouble. We recommend the following:

• Do not use labels that are the same as operation codes or other mnemonics. Most assemblers

will not allow this usage; others will, but it is confusing.

• Do not use labels that are longer than the assembler recognises. Assemblers have various

rules, and often ignore some of the characters at the end of a long label.

• Avoid special characters (non-alphabetic and non-numeric) and lower-case letters. Some
assemblers will not permit them; others allow only certain ones. The simplest practice is to
stick to capital letters and numbers.

• Start each label with a letter. Such labels are always acceptable.

• Do not use labels that could be confused with each other. Avoid the letters I, O, and Z and
the numbers 0, 1, and 2. Also avoid things like XXXX and XXXXX. Assembly programming is
difficult enough without tempting fate or Murphy’s Law.

• When you are not sure if a label is legal, do not use it. You will not get any real benefit

from discovering exactly what the assembler will accept.

These are recommendations, not rules. You do not have to follow them but don’t blame us if you
waste time on unnecessary problems.

14

CHAPTER 2. ASSEMBLERS

2.2 Operation Codes (Mnemonics)

One main task of the assembler is the translation of mnemonic operation codes into their binary
equivalents. The assembler performs this task using a fixed table much as you would if you were
doing the assembly by hand.

It must also
The assembler must, however, do more than just translate the operation codes.
somehow determine how many operands the instruction requires and what type they are. This
may be rather complex — some instructions (like a Stop) have no operands, others (like a Jump
instruction) have one, while still others (like a transfer between registers or a multiple-bit shift)
require two. Some instructions may even allow alternatives; for example, some computers have
instructions (like Shift or Clear) which can either apply to a register in the CPU or to a memory
location. We will not discuss how the assembler makes these distinctions; we will just note that it
must do so.

2.3 Directives

Some assembly language instructions are not directly translated into machine language instruc-
tions. These instructions are directives to the assembler; they assign the program to certain areas
in memory, define symbols, designate areas of memory for data storage, place tables or other fixed
data in memory, allow references to other programs, and perform minor housekeeping functions.

To use these assembler directives or pseudo-operations a programmer places the directive’s mnemonic
in the operation code field, and, if the specified directive requires it, an address or data in the
address field.

The most common directives are:

DEFINE CONSTANT (Data)
EQUATE (Define)
AREA
DEFINE STORAGE (Reserve)

Different assemblers use different names for those operations but their functions are the same.
Housekeeping directives include:

END

LIST

FORMAT

TTL

PAGE

INCLUDE

We will discuss these pseudo-operations briefly, although their functions are usually obvious.

2.3.1 The DEFINE CONSTANT (Data) Directive

The DEFINE CONSTANT directive allows the programmer to enter fixed data into program
memory. This data may include:

• Names
• Messages
• Commands
• Tax tables
• Thresholds
• Test patterns
• Lookup tables
• Standard forms
• Masking patterns
• Weighting factors

• Conversion factors
• Key identifications
• Subroutine addresses
• Code conversion tables
• Identification patterns
• State transition tables
• Synchronisation patterns
• Coefficients for equations
• Character generation patterns
• Characteristic times or frequencies

2.3. DIRECTIVES

15

The define constant directive treats the data as a permanent part of the program.

The format of a define constant directive is usually quite simple. An instruction like:

DZCON

DCW

12

will place the number 12 in the next available memory location and assign that location the name
DZCON. Every DC directive usually has a label, unless it is one of a series. The data and label may
take any form that the assembler permits.

More elaborate define constant directives that handle a large amount of data at one time are
provided, for example:

EMESS
SQRS

DCB
DCW

’ERROR’
1,4,9,16,25

A single directive may fill many bytes of program memory, limited perhaps by the length of a
line or by the restrictions of a particular assembler. Of course, you can always overcome any
restrictions by following one define constant directive with another:

MESSG

DCB
DCB
DCB
DCB
DCB
DCB

"NOW IS THE "
"TIME FOR ALL "
"GOOD MEN "
"TO COME TO THE "
"AID OF THEIR "
"COUNTRY", 0 ;note the ’0’ terminating the string

Microprocessor assemblers typically have some variations of standard define constant directives.
Define Byte or DCB handles 8-bit numbers; Define Word or DCW handles 32-bit numbers or addresses.
Other special directives may handle character-coded data. The ARM assembler also defines DCD
to (Define Constant Data) which may be used in place of DCW.

2.3.2 The EQUATE Directive

The EQUATE directive allows the programmer to equate names with addresses or data. This
pseudo-operation is almost always given the mnemonic EQU. The names may refer to device ad-
dresses, numeric data, starting addresses, fixed addresses, etc.

The EQUATE directive assigns the numeric value in its operand field to the label in its label field.
Here are two examples:

TTY
LAST

EQU
EQU

5
5000

Most assemblers will allow you to define one label in terms of another, for example:

LAST
ST1

EQU
EQU

FINAL
START+1

The label in the operand field must, of course, have been previously defined. Often, the operand
field may contain more complex expressions, as we shall see later. Double name assignments (two
names for the same data or address) may be useful in patching together programs that use different
names for the same variable (or different spellings of what was supposed to be the same name).

Note that an EQU directive does not cause the assembler to place anything in memory. The as-
sembler simply enters an additional name into a table (called a “symbol table”) which the assembler
maintains.

When do you use a name? The answer is: whenever you have a parameter that you might want to
change or that has some meaning besides its ordinary numeric value. We typically assign names to

16

CHAPTER 2. ASSEMBLERS

time constants, device addresses, masking patterns, conversion factors, and the like. A name like
DELAY, TTY, KBD, KROW, or OPEN not only makes the parameter easier to change, but it also adds to
program documentation. We also assign names to memory locations that have special purposes;
they may hold data, mark the start of the program, or be available for intermediate storage.

What name do you use? The best rules are much the same as in the case of labels, except that
here meaningful names really count. Why not call the teletypewriter TTY instead of X15, a bit
time delay BTIME or BTDLY rather than WW, the number of the “GO” key on a keyboard GOKEY
rather than HORSE? This advice seems straightforward, but a surprising number of programmers
do not follow it.

Where do you place the EQUATE directives? The best place is at the start of the program, under
appropriate comment headings such as i/o addresses, temporary storage, time constants,
or program locations. This makes the definitions easy to find if you want to change them.
Furthermore, another user will be able to look up all the definitions in one centralised place.
Clearly this practice improves documentation and makes the program easier to use.

Definitions used only in a specific subroutine should appear at the start of the subroutine.

2.3.3 The AREA Directive

The AREA directive allows the programmer to specify the memory locations where programs,
subroutines, or data will reside. Programs and data may be located in different areas of memory
depending on the memory configuration. Startup routines interrupt service routines, and other
required programs may be scattered around memory at fixed or convenient addresses.

The assembler maintains a location counter (comparable to the computer’s program counter) which
contains the location in memory of the instruction or data item being processed. An area directive
causes the assembler to place a new value in the location counter, much as a Jump instruction
causes the CPU to place a new value in the program counter. The output from the assembler
must not only contain instructions and data, but must also indicate to the loader program where
in memory it should place the instructions and data.

Microprocessor programs often contain several AREA statements for the following purposes:

• Reset (startup) address
• Interrupt service addresses
• Trap (software interrupt) addresses
• RAM storage

• Stack
• Main program
• Subroutines
• Input/Output

Still other origin statements may allow room for later insertions, place tables or data in memory,
or assign vacant memory space for data buffers. Program and data memory in microcomputers
may occupy widely separate addresses to simplify the hardware. Typical origin statements are:

AREA
AREA
AREA

RESET

$1000
INT3

The assembler will assume a fake address if the programmer does not put in an AREA statement.
The AREA statement at the start of an ARM program is required, and its absence will cause the
assembly to fail.

2.3.4 Housekeeping Directives

There are various assembler directives that affect the operation of the assembler and its program
listing rather than the object program itself. Common directives include:

2.4. OPERANDS AND ADDRESSES

17

END, marks the end of the assembly language source program. This must appear in the file or a

“missing END directive” error will occur.

INCLUDE will include the contents of a named file into the current file. When the included file
has been processed the assembler will continue with the next line in the original file. For
example the following line

INCLUDE

MATH.S

will include the content of the file math.s at that point of the file.

You should never use a lable with an include directive. Any labels defined in the included file
will be defined in the current file, hence an error will be reported if the same label appears
in both the source and include file.

An include file may itself include other files, which in turn could include other files, and so
on, however, the level of includes the assembler will accept is limited. It is not recommended
you go beyond three levels for even the most complex of software.

2.3.5 When to Use Labels

Users often wonder if or when they can assign a label to an assembler directive. These are our
recommendations:

1. All EQU directives must have labels; they are useless otherwise, since the purpose of an EQU

is to define its label.

2. Define Constant and Define Storage directives usually have labels. The label identifies the

first memory location used or assigned.

3. Other directives should not have labels.

2.4 Operands and Addresses

The assembler allow the programmer a lot of freedom in describing the contents of the operand or
address field. But remember that the assembler has built-in names for registers and instructions
and may have other built-in names. We will now describe some common options for the operand
field.

2.4.1 Decimal Numbers

The assembler assume all numbers to be decimal unless they are marked otherwise. So:

ADD

100

means “add the contents of memory location 10010 to the contents of the Accumulator.”

2.4.2 Other Number Systems

The assembler will also accept hexadecimal entries. But you must identify these number systems
in some way: for example, by preceding the number with an identifying character.

18

CHAPTER 2. ASSEMBLERS

2_nnn Binary
8_nnn Octal
nnn
0xnnn

Base 2
Base 8
Base 10
Decimal
Hexadecimal Base 16

It is good practice to enter numbers in the base in which their meaning is the clearest: that is,
decimal constants in decimal; addresses and BCD numbers in hexadecimal; masking patterns or
bit outputs in hexadecimal.

2.4.3 Names

Names can appear in the operand field; they will be treated as the data that they represent.
Remember, however, that there is a difference between operands and addresses.
In an ARM
assembly language program the sequence:

FIVE

EQU
ADD

5
R2, #FIVE

will add the contents of memory location FIVE (not necessarily the number 5) to the contents of
data register R2.

2.4.4 Character Codes

The assembler allows text to be entered as ASCII strings. Such strings must be surrounded with
double quotation marks, unless a single ASCII character is quoted, when single qoutes may be
used exactly as in ’C’. We recommend that you use character strings for all text. It improves the
clarity and readability of the program.

2.4.5 Arithmetic and Logical Expressions

Assemblers permit combinations of the data forms described above, connected by arithmetic,
logical, or special operators. These combinations are called expressions. Almost all assemblers
allow simple arithmetic expressions such as START+1. Some assemblers also permit multiplication,
division, logical functions, shifts, etc. Note that the assembler evaluates expressions at assembly
if a symbol appears in an expression, the address is used (i.e., the location counter or
time;
EQUATE value).

Assemblers vary in what expressions they accept and how they interpret them. Complex expres-
sions make a program difficult to read and understand.

2.4.6 General Recommendations

We have made some recommendations during this section but will repeat them and add others
here. In general, the user should strive for clarity and simplicity. There is no payoff for being an
expert in the intricacies of an assembler or in having the most complex expression on the block.
We suggest the following approach:

• Use the clearest number system or character code for data.

• Masks and BCD numbers in decimal, ASCII characters in octal, or ordinary numerical

constants in hexadecimal serve no purpose and therefore should not be used.

• Remember to distinguish data from addresses.

2.5. COMMENTS

19

• Don’t use offsets from the location counter.

• Keep expressions simple and obvious. Don’t rely on obscure features of the assembler.

2.5 Comments

All assemblers allow you to place comments in a source program. Comments have no effect on the
object code, but they help you to read, understand, and document the program. Good commenting
is an essential part of writing computer programs, programs without comments are very difficult
to understand.

We will discuss commenting along with documentation in a later chapter, but here are some
guidelines:

• Use comments to tell what application task the program is performing, not how the micro-

computer executes the instructions.

• Comments should say things like “is temperature above limit?”, “linefeed to TTY,” or “ex-

amine load switch.”

• Comments should not say things like “add 1 to Accumulator,” “jump to Start,” or “look at
carry.” You should describe how the program is affecting the system; internal effects on the
CPU should be obvious from the code.

• Keep comments brief and to the point. Details should be available elsewhere in the docu-

mentation.

• Comment all key points.

• Do not comment standard instructions or sequences that change counters or pointers; pay

special attention to instructions that may not have an obvious meaning.

• Do not use obscure abbreviations.

• Make the comments neat and readable.

• Comment all definitions, describing their purposes. Also mark all tables and data storage

areas.

• Comment sections of the program as well as individual instructions.

• Be consistent in your terminology. You can (should) be repetitive, you need not consult a

thesaurus.

• Leave yourself notes at points that you find confusing: for example, “remember carry was set
by last instruction.” If such points get cleared up later in program development, you may
drop these comments in the final documentation.

A well-commented program is easy to use. You will recover the time spent in commenting many
times over. We will try to show good commenting style in the programming examples, although
we often over-comment for instructional purposes.

2.6 Types of Assemblers

Although all assemblers perform the same tasks, their implementations vary greatly. We will not
try to describe all the existing types of assemblers, we will merely define the terms and indicate
some of the choices.

20

CHAPTER 2. ASSEMBLERS

A cross-assembler is an assembler that runs on a computer other than the one for which it assembles
object programs. The computer on which the cross-assembler runs is typically a large computer
with extensive software support and fast peripherals. The computer for which the cross-assembler
assembles programs is typically a micro like the 6809 or MC68000.

When a new microcomputer is introduced, a cross-assembler is often provided to run on existing
development systems. For example, ARM provide the ’Armulator’ cross-assembler that will run
on a PC development system.

A self-assembler or resident assembler is an assembler that runs on the computer for which it
assembles programs. The self-assembler will require some memory and peripherals, and it may
run quite slowly compared to a cross-assembler.

A macroassembler is an assembler that allows you to define sequences of instructions as macros.

A microassembler is an assembler used to write the microprograms which define the instruction
set of a computer. Microprogramming has nothing specifically to do with programming micro-
computers, but has to do with the internal operation of the computer.

A meta-assembler is an assembler that can handle many different instruction sets. The user must
define the particular instruction set being used.

A one-pass assembler is an assembler that goes through the assembly language program only
once. Such an assembler must have some way of resolving forward references, for example, Jump
instructions which use labels that have not yet been defined.

A two-pass assembler is an assembler that goes through the assembly language source program
twice. The first time the assembler simply collects and defines all the symbols; the second time
it replaces the references with the actual definitions. A two-pass assembler has no problems with
forward references but may be quite slow if no backup storage (like a floppy disk) is available;
then the assembler must physically read the program twice from a slow input medium (like a
teletypewriter paper tape reader). Most microprocessor-based assemblers require two passes.

2.7 Errors

Assemblers normally provide error messages, often consisting of an error code number. Some
typical errors are:

Undefined name

Illegal character

Illegal format

Often a misspelling or an omitted definition

Such as a 2 in a binary number

A wrong delimiter or incorrect operands

Invalid expression

for example, two operators in a row

Illegal value

Usually the value is too large

Missing operand

Pretty self explanatory

Double definition

Two different values assigned to one name

Illegal label

Missing label

Undefined operation code

Such as a label on a pseudo-operation that cannot have one

Probably a miss spelt lable name

In interpreting assembler errors, you must remember that the assembler may get on the wrong
track if it finds a stray letter, an extra space, or incorrect punctuation. The assembler will
then proceed to misinterpret the succeeding instructions and produce meaningless error messages.

2.8. LOADERS

21

Always look at the first error very carefully; subsequent ones may depend on it. Caution and
consistent adherence to standard formats will eliminate many annoying mistakes.

2.8 Loaders

The loader is the program which actually takes the output (object code) from the assembler and
places it in memory. Loaders range from the very simple to the very complex. We will describe a
few different types.

A bootstrap loader is a program that uses its own first few instructions to load the rest of itself
or another loader program into memory. The bootstrap loader may be in ROM, or you may have
to enter it into the computer memory using front panel switches. The assembler may place a
bootstrap loader at the start of the object program that it produces.

It typically loads each program
A relocating loader can load programs anywhere in memory.
into the memory space immediately following that used by the previous program. The programs,
however, must themselves be capable of being moved around in this way; that is, they must be
relocatable. An absolute loader, in contrast, will always place the programs in the same area of
memory.

A linking loader loads programs and subroutines that have been assembled separately; it resolves
cross-references — that is, instructions in one program that refer to a label in another program.
Object programs loaded by a linking loader must be created by an assembler that allows external
references. An alternative approach is to separate the linking and loading functions and have the
linking performed by a program called a link editor and the loading done by a loader.

22

CHAPTER 2. ASSEMBLERS

3 ARM Architecture

This chapter outlines the ARM processor’s architecture and describes the syntax rules of the ARM
assembler. Later chapters of this book describe the ARM’s stack and exception processing system
in more detail.

Figure 3.1 on the following page shows the internal structure of the ARM processor. The ARM
is a Reduced Instruction Set Computer (RISC) system and includes the attributes typical to that
type of system:

• A large array of uniform registers.

• A load/store model of data-processing where operations can only operate on registers and not
directly on memory. This requires that all data be loaded into registers before an operation
can be preformed, the result can then be used for further processing or stored back into
memory.

• A small number of addressing modes with all load/store addresses begin determined from

registers and instruction fields only.

• A uniform fixed length instruction (32-bit).

In addition to these traditional features of a RISC system the ARM provides a number of additional
features:

• Separate Arithmetic Logic Unit (ALU) and shifter giving additional control over data pro-

cessing to maximize execution speed.

• Auto-increment and Auto-decrement addressing modes to improve the operation of program

loops.

• Conditional execution of instructions to reduce pipeline flushing and thus increase execution

speed.

3.1 Processor modes

The ARM supports the seven processor modes shown in table 3.1.

Mode changes can be made under software control, or can be caused by external interrupts or
exception processing.

Most application programs execute in User mode. While the processor is in User mode, the
program being executed is unable to access some protected system resources or to change mode,
other than by causing an exception to occur (see 3.4 on page 29). This allows a suitably written
operating system to control the use of system resources.

The modes other than User mode are known as privileged modes. They have full access to system
resources and can change mode freely. Five of them are known as exception modes: FIQ (Fast
Interrupt), IRQ (Interrupt), Supervisor, Abort, and Undefined. These are entered when specific

23

24

CHAPTER 3. ARM ARCHITECTURE

Figure 3.1: ARM Block Diagram

3.2. REGISTERS

25

Processor mode Description
User
FIQ
IRQ
Supervisor
Abort
Undefined
System

Normal program execution mode
Fast Interrupt for high-speed data transfer
Used for general-purpose interrupt handling
A protected mode for the operating system
Implements virtual memory and/or memory protection
Supports software emulation of hardware coprocessors
Runs privileged operating system tasks

usr
fiq
irq
svc
abt
und
sys

Table 3.1: ARM processor modes

exceptions occur. Each of them has some additional registers to avoid corrupting User mode state
when the exception occurs (see 3.2 for details).

The remaining mode is System mode, it is not entered by any exception and has exactly the same
registers available as User mode. However, it is a privileged mode and is therefore not subject to
the User mode restrictions. It is intended for use by operating system tasks which need access to
system resources, but wish to avoid using the additional registers associated with the exception
modes. Avoiding such use ensures that the task state is not corrupted by the occurrence of any
exception.

3.2 Registers

The ARM has a total of 37 registers. These comprise 30 general purpose registers, 6 status registers
and a program counter. Figure 3.2 illustrates the registers of the ARM. Only fifteen of the general
purpose registers are available at any one time depending on the processor mode.

There are a standard set of eight general purpose registers that are always available (R0 – R7 ) no
matter which mode the processor is in. These registers are truly general-purpose, with no special
uses being placed on them by the processors’ architecture.

A few registers (R8 – R12 ) are common to all processor modes with the exception of the fiq
mode. This means that to all intent and purpose these are general registers and have no special
use. However, when the processor is in the fast interrupt mode these registers and replaced with
different set of registers (R8_fiq - R12_fiq). Although the processor does not give any special
purpose to these registers they can be used to hold information between fast interrupts. You can
consider they to be static registers. The idea is that you can make a fast interrupt even faster
by holding information in these registers.

The general purpose registers can be used to handle 8-bit bytes and 32-bit words1. When we use
a 32-bit register in a byte instruction only the least significant 8 bits are used. Figure 3.3 on the
following page demonstrates this.

The remaining registers (R13 – R15 ) are special purpose registers and have very specific roles:
R13 is also known as the Stack Pointer, while R14 is known as the Link Register, and R15 is
the Program Counter. The “user” (usr) and “System” (sys) modes share the same registers. The
exception modes all have their own version of these registers. Making a reference to register R14
will assume you are referring to the register for the current processor mode. If you wish to refer
to the user mode version of this register you have refer to the R14_usr register. You may only
refer to register from other modes when the processor is in one of the privileged modes, i.e., any
mode other than user mode.

1 Later revisions of the ARM architecture are also able to handle 16-bit half-words.

26

CHAPTER 3. ARM ARCHITECTURE

Modes
Privileged Modes

Exception Modes

User
R0
R1
R2
R3
R4
R5
R6
R7
R8
R9
R10
R11
R12
R13
R14
PC

System Supervisor

R0
R1
R2
R3
R4
R5
R6
R7
R8
R9
R10
R11
R12
R13
R14
PC

R0
R1
R2
R3
R4
R5
R6
R7
R8
R9
R10
R11
R12
R13_svc
R14_svc
PC

Abort
R0
R1
R2
R3
R4
R5
R6
R7
R8
R9
R10
R11
R12
R13_abt
R14_abt
PC

Undefined
R0
R1
R2
R3
R4
R5
R6
R7
R8
R9
R10
R11
R12
R13_und
R14_und
PC

Interrupt
R0
R1
R2
R3
R4
R5
R6
R7
R8
R9
R10
R11
R12
R13_irq
R14_irq
PC

Fast Interrupt
R0
R1
R2
R3
R4
R5
R6
R7
R8_fiq
R9_fiq
R10_fiq
R11_fiq
R12_fiq
R13_fiq
R14_fiq
PC

CPSR CPSR

CPSR
SPSR_svc

CPSR
SPSR_abt

CPSR
SPSR_und

CPSR
SPSR_irq

CPSR
SPSR_fiq

Figure 3.2: Register Organization

Bit:

31

· · ·

23

24

· · ·

16

15

· · ·

8

0
· · ·
7
8-Bit Byte

32-Bit Word

Figure 3.3: Byte/Word

There are also one or two status registers depending on which mode the processor is in. The Cur-
rent Processor Status Register (CPSR) holds information about the current status of the processor
(including its current mode). In the exception modes there is an additional Saved Processor Status
Register (SPSR) which holds information on the processors state before the system changed into
this mode, i.e., the processor status just before an exception.

3.2.1 The stack pointer, SP or R13

Register R13 is used as a stack pointer and is also known as the SP register. Each exception mode
has its own version of R13 , which points to a stack dedicated to that exception mode.

The stack is typically used to store temporary values. It is normal to store the contents of any
registers a function is going to use on the stack on entry to a subroutine. This leaves the register
free for use during the function. The routine can then recover the register values from the stack
on exit from the subroutine. In this way the subroutine can preserve the value of the register and
not corrupt the value as would otherwise be the case.

See Chapter 14 for more information on using the stack.

3.2. REGISTERS

27

3.2.2 The Link Register, LR or R14

Register R14 is also known as the Link Register or LR.

It is used to hold the return address for a subroutine. When a subroutine call is performed via a
BL instruction, R14 is set to the address of the next instruction. To return from a subroutine you
need to copy the Link Register into the Program Counter. This is typically done in one of the two
ways:

• Execute either of these instructions:

MOV

PC, LR

or

BAL

LR

• On entry to the subroutine store R14 to the stack with an instruction of the form:

STMIA

SP!,{(cid:104)registers(cid:105), LR}

and use a matching instruction to return from the subroutine:

LDMIA

SP!,{(cid:104)registers(cid:105), PC}

This saves the Link Register on the stack at the start of the subroutine. On exit from the
subroutine it collects all the values it placed on the stack, including the return address that
was in the Link Register, except it returns this address directly into the Program Counter
instead.

See Chapter 14 for further details on subroutines and using the stack.

When an exception occurs, the exception mode’s version of R14 is set to the address after the
instruction which has just been completed. The SPSR is a copy of the CPSR just before the
exception occurred. The return from an exception is performed in a similar way to a subroutine
return, but using slightly different instructions to ensure full restoration of the state of the program
that was being executed when the exception occurred. See 3.4 on page 29 for more details.

3.2.3 The program counter, PC or R15

Register R15 holds the Program Counter known as the PC. It is used to identify which instruction
is to be preformed next. As the PC holds the address of the next instruction it is often referred
to as an instruction pointer. The name “program counter” dates back to the times when program
instructions where read in off of punched cards, it refers to the card position within a stack of
cards. In spite of its name it does not actually count anything!

Reading the program counter

When an instruction reads the PC the value returned is the address of the current instruction plus
8 bytes. This is the address of the instruction after the next instruction to be executed2.

This way of reading the PC is primarily used for quick, position-independent addressing of nearby
instructions and data, including position-independent branching within a program.

An exception to this rule occurs when an STR (Store Register) or STM (Store Multiple Registers)
instruction stores R15 . The value stored is UNKNOWN and it is best to avoid the use of these
instructions that store R15 .

2 This is caused by the processor having already fetched the next instruction from memory while it is decoding
the current instruction. Thus the PC is still the next instruction to be executed, but that is not the instruction
immediately after the current one.

28

CHAPTER 3. ARM ARCHITECTURE

Writing the program counter

When an instruction writes to R15 the normal result is that the value written is treated as an
instruction address and the system starts to execute the instruction at that address3.

3.2.4 Current Processor Status Registers: CPSR

Rather surprisingly the current processor status register (CPSR) contains the current status of the
processor. This includes various condition code flags, interrupt status, processor mode and other
status and control information.

The exception modes also have a saved processor status register (SPSR), that is used to preserve
the value of the CPSR when the associated exception occurs. Because the User and System modes
are not exception modes, there is no SPSR available.

Figure 3.4 shows the format of the CPSR and the SPSR registers.

31
N

30
Z

29
28
C V

27

· · ·
SBZ

8

6

7
I F SBZ

5

4

0

· · ·
Mode

Figure 3.4: Structure of the Processor Status Registers

The processors’ status is split into two distinct parts: the User flags and the Systems Control
flags. The upper byte is accessible in User mode and contains a set of flags which can be used to
effect the operation of a program, see section 3.3. The lower byte contains the System Control
information.

Any bit not currently used is reserved for future use and should be zero, and are marked SBZ in
the figure. The I and F bits indicate if Interrupts (I) or Fast Interrupts (F) are allowed. The Mode
bits indicate which operating mode the processor is in (see 3.1 on page 23).

The system flags can only be altered when the processor is in protected mode. User mode programs
can not alter the status register except for the condition code flags.

3.3 Flags

The upper four bits of the status register contains a set of four flags, collectively known at the
condition code. The condition code can be used to control the flow of the program execution. The
is often abbreviated to just (cid:104)cc(cid:105). The condition code flags are:

N The Negative (sign) flag takes on the value of the most significant bit of a result. Thus when
an operation produces a negative result the negative flag is set and a positive result results
in a the negative flag being reset. This assumes the values are in standard two’s complement
form. If the values are unsigned the negative flag can be ignored or used to identify the value
of the most significant bit of the result.

Z The Zero flag is set when an operation produces a zero result. It is reset when an operation

produces a non-zero result.

C The Carry flag holds the carry from the most significant bit produced by arithmetic operations
or shifts. As with most processors, the carry flag is inverted after a subtraction so that the
flag acts as a borrow flag after a subtraction.

3 As the processor has already fetched the instruction after the current instruction it is required to flush the

instruction cache and start again. This will cause a short, but not significant, delay.

3.4. EXCEPTIONS

29

V The Overflow flag is set when an arithmetic result is greater than can be represented in a

register.

Many instructions can modify the flags, these include comparison, arithmetic, logical and move
instructions. Most of the instructions have an S qualifier which instructs the processor to set the
condition code flags or not.

3.4 Exceptions

Exceptions are generated by internal and external sources to cause the processor to handle an event,
such as an externally generated interrupt or an attempt to execute an undefined instruction. The
ARM supports seven types of exception, and a provides a privileged processing mode for each
type. Table 3.2 lists the type of exception and the processor mode associated with it.

When an exception occurs, some of the standard registers are replaced with registers specific to the
exception mode. All exception modes have their own Stack Pointer (SP) and Link (LR) registers.
The fast interrupt mode has more registers (R8_fiq – R12_fiq) for fast interrupt processing.

Processor Mode
Exception Type
Supervisor
Reset
Software Interrupt
Supervisor
Undefined Instruction Undefined
Prefetch Abort
Data Abort
Interrupt
Fast Interrupt

svc
svc
und
abt
abt
irq
fiq

Abort
Abort
IRQ
FIQ

Table 3.2: Exception processing modes

The seven exceptions are:

Reset when the Reset pin is held low, this is normally when the system is first turned on or when

the reset button is pressed.

Software Interrupt is generally used to allow user mode programs to call the operating system.
The user program executes a software interrupt (SWI, A.26 on page 152) instruction with a
argument which identifies the function the user wishes to preform.

Undefined Instruction is when an attempt is made to preform an undefined instruction. This
normally happens when there is a logical error in the program and the processor starts to
execute data rather than program code.

Prefetch Abort occurs when the processor attempts to access memory that does not exist.

Data Abort occurs when attempting to access a word on a non-word aligned boundary. The

lower two bits of a memory must be zero when accessing a word.

Interrupt occurs when an external device asserts the IRQ (interrupt) pin on the processor. This
can be used by external devices to request attention from the processor. An interrupt can
not be interrupted with the exception of a fast interrupt.

Fast Interrupt occurs when an external device asserts the FIQ (fast interrupt) pin. This is
designed to support data transfer and has sufficient private registers to remove the need for
register saving in such applications. A fast interrupt can not be interrupted.

When an exception occurs, the processor halts execution after the current instruction. The state
of the processor is preserved in the Saved Processor Status Register (SPSR) so that the original

30

CHAPTER 3. ARM ARCHITECTURE

program can be resumed when the exception routine has completed. The address of the instruction
the processor was just about to execute is placed into the Link Register of the appropriate processor
mode. The processor is now ready to begin execution of the exception handler.

The exception handler are located a pre-defined locations known as exception vectors. It is the
responsibility of an operating system to provide suitable exception handling.

3.5 Register Transfer Language

Before continuing, we need to develop an unambiguous notation to help us describe the way in
which information moves around the processor (see figure 3.1 on page 24). The register transfer
language (RTL) is just such a notation.

Each component of the processor is given a name or an abbreviation, for example the Memory
Address Register is known as the MAR, and the Program Counter is refereed to as PC. A left-,
or back-arrow (←) indicates the transfer of data from one component to another. Thus the RTL
expression:

MAR ← PC

means that the contents of the program counter are transferred (i.e. copied into) the memory
address register. A comment can be added to the line by placing the text after a semi-colon (;)
following the expression.

In addition to accessing a component directly we can also refer to a particular field, or part, of
a device by placing the name of the field in parentheses after the device name. For example, the
Instruction Register (IR) is split into a number of fields including the operation code (or op-code)
field (see section 3.6.2 for a further description of the IR fields). In order to access the op-code
field we would need to write:

IR(op-code)

A field is not always given a name, so we need to indicate the field by specifying which bits have
been grouped to provide the field. This is known as a bit field which we denote by giving the lower
and upper bits, separated by a colon as the field name. Thus to select the upper four bits (bits
28 to 31 inclusive) of register R4 we would write:

R4 (28:31)

Finally, we also have the notion of a guard. This is a condition which must be true before the
expression can be evaluated. The guard is written before the RTL expression it is guarding and
is separated from that expression with a colon (:). Normally the guard is a test for an optional
item in the instruction. For example, there is a version of the MOV instruction (MOVS) which sets
the CPSR flags N and Z. We can place a guard, which test for the extra S like this:

(cid:104)S (cid:105): CPSR ← ALU(Flags)

When more than one guard is required we simply list the guards next to each other in sequence:

(cid:104)cc(cid:105)(cid:104)S (cid:105): CPSR ← ALU(Flags)

indicates the process must be in the current condition, as indicated by the (cid:104)cc(cid:105) guard (see sec-
tion 4.1.2 on page 42), and it must be the S form of the instruction before the RTL expression is
be executed.

3.5. REGISTER TRANSFER LANGUAGE

31

3.5.1 Memory

When accessing external random access memory the processor must go though the memory device,
indicated by the device name M. The processor must first place the address, or location, it intends
to access in the Memory Address Register (MAR). When writing to external memory the value
to be written should then be copied into the Memory Buffer Register (MBR). When reading from
memory the value made available in the MBR.

The following RTL demonstrates how the system accesses memory. The location we wish to access
is held in the register R12 . In the first example we are reading the value from memory into register
R0 while the second example writes the value in register R1 to memory.

Read

MAR ← R12
MBR ← M(MAR)
R0 ← MBR

Write

MAR ← R12
MBR ← R1
M(MAR) ← MBR

In particular you should note the two lines which include the item M(MAR). This is where the
data is actually transferred between the processor and external memory.

Although this is the correct way of accessing external memory, it is rather tiresome. To overcome
this we abuse the notation slightly by placing the location we wish to access as a field to the
memory device, and read/write to it directly. Thus we can write the above examples as:

R0 ← M(R12 )

M(R12 ) ← R1

In this way we hide the reference to the MAR and MBR inside the reference to the memory device
M(. . . ).

3.5.2 Arithmetic and Logic Unit

In a similar manner the ALU has a number of rules for its use. The ALU has a number of parts,
or registers. Normally the ALU requires two operands (arguments) and a command.

A

B

Cmd

The first operand (or argument) goes into ALU register A, written as ALU(A). The
design of the processor means that only a register may be moved into the A register.

The second operand (or argument) should be placed in the ALU’s B register, denoted
by ALU(B). The value for the B register may come from a number of different sources,
normally a register. As the B register is connected to the Barrel Shift component the
value can be modified (shifted) as it is copied into the B register, this is discussed later.

Once the two operands have been set up, copied into registers A and B, the ALU needs
to know what to do with them. Thus we also have a Command (Cmd) register. We
should write this as ALU(Cmd) but we normally miss off the register name when writing
to the ALU.

The commands the ALU can process are:

R = A + B Add A to B

add
subtract R = A − B Subtract B from A
and
or
eor
not

R = A ∧ B Bitwise AND of A and B
R = A ∨ B Bitwise OR of A and B
R = A ⊕ B Exclusive OR of A and B
Logical complement of B
R = B

R

Having preformed some operation the result of that operation is placed in the Result
(R) register. As with the command register we should refer to this register as ALU(R),

32

CHAPTER 3. ARM ARCHITECTURE

however, when reading from the ALU it can be assumed we are referring to the Result
register, so we miss it off.

Flags

Finally we have the flags register. This includes the N, Z, C and V flags after the
operation. The state of these flags can be copied into the Current Process Status Register
(CPSR) for use with conditional execution, discussed in section 4.1.2 on page 42.

To demonstrate the way the ALU works, let us look at how it would add two registers together.
The instruction ADD
R0, R1, R2 would add the content of register R1 to the content of register
R2 placing the result in the register R0 . The register transfer language to describe this would be:

ALU(A) ← R1
ALU(B) ← R2
ALU ← add

; First operand
; Second operand
; ALU command

R0 ← ALU ; Read result

Note the lack of field in the last two lines. We are using the defaults for the write (cmd) and read
(R) operations. As with the memory operations we tend to abuse the notation by writing a single
line which summarised this operation:

R0 ← R1 + R2

The convention is that the item before the operator is placed in the ALU register A and the item
after the operator is placed in register B. In our example the operator is the plus sign (+), while
the register R1 would be copied into ALU(A) and R2 would be copied into ALU(B).

The data paths leading to the A and B registers pass though two additional components that
preform operations which traditionally form part of the Arithmetic and Logic Unit. These are the
Booth Multiplier and Barrel Shifter respectively.

Booth Multiplier

The Booth multiplier is a hardware component that is capable of multiplying two signed numbers.
It can take two 16-bit values and produces a single 32-bit result. As with the ALU, the Booth
Multiplier (BM) has two input registers (A and B) and one results register (R). The instruction:

MUL

R0, R1, R2

will multiply the value in register R1 with that in register R2 , placing the result in R0 . The RTL
for this would be:

BM(A) ← R1 (0:15)
BM(B) ← R2 (0:15)
ALU(A) ← BM(R)

R0 ← ALU

; First operand
; Second operand
; Result gos on to the ALU
; Read result

There are two points to note here, firstly only the lower 16-bits (or halfwords) of the operands
are used in the multiply operation. The second point to note is that the result must go on to the
ALU before it can be copied into the destination register.

You may be surprised to discover that whilst this is what actually happens inside the processor
we tend to write it differently:

R0 ← R1 × R2

The Booth Multiplier is named after Andrew D. Booth who first suggested a method of multiplying
two numbers together that could be implemented as a hardware component4. See Chapter 12 for
a discussion of the Multiply instructions.

4Booth’s paper A signed binary multiplication technique was first published in the Quarterly Journal of Me-

chanics and Applied Mathematics, 4:2 (236–240) in 1951.

3.6. CONTROL UNIT

33

Shift Method
RTL
Logical Shift Left
A << B
A >> B
Logical Shift Right
Arithmetic Shift Right A +>> B
A >>> B
Rotate Right
A >>> B
Rotate Right Extend

Section Page

5.1.2
5.1.3
5.1.4
5.1.5
5.1.6

51
52
52
53
54

Table 3.3: Barrel Shifter Operations

Barrel Shifter

The Barrel Shift unit allows the ARM to manipulate the second operand, leading to register B
of the ALU, before it actually reaches it. This is an advantage when dealing with bit-orientated
operations (Chapter 8) and data structures (Chapter 13). For a discussion on the use of the Barrel
Shift we refer you to Chapter 5. Particularly the discussion of the data addressing mode ((cid:104)op1 (cid:105))
in section 5.1 on page 51.

Table 3.3 above shows how we write the five different ways in which data can be manipulated by
the barrel shift, and which section of Chapter 5 discusses the shift method.

For example, if we wish to add the value of register R2 shifted left by 4 bits (effectively multiplied
by 16) to the register R1 leaving the result in register R0 , we would give the instruction:

ADD

R0, R1, R2, LSL #4

Which would be represented in RTL as:

R0 ← R1 + R2 (cid:28) 4

Note the use of the shorthand form of the ALU operation + (add). As the shift is on the right
hand side of the operator, the result must be placed in register B of the ALU. It is only data going
to register B which can be shifted in the manner anyhow so that works out.

You should also note that most processes require a separate instruction to preform these operations.

3.6 Control Unit

The Control Unit is the most complex part of the processor. This controls the overall operation
of the processor. It sends control signals to the other devices prompting them to place data on
one of the buses or take data from the bus. The control unit is the device which actually executes
the RTL we have been looking at. Indeed the purpose of this book is to describe the operation of
this unit.

So what happens in the Control Unit? In essence it is quite simple, it reads a machine instruction
from memory and then preforms the operation described by the instruction. It does this though
the now famous fetch/execute cycle. This starts with the fetch phase where the control unit will
fetch the next instruction from memory into the Instruction Register (IR).

In older microprocessors the execute phase was a simple task. When processors become more
complex a micro-code system was introduced where the execute phase consisted of executing a
series of RTL like instructions within the control unit itself. Such systems are refereed to as
Complex Instruction Set Computers (CISC). Such systems include the Motorola MC68000, and
the Intel 80x86 series.

34

CHAPTER 3. ARM ARCHITECTURE

The ARM however, is a Reduced Instruction Set Computer (RISC) system, which means the
designers chose to use a larger instruction size (32-bit) in exchange for making the control unit
simple, well simpler. Other RISC processors include the SPARC, and the PowerPC range (Gn).

A RISC processor has a number of stages to the execute part of the fetch/execute cycle:

Instruction Fetch

Fetch the instruction from memory into the instruction register.

Instruction Decode Decode the instruction in the instruction register working out what we

are supposed to do next.

Operand Fetch

Fetch the source operands for the task, this may involve the use of the
data addressing mode (see 5.1 on page 51) or a memory addressing
mode (see 5.2 on page 54).

Execute

Perform the requested operation.

Operand Store

Save the result someplace, this will almost always be a register.

The ARM have been designed to perform these stages simultaneously. So whilst it is decoding one
instruction it can be fetching the next instruction from memory. This is known as the instruction
pipeline.

The way in which the pipeline works can best be seen by examining the RTL the control unit will
generate when processing the instruction

SUBS

R0, R0, #1

This instruction will subtract 1 from the content of the register R0 , and set the Z flag should it
become zero. This is a particularly useful instruction as we will see in chapter 9.

It should be noted that while we discuss a five stage instruction pipeline, different variants of the
ARM have a different stages. The version of the ARM we are using has a three stage pipeline:
Instruction Fetch and Decode; Operand Fetch; Execute and Operand Store.

3.6.1

Instruction Fetch

The first step is to copy the memory location contained in the program counter into the memory
address register.

MAR ← PC

The program counter is badly named as it does not count programs, or anything else for that
matter, but contains the address of the next instruction in memory to be executed. In some system
it is called the instruction pointer, or IP. Once the MAR has the location of the next instruction,
the contents of the program counter are incremented (moved on to the next instruction) with a
special address incrementer (INC) circuit and moved back to the program counter. In this way, the
program counter is pointing to the next instruction while the current instruction is being executed.

INC ← MAR
PC ← INC

The MAR now contains a copy of the contents of the PC, so that the instruction to be executed
is read from the memory and transferred to the memory buffer register (MBR). Once in the MBR
the instruction can then be copied into the instruction register (IR).

MBR ← M(MAR)
IR ← MBR

By abusing the notation we can reduce these five lines to just two:

3.6. CONTROL UNIT

35

IR ← M(PC)
PC ← INC(PC)

; Read value at PC into the Instruction Register
; Move Program Counter to next instruction

3.6.2

Instruction Decode

Now the instruction is in the instruction register it is necessary to decode it. The IR is divided
into a number of fields or parts.

op-code

This is the operation code or binary instruction which tells the control unit which function
to preform.

condition code.

This is a general guard for the whole instruction. The instruction will only be preformed if
the process is in this state or the condition code has been set to always, the default. See
section 4.1.2 on page 42 for more information. Our example instruction has the default
setting, so there is not guard and the instruction will always be executed.

Set flags.

Most of the data processing instructions include a S variation. Such as our SUBS instruction.
The (cid:104)S (cid:105) field used to indicate whether the instruction should set the CPSR flags (as in our
example) or not.

destination register.

All of the instructions take one or two source values, and preform some operation on them,
placing a result in the destination register. With the exception of the store instruction it is
always a register. In our example this would be the register R0 .

source register.

The majority of instructions require one or two source values to operate on. The (cid:104)source(cid:105)
register indicates which register holds the source value for the instruction/operation. In our
example this would also be the register R0 .

op1 The data processing instructions all require a source value which can be calculated as part
of the instruction. This is known as a (cid:104)op1 (cid:105) value. See section 5.1 on page 51 for a full
discussion of the possible values for (cid:104)op1 (cid:105). Our subtract instruction is a data processing
instruction and (cid:104)op1 (cid:105) is an immediate value, so it will take the value from the (cid:104)value(cid:105) part
of the instruction register.

op2 The memory based instructions require a memory location to work with, this is specified in
an effective address, known as an (cid:104)op2 (cid:105) operand. Section 5.2 on page 54 goes into the details
of the (cid:104)op2 (cid:105) field.

value

Holds a small value as part of the binary instruction. This is normally used when calculating
the second operand, either (cid:104)op1 (cid:105) or (cid:104)op2 (cid:105). As we are using immediate addressing in our
example instruction (the #1), the value (1) will be in the (cid:104)value(cid:105) field.

offset

This is similar to (cid:104)value(cid:105) except the value is larger. This field is only used by the branch
instructions.

Whilst all of the fields are available, they only have any real meaning in the context of the
instruction. The only fields that have any real value are the op-code and condition fields. Even
then not all instructions have an condition field.

The instruction decode stage of the fetch/execute cycle does not actually produce any RTL.
However, for our example instruction the IR will be broken down as follows:

36

CHAPTER 3. ARM ARCHITECTURE

op-code
subtract

condition
always

set
true

destination
R0

source
R0

op1
immediate

value
1

3.6.3 Operand Fetch

The operand fetch stage will prepare the ALU for the execution phase by reading the operands
into the ALU registers.

In our example we read the (cid:104)source(cid:105) register into the ALU’s A register.

ALU(A) ← R0

At the same time we can also copy the second operand into register B of the ALU. As this is
a data processing instruction the control unit will analyse IR(op1) and see that we are using an
immediate value. Thus it will copy the value field into the ALU register.

ALU(B) ← IR(value)

As the (cid:104)op1 (cid:105) value passes through the Barrel Shifter we can preform a shift operation on the value
as it passes into the ALU.

For the memory load instruction, the operand fetch stage will read a value from external memory
as specified by (cid:104)op2 (cid:105).

3.6.4 Execute

Now that the ALU has been configured, both registers have been loaded with the appropriate
operands (values), we can now instruct the ALU to subtract the second operand from the first.
This is done by simply sending a subtract message to the ALU.

ALU ← subtract

3.6.5 Operand Store

Finally we need to store the result of this operation by copying the value from the ALU to the
destination register.

R0 ← ALU

However we also have the (cid:104)S (cid:105) flag set, so we must copy the flags from the ALU back into the
CPSR in the control unit.

CPSR ← ALU(flags)

For the memory store instruction the destination is a memory location specified by (cid:104)op2 (cid:105). This is
the only instruction that does not use a register as the destination. This will cause the system to
write the result to memory.

3.6.6 Summary

We have just looked at the fetch/execute cycle and the way the system actually processes an
instruction. In particular we looked at the processing of a specific instruction

SUBS

R0, R0, #1

3.6. CONTROL UNIT

37

which subtracts 1 from the content of the register R0 , and sets the Z flag should it become zero.

Here we will list the RTL that was produced without all the bothersome interpretation.

MAR ← PC

INC ← MAR
PC ← INC

MBR ← M(MAR)
IR ← MBR

ALU(A) ← R0
ALU(B) ← IR(value)

; Instruction Fetch

; Instruction Decode

; Operand Fetch

ALU ← subtract

; Execute the instruction

R0 ← ALU

; Operand Store

CPSR ← ALU(flags)

If we abuse the notation, as discussed earlier in sections 3.5.2 and 3.6.1, we can reduce this from
10 instructions to just four.

IR ← M(PC)
PC ← INC(PC)
R0 ← R0 − IR(value)

CSPR ← ALU(flags)

; Fetch the instruction
; Move on to the next instruction
; Execute the subtract
; Save the flags

Although this might be smaller and easier to read. Unfortunately it does not describe the data
flow correctly, the longer version is more correct.

38

CHAPTER 3. ARM ARCHITECTURE

4 Instruction Set

Why are a microprocessor’s instructions referred to as an instruction set? Because the micropro-
cessor designer selects the instruction complement with great care; it must be easy to execute
complex operations as a sequence of simple events, each of which is represented by one instruction
from a well-designed instruction set.

Assembler often frighten users who are new to programming. Yet taken in isolation, the operations
involved in the execution of a single instruction are usually easy to follow. Furthermore, you need
not attempt to understand all the instructions at once. As you study each of the programs in this
book you will learn about the specific instructions involved.

From the advent of the microprocessor in the early 1970s there has been a trend for instruction
sets to become more sophisticated and complex. In the early 1980s, a different approach to the
instruction set emerged, that of the Reduced Instruction Set Computer or RISC, which attempted
to produce a ‘striped down’ processor, or a ‘supercharged’ one if you prefer.

When David Patterson and David Ditzel1 first proposed the idea, they analysed a large number
of programs to find out how developers where actually using the processor. They came up with
three interesting observations. The first of which can be seen in table 4.1 below.

Instruction Grouping
1 Data Movement
Flow Control
2
3 Arithmetic
4 Comparison

Usage
45.28%
28.73%
10.75%
5.92%

Logical
Shift

Instruction Grouping Usage
3.91%
2.93%
2.04%
0.44%

5
6
7 Bit Manipulation
8

Input/Output and
Miscellaneous

Table 4.1: Instruction Group Average Usage

This shows that most programs spend over 80% of the time executing instructions from the
Data Movement, Flow Control or Arithmetic instruction groups. If they could find a method of
increasing the speed of these instructions, the program would execute that much faster.

There second observation was in the use of constant values. Some 56% of constant values where
within the range of ±15, while 98% of constants where in the range ±511. As the most constant
values are small enough (just 5 bits) they could be included as part of the instruction, rather than
forcing yet another memory access for it.

The final observation was that the number of parameters (or arguments) passed between subrou-
tines was normally less than six. Remember this research was done before the advent of Microsoft
Windows and the 13-parameter API call which nobody can ever remember. Thus if the processor
had sufficient registers, it would not be necessary to use external (off-chip) memory for a stack
frame. See Chapter ?? for a discussion of the stack frame.

1David A. Patterson and David R. Ditzel, The case for the reduced instruction set computer published in

Computer Architecture News, volume 9(3) in 1980.

39

40

CHAPTER 4.

INSTRUCTION SET

Mnemonic

Instruction

Mnemonic

Instruction

MOV

ADD
SUB
RSB
MUL

CMP
TEQ
BAL
BLAL

LDR
LDRB
LDM

AND
ORR
MVN

MRS
SWP
SWI

Move

Add
Subtract
Reverse Subtract
Multiply

Data Movement

Arithmetic

ADC
SBC
RSC
MLA

Add with Carry
Subtract with Carry
Reverse Subtract with Carry
Multiply Accumulate

Compare
Test Equivalence
Branch Always
Branch and Link Always

Flow Control
CMN
TST
B(cid:104)cc(cid:105)
BL(cid:104)cc(cid:105)

Load Register
Load Register Byte
Load Multiple

Memory Access
STR
STRB
STM

Compare Negative
Test
Conditional Branch
Conditional Branch and Link

Store Register
Store Register Byte
Store Multiple

Logical and Bit Manipulation

Bitwise AND
Bitwise (inclusive) OR
Move Negative

BIC
EOR

Bit Clear
Bitwise Exclusive OR

Move to Register from Status
Swap
Software Interrupt

System Control
MSR
SWPB

Move to Status from Register
Swap Byte

Table 4.2: Instruction Mnemonics

These three observations form the basis of what has become know as the Berkeley RISC architec-
ture, and subsequently formed the principles upon which the ARM instruction set was designed.

The mnemonics for the ARM instruction set are listed in table 4.2 above. This provides a survey
of the processors capabilities, and will also be useful when you need a certain kind of operation
but are either unsure of the specific mnemonics or not yet familiar with what instructions are
available.

The instruction mnemonics are supposed to provide you with an easy way to remember the in-
structions set. Although, despite having developed in assembler on ten different microprocessors,
we have yet to find a set of mnemonics which are truly mnemonic. This is one of the reasons people
new to assembler find it so daunting, attempting to remember all of these so called mnemonics. It
is probably easier to remember the instruction you want, and then work out the mnemonic for it.

There are generally three types of developer. One who can understand what each instruction does
but is not able to put them together to form a program, chapters 7 through to 14 are intended to
help these people. The second is one who can understand the principles behind the instructions
and is able to put it all together to form a program, but can never remember the precise details
of the instructions, appendix A is intended for these people. The final type, is the most annoying,
those who can not only put it all together in the form of a program, in addition they are also able
to remember each and every instruction in minute detail.

The ARM instruction set can be divided into six broad classes of instruction.

• Data Movement
• Arithmetic
• Flow Control

• Memory Access
• Logical and Bit Manipulation
•

System Control

4.1. DATA MOVEMENT

4.1 Data Movement

41

There is only one instruction which falls into this category, which is the MOV (Move) Instruction.
This is however probably the most important instruction, as over 40% of a program will consists
of this instructions. It is worth our while looking at this instruction in more detail as it shares
two attributes with most of the other instructions, namely the set flags variant and conditional
execution.

Section A.13 on page 144 gives the syntax for this instruction as:

MOV(cid:104)cc(cid:105)(cid:104)S (cid:105) Rd , (cid:104)op1 (cid:105)

which we interpret as:

MOV

(cid:104)cc(cid:105)

(cid:104)S (cid:105)

Rd

this is the mnemonic for the instruction which indicates which operation we would like
to preform. See table 4.2 on the preceding page for a list of mnemonics, and Appendix
A for a detailed discussion of each instruction. The MOV mnemoic is a bit of a misnomer
as the original value is not destroyed, but copied.

indicates conditional execution, where the instruction is only executed under the specified
condition. This is either a two letter condition code (see table 4.3 on the following page)
or missing altogether, in which case it is assumed the instruction should always be
executed. This is discussed in more detail later (in section 4.1.2).

indicates the operation should set the condition code. This is to be used in conjunction
with the conditional execution and is discussed in more detail later.

indicates the destination register. This is always a register and may be any register
available in the current processor mode.

(cid:104)op1 (cid:105)

indicates the source of the data. The MOV instruction is unusual in that it only has the
one source. Most instructions require two operands, thus they require two sources. The
first is normally a source register (Rs).

The source register (if present) is copied into the register A of the ALU. The (cid:104)op1 (cid:105)
source is copied into register B of the ALU, which means it has to pass though the
Barrel Shifter. This allows the programmer to preform an arithmetic or logical shift of
the data en route. As a result there is no requirement for the dedicated shift instructions
found in other (CISC) processors. See section 5.1 for a full discussion of what can be
done with an (cid:104)op1 (cid:105) value.

With this in mind, the register transfer language notation describing a general MOV instruction
would be:

(cid:104)cc(cid:105): ALU(B) ← (cid:104)op1 (cid:105)
(cid:104)cc(cid:105):
(cid:104)cc(cid:105)(cid:104)S (cid:105):

Rd ← ALU

CSPR ← ALU(flags)

Where the (cid:104)cc(cid:105) guard indicates the conditional execution, and the (cid:104)S (cid:105) guard indicates the Set
flags variant of the instruction.

4.1.1 Set Flags Variant

Most of the data processing instructions, this includes the Arithmetic, and Logic instruction
groups, have a variant which allows them to sets the condition code flags. This variant is indicated
by an (cid:104)S (cid:105) in the syntax definition of the instruction (in Appendix A) and by adding the letter S
to the end of the mnemonic.

42

CHAPTER 4.

INSTRUCTION SET

General

Signed

Unsigned

C arry S et

GT Greater T han
CS
GE Greater Than or E qual HS H igher or S ame (aka CS)
CC C arry C lear
EQ Equal (Zero Set)
LT
NE N ot E qual (Zero Clear) LE
VS Ov erflow S et
VC Ov erflow C lear

MI Minus (Negative)
PL

Less T han
Less Than or E qual

LO Lower Than
LS

Lower or S ame (aka CC)

Pl us (Positive)

Higher Than

HI

Table 4.3: (cid:104)cc(cid:105) (Condition code) Mnemonics

The flow control instructions will always effect the processor status, thus they do not need a
variant. The Memory Access instructions do not pass the data via the ALU and so can not set
the flags. This is quite a pain, as it means we have to add an extra instruction to check if the
value we have just read from memory is a zero or not.

The MOVS instruction will pass the value though the ALU in order to generate the condition codes.
In particular it can only set the Zero and Negative flags. This variant can be combined with the
flow control and conditional execution feature to provide fast efficient code. Chapter 9 provides a
number of examples.

We can indicate the action taken by a Set Flags variant by using the (cid:104)S (cid:105) guard in the RTL
description of the instruction:

(cid:104)S (cid:105): CSPR ← ALU(flags)

4.1.2 Conditional Execution: (cid:104)cc(cid:105)

The vast majority of instructions include the condition code ((cid:104)cc(cid:105)) attribute which allows the
instruction to be executed conditionally dependent on the current status of the processor (condition
code flags, see section 3.3 on page 28). If the flags indicate that the condition is true, the instruction
is executed, otherwise the instruction is ignored, and the processor simply moves on to the next
instruction.

A list of the two letter condition codes are shown in table 4.3 above. To indicate that an instruction
is conditional we simply place the two letter mnemonic for the condition after the mnemonic for
the instruction, but before the (cid:104)S (cid:105) if present. If no condition code mnemonic is used the instruction
will always be executed, there is even a two letter mnemonic to indicate always (AL) if you prefer.

Note that the Greater and the Less conditions are for use with signed numbers while the Higher
and Lower conditions are for use with unsigned numbers. These condition codes only really make
sense after a comparison instruction (see 4.3).

For example, the instruction

MOVCC

R0, R1

will copy the value in the register R1 into the R0 register, only when the Carry flag is clear (or
not set, this is the CC condition code), R0 will remain unaffected if the C flag is set.

Conditional execution can be combined with the (cid:104)S (cid:105) variant instruction. For example, what would
the following code fragment do?

MOVS
MOVEQS
MOVEQ

R0, R1
R0, R2
R0, R3

1. The first instruction will always copy R1 into R0 , but it will also set the N and Z flags

accordingly.

4.2. ARITHMETIC

43

2. The second instruction is only executed if the Z flag is set, i.e., the value of R1 was zero. If
the value of R1 was not zero the instruction is skipped. If the second instruction is executed
it will copy the value of R2 into R0 and it will also set the N and Z flags according to the
value of R2 .

3. The third instruction is only executed if both R1 and R2 are both zero.

4.2 Arithmetic

The eight instructions in this group all preform an arithmetic operation on two (and in one case
three) source operands, one of which must be a register (Rn) and another value ((cid:104)op1 (cid:105)) producing a
result which is placed in a destination register (Rd ). They can also optionally update the condition
code flags based on the result.

4.2.1 Addition

There are only two addition instructions: ADD (Add) and ADC (Add with Carry). These instructions
both add two 32-bit numbers, but they can be combined to add 64-bit or larger numbers. Let
registers R0 , R1 and R2 , R3 contain two 64-bit numbers, with the lower word in registers R0 and
R2 respectively. We can add these 64-bit numbers to produce a new 64-bit number in R4 , R5 :

ADDS
ADC

R4, R0, R2
R5, R1, R3

; Add lower words
; Add upper words

The ADDS not only adds the two lower words together, but also sets the carry flag if there is a
carry over, and clears it if there is no carry over. The ADC not only adds the upper words, but
also adds in the carry over from the lower words.

4.2.2 Subtraction

There are four subtraction instructions. The SUB (Subtract) and SBC (Subtract with Carry)
instructions are the equivalent of the ADD and ADC instructions above.

Unlike addition, subtraction is not commutative. This means that while the result of adding 4 + 2
is the same as that of adding 2+4, the result of subtracting 4−2 (2) is not the same as subtracting
2 − 4 (−2). To allow for this we have two additional “reverse” instructions: RSB (Reverse Subtract)
and RSC (Reverse Subtract with Carry) which subtract the value in a register from another value.

In subtraction, the carry flag records the fact that a subtraction had to borrow from the next
(32nd) bit. Thus the SUB and SBC instructions can be combined in the same way as the ADD and
ADC instructions to form a 64-bit subtraction. The RSB and RSC instruction can also be combined
in this manner.

4.2.3 Multiplication

There are two multiplication instructions, MUL (Multiply) and MLA (Multiply Accumulate). These
instructions are different from the others in this group, in that they can only operate on regis-
ters. The second operand must be a register, rather than an (cid:104)op1 (cid:105) value, as used by the other
instructions in the arithmetic group.

The MUL instruction will use the Booth Multiplier to simply multiply two 32-bit numbers, giving
a 32-bit result. This causes a small problem, as multiplying two 32-bit numbers should produces

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CHAPTER 4.

INSTRUCTION SET

a 64-bit result. So the multiply instructions should be used with caution. It does, however, mean
that we can multiply to signed 16-bit numbers, and receive a correct signed 32-bit result.

The MLA instruction the same as the MUL instruction with the exception that it will add the result
of the multiplication to the value in a third register. This can be used for calculating running
totals.

4.2.4 Division

If you have been paying attention, you will have realised that we have already discussed the eight
instructions in this group. This means that there are no instructions to support division.

Division is performed considerably less frequently than the other arithmetic operations.
It is,
however, possible to perform division by repeatedly subtracting the divisor from the dividend until
the result is either zero or less than the divisor. The number of times the divisor is subtracted is
known as the quotient, and the value left after the final subtraction is the remainder. That is:

Let us consider the following equation:

dividend = quotient × divisor + remainder

10
7

= 1 remainder 3

in this equation the dividend is 10, the divisor is 7, the result is in two parts the quotient (1) and
the remainder (3). This is all very well, but what happens when the numbers are signed, do we
use floored or symmetric division?

Floored division is where the quotient is rounded down toward negative infinity and the remainder
is used to correct the rounding. Thus if we divide −10 by 7, we would have a quotient of −2 and
a remainder of 4 (7 × −2 = −14 + 4 = −10). In symmetric division the rounding is done the other
way, towards zero. Thus −10 divided by 7 would have a quotient of −1 with a remainder of −3
(7 × −1 = −7 + −3 = −10).

Just to complicate matter more, like subtraction, division is not commutative. Given that division
is the least used arithmetic operation is it any wonder there are no instructions to support (integer)
division.

4.3 Flow Control

The flow control instructions fall into two sub-groups, those that compare two values and set the
condition codes, and those that which change the flow of execution (branch).

4.3.1 Comparisons

There are four comparison instructions which use the same format as the arithmetic instructions.
These perform an arithmetic or logical operation on two source operands, one of which must be
a register, and the other can be any (cid:104)op1 (cid:105) value. The instructions do not write the result to a
register, but use it to update the condition flags.

The CMP (Compare) instructions compares two numbers. It will subtract the second operand (the
(cid:104)op1 (cid:105) value) from the first (the register), setting all of the status flags accordingly:

4.3. FLOW CONTROL

45

Zero
Negative
Carry
oVerflow Set if there was a signed overflow

Set if the two number are equal, otherwise it is clear.
Set should the second operand be larger than the first, and clear if it is smaller.
Set if the subtraction had to borrow, (unsigned underflow)

This will allow the full set of condition codes, general, signed, and unsigned, to be used in the
instructions which follow.

The CMN (Compare Negative) instruction compares the register value with the negative of the
second value. Thus it can be used to change the sign of the second value. Although this may
cause difficulties if the value is zero.

The TEQ (Test Equivalence) instruction checks to see if the two values are equal. It will set the Z
flag if they are, so the EQ (Equal) and NE (Not Equal) condition codes can be used. It can also
be used to check if two values have the same sign. The N flag will be clear should the two values
have the same sign, so the PL (Plus) condition can be used. Should the two values have different
signs the flag will be clear and the MI (Minus) condition code could be used.

The final instruction in this sub-group is the TST (Test) instruction, which is used to discover
whether a single bit of the register is set or not. The Z flag will be set if the bit in the register
is clear and the EQ (Equal) condition code can be used. If the bit in the register was set, the flag
will be clear and the NE (Not Equal) condition code can be used.

The TST can be used to test a collection of bits. However, all of the bits must be clear for the Z
flag to be set. If any of the register bits are set, the flag will be cleared.

4.3.2 Branching

The B(cid:104)cc(cid:105) (Branch) instruction is used to change the control flow by adding an offset to the current
Program Counter. The offset is a signed number, allowing forward and backward branches of up
to 32MB. In assembler we place a label after the B(cid:104)cc(cid:105) (Branch) instruction, and the assembler
will calculate the offset for us.

The Branch instruction can be used in two ways, conditionally and unconditionally. An uncon-
ditional branch will always branch no matter what the state of the flags, conventionally this is
written as BAL (Branch Always). For a conditional branch, one of the condition codes, listed in
table 4.3 on page 42, is given immediately after the B. Thus the BEQ instruction is used to branch
on equal condition (the Z flag is set).

There is also the BL(cid:104)cc(cid:105) (Branch and Link) instruction which preserves the address of the instruc-
tion immediately after the branch in the Link Register (LR or R14 ). This allows for a subroutine
call. The code which has been called (the target of the branch) can return to the next instruction,
after the branch by copying the Link Register (LR) into the Program Counter (PC). Chapters ??
and 14 go into detail over the use of subroutines.

4.3.3 Jumping

All of the instructions in the Arithmetic, Logical, and Data Movement groups can use the Program
Counter (PC or R15 ) as a destination register. This allows them to alter the location from which
the next instruction is fetched. The direct manipulation of the PC in this way is known as a
jump. Any 32-bit value can be copied into the PC, thus a jump can be to any location in the 4GB
memory space.

This is a particularly useful feature when the location of the next instruction can be calculated,
via a jump table (see 13.1.5 on page 121 for further discussion).

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CHAPTER 4.

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4.4 Memory Access

The memory access instructions naturally fall into three sub-groups: Register; Register Byte; and
Multiple Registers. Each of which has a Load and a Store instruction.

4.4.1 Load and Store Register

The LDR (Load Register) instruction can load a 32-bit word from memory into a register. While,
surprisingly, the STR (Store Register) instructions will store a 32-bit word from a register to
memory.

The Load and Store Register instructions use an (cid:104)op2 (cid:105) value to identify the memory to use. All
of the (cid:104)op2 (cid:105) values use a base register and an offset of some form specified by the instruction:

Offset addressing is where the memory address is formed by adding (or subtracting) an offset to

(or from) the value held in the base register.

Pre-indexed addressing is where the memory address is formed in the same way as for offset
addressing. As a added extra the memory address is also written back into the base register.
This is quite useful for looping though data, see Chapter 9 for more details.

Post-indexed addressing is where the memory address is taken from the base register directly,
without any modification. This time the base register is modified (the offset is added or
subtracted) after the load or store has occurred.

See section 5.2 for a full discussion of the values available for an (cid:104)op2 (cid:105) value.

4.4.2 Load and Store Register Byte

Rather like the Load and Store Register instruction the LDRB (Load Register Byte) and STRB
(Store Register Byte) instructions can load an 8-bit byte from memory, or store one respectively.
The memory location is identified by an (cid:104)op2 (cid:105) value in the same way.

The load byte instruction, will load the byte into the lower 8-bits of the register. The upper 24-bit
of the register are cleared, set to zero. The byte can not be signed unless additional steps are
taken.

When storing a byte, only the lower 8-bits of the register are stored. The remaining 24-bits are
ignored.

4.4.3 Load and Store Multiple registers

The LDM (Load Multiple) and STM (Store Multiple) instructions allow a block transfer of any
number of registers to or from memory. The memory location must be in a base register, which
can be optionally updated after the transfer.

A list of registers to transfer is given, this can include the Program Counter. Each register is
transferred to or from memory in turn starting with the lowest register (R0 if given) and ending
with the highest register (R15 if given in the register list). The base register is modified with
transfer, so the system knows the memory location for the next transfer.

The base register is modified according to one of four transfer modes:

4.5. LOGICAL AND BIT MANIPULATION

47

A B C
0
0
0
0
1
0
0
0
1
1
1
1

A B C
0
0
0
0
1
0
1
0
1
0
1
1

A B C
0
0
0
1
1
0
1
0
1
0
1
1

A B C
0
0
0
1
1
0
1
0
1
1
1
1

C = A AND B

C = A BIC B

C = A EOR B

C = A ORR B

Table 4.4: True tables for Logical operations

Increment Before
Increment After

IB
IA
DB Decrement Before
DA Decrement After

The Load and Store Multiple instructions are particularly well suited for preserving registers during
a subroutine call. A Store Multiple allows registers to be saved at the start of the subroutine,
while a Load Multiple can return the registers back to their original state at the end of the routine.
See Chapter 14 for a detailed discussion of subroutines.

4.5 Logical and Bit Manipulation

The five instructions in this group all preform logical operation on two (or in one case, one) source
operands, one of which must be a register (Rn) and the other is an (cid:104)op1 (cid:105) value, producing a result
which is placed in a destination register. They can also optionally update the condition code flags
based on the result.

An (cid:104)op1 (cid:105) value must pass through the Barrel Shift before it can be used. This means that any
instruction which uses an (cid:104)op1 (cid:105) value can preform an arithmetic or logical shift. This includes the
MOV (Move) instruction on page 41. As a result of this, there is no requirement for the dedicated
shift instructions found on Complex instruction Set Computers.

The AND (Bitwise AND) instruction will AND the bits of the first operand (the register) the those
of the second operand (the (cid:104)op1 (cid:105) value). In other words, the bit will be set in the result only if it
is set in both the first and second operands, otherwise the bit will be clear. The AND is frequently
used to mask out, or clear, bits we are not interested in.

The BIC (Bit Clear) instruction is intended to take over the masking function from the AND
instruction. With this instruction a bit in the result is clear if the bit in the second operand is
set, no matter what the bit is set to in the first operand. The remanding bits are left unchanged.

The EOR (Exclusive OR) instruction will OR the bits of the first operand with those of the second
operand, except if both bits are set the resulting bit will be clear. This is known as an Exclusive
OR, as it allows a bit from the first operand or the second operand, but not from both. This can
be used to invert selected bits. By setting a bit in the second operand the corresponding bit in
the first operand will be changed, if it was set it will become clear, if it was clear it will now be
set.

The ORR (Bitwise OR) instruction will simply OR the bits of the first and second operands. The
bit in the result will be set if the corresponding bit is set in either of the two operands. If the
bit is clear in the second operand, the resulting bit will be the same as the first operand. This
is known as an inclusive OR, as it include a bit set in not only the first operand, or the second
operand, but also in both operands. This is frequently used to for a bit to be set.

Finally the MVN (Move Not) instruction is slightly different in that it only uses one operand. The
result is a complete inversion of the operands value. That is each bit is switch from set to clear and

48

CHAPTER 4.

INSTRUCTION SET

clear to set. This can be useful for making a bit-mask for use with one of the other instructions.

4.6 System Control / Privileged

4.6.1 Software Interrupt

The SWI (Software Interrupt) instructions cause a software Interrupt exception to occur. These
are normally used to request a service from an operating system. The exception causes processor
to change into a privileged mode, thus allowing a user mode task to gain access to privileged
functions, but only under the supervision of an operating system.

There are three methods an operating system could use to identify which service is required.

1. A service number is provided as an immediate value associated with the SWI. Any additional
parameters are passed between the operating system and the task though general-purpose
registers. This is the most common form of communication.

2. The service number is placed in a general-purpose register. Any additional parameters are

also passed in general-purpose registers.

3. The service number is placed in a data structure which follows immediately after the instruc-
tion. Additional parameters may be passed as part of the data structure, or in registers.

The precise details of the SWI instruction are dependent on the operating system or environment.
This should be described in the developers documentation provided with the environment.

The only function used in this book is function number 17 (or &11). In the system we are using
this represents a the exit() function, and is used to exit a program.

4.6.2 Semaphores

When there are two or more processes running on one or more processors they need a way of
communicating in order to share common resources. This can be provided by a semaphore. Two
instructions are provided that would help an Operating System to provide a semaphore.

The SWP (Swap) instruction reads a word value from memory (into a destination register) replacing
it with another value (from a source register). The memory location is specified in yet another
register. The same register can be given for the source and destination in which case the value at
the memory location and the value is the register are simply exchanged.

There is also a SWPB (Swap Byte) instruction which preforms the same operation for a single byte
rather than a full word.

4.6.3 Status Register Access

The status register can be accessed via two instructions. The MRS (Move to Register from Status)
instruction will copy the current process status register (CPSR) into a general-purpose register.
There is only one real reason for reading the status register, and that is to change one of the fields.

There is also a corresponding MSR (Move to Status from Register) instruction which replaces
specific field(s) in the current status register. Access to the control fields (process mode and
interrupt enable flags) is restricted to privileged mode. Only the condition code flags may be
overwritten in User mode, this is to stop the wayward User mode program from entering into a
privileged mode.

4.6. SYSTEM CONTROL / PRIVILEGED

49

The MRS and MSR instructions may also access the saved process status register (SPSR). As the
User and System modes do not have a SPSR register and the privileged modes do, these instruction
are limited to privileged mode only.

The Data Movement, Arithmetic, and Logical instruction groups all allow the Program Counter as
a destination register. This allows them to calculate the address of the next instruction. In addition
to this, when the PC is the destination register and the (cid:104)S (cid:105) flag is given, these instructions will
also copy the SPSR for the current mode into the CPSR. Obviously such instructions are restricted
to privileged modes, as they actually have a SPSR. This is a rather peculiar special effect which
has been provided to allow exception handlers to return the processor back to the same state it
was in before the exception occurred.

4.6.4 Coprocessor

Up to 16 coprocessors can be added on to the processor to perform additional operations. Such
coprocessors could include a Memory Management Unit (MMU), a Floating Point Unit (FPU),
and a Digital Signal Processor (DSP) Unit.

Different manufactures extend or customise the ARM by the addition of various coprocessors.
A discussion of such coprocessors is well beyond the scope of this book.
It is however worth
noting that there are five instructions for communicating with the coprocessor which fall into
three sub-groups:

Data-processing instructions which start a coprocessor specific operation, CDP (Coprocessor

Data Operation).

Data transfer instructions which transfer data between the coprocessor and memory. The LDC

(Load Coprocessor) and STC (Store Coprocessor) instructions.

Register transfer instructions to transfer data between a register and the coprocessor. In par-
ticular the MRC (Move from Register to Coprocessor) and MCR (Move from Coprocessor to
Register) instructions.

A manufacture may provide additional pseudo instructions which hide the use of the coprocessor
from you.

4.6.5 Privileged Memory Access

The Memory Access instructions (4.4 on page 46) provide access to memory in the current processor
mode. Thus if used when the processor is in a privileged mode, the memory access will also be
privileged. The could cause difficulty if the memory being accessed is only available in User mode.

To cater for this difficulty a set of memory access instructions have been provided which access
the memory in User mode, without effecting the current processor mode. These are known as the
“with Translation” instructions.

LDRT (Load Register with Translation) and STRT (Store Register with Translation) provide 32-
bit memory access with User mode Translation. The corresponding LDRBT (Load Register Byte
with Translation) and STRBT (Store Register Byte with Translation) instructions provide the 8-bit
memory access with User mode Translation.

As these instructions are used when an exception handler is required to access User-mode memory,
any further discussion is beyond the scope of an introductory text.

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CHAPTER 4.

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4.6.6 Undefined Instructions

There are many instruction words (values) which have yet to be defined. There are even some
which have been explicitly left undefined. Attempting to perform one of these undefined words
will cause an Undefined Instruction exception to occur.

An Operating System can use this exception to provide additional services, and instructions which
are specific to that operating environment. Any such additional instructions should be documented
in the developers documented provided with the environment.

Should the environment not provide any additional services in this manner it should produce a
suitable error response.

5 Addressing Modes

5.1 Data Processing Operands: (cid:104)op1 (cid:105)

The majority of the instructions relate to data processing of some form. One of the operands
to these instructions is routed through the Barrel Shifter. This means that the operand can be
modified before it is used. This can be very useful when dealing with lists, tables and other
complex data structures. We denote instructions of this type as taking one of its arguments from
(cid:104)op1 (cid:105).

An (cid:104)op1 (cid:105) argument may come from one of two sources, a constant value or a register, and be
modified in five different ways.

5.1.1 Unmodified Value

You can use a value or a register unmodified by simply giving the value or the register name. For
example the following instructions will demonstrate the two methods:

Immediate

MOV

R0, #1234

Will move the immediate constant value 123410 into the reg-
ister R0 .

R0 ← IR(value)

Register

MOV

R0, R1

Will move the value in the register R1 into the register R0 .

R0 ← R1

5.1.2 Logical Shift Left

This will take the value of a register and shift the value up, towards the most significant bit, by n
bits. The number of bits to shift is specified by either a constant value or another register. The
lower bits of the value are replaced with a zero. This is a simple way of performing a multiply by
a power of 2 (×2n).

51

52

CHAPTER 5. ADDRESSING MODES

Logical Shift Left Immediate

MOV

R0, R1, LSL #2

R0 will become the value of R1 shifted left by 2 bits. The
value of R1 is not changed.

R0 ← R1 << IR(value)

Logical Shift Left Register

MOV

R0, R1, LSL R2

R0 will become the value of R1 shifted left by the number
of bits specified in the R2 register. R0 is the only register to
change, both R1 and R2 are not effected by this operation.

R0 ← R1 << R2 (7:0)

If the instruction is to set the status register, the carry flag (C) is the last bit that was shifted out
of the value.

5.1.3 Logical Shift Right

Logical Shift Right is very similar to Logical Shift Left except it will shift the value to the right,
towards the lest significant bit, by n bits. It will replace the upper bits with zeros, thus providing
an efficient unsigned divide by 2n function (| ÷ 2n|). The number of bits to shift may be specified
by either a constant value or another register.

Logical Shift Right Immediate

MOV

R0, R1, LSR #2

R0 will take on the value of R1 shifted to the right by 2 bits.
The value of R1 is not changed.

R0 ← R1 >> IR(value)

Logical Shift Right Register

MOV

R0, R1, LSR R2

As before R0 will become the value of R1 shifted to the right
by the number of bits specified in the R2 register. R1 and
R2 are not altered by this operation.

R0 ← R1 >> R2

If the instruction is to set the status register, the carry flag (C) is the last bit to be shifted out of
the value.

5.1.4 Arithmetic Shift Right

5.1. DATA PROCESSING OPERANDS: (cid:104)OP1 (cid:105)

53

The Arithmetic Shift Right is rather similar to the Logical Shift Right, but rather than replacing
the upper bits with a zero, it maintains the value of the most significant bit. As the most significant
bit is used to hold the sign, this means the sign of the value is maintained, thus providing a signed
divide by 2n operation (÷2n).

Arithmetic Shift Right Immediate

MOV

R0, R1, ASR #2

Register R0 will become the value of register R1 shifted to
the right by 2 bits, with the sign maintained.

R0 ← R1 +>> IR(value)

Arithmetic Shift Right Register

MOV

R0, R1, ASR R2

Register R0 will become the value of the register R1 shifted
to the right by the number of bits specified by the R2 register.
R1 and R2 are not altered by this operation.

R0 ← R1 +>> R2

Given the distinction between the Logical and Arithmetic Shift Right, why is there no Arithmetic
Shift Left operation?

As a signed number is stored in two’s complement the upper most bits hold the sign of the number.
These bits can be considered insignificant unless the number is of a sufficient size to require their
use. Thus an Arithmetic Shift Left is not required as the sign is automatically preserved by the
Logical Shift.

5.1.5 Rotate Right

In the Rotate Right operation, the lest significant bit is copied into the carry (C) flag, while the
value of the C flag is copied into the most significant bit of the value. In this way none of the bits
in the value are lost, but are simply moved from the lower bits to the upper bits of the value.

Rotate Right Immediate

MOV

R0, R1, ROR #2

This will rotate the value of R1 by two bits. The most signif-
icant bit of the resulting value will be the same as the least
significant bit of the original value. The second most signifi-
cant bit will be the same as the Carry flag. In the S version
the Carry flag will be set to the second least significant bit
of the original value. The value of R1 is not changed by this
operation.

R0 ← R1 >>> IR(value)

54

CHAPTER 5. ADDRESSING MODES

Rotate Right Register

MOV

R0, R1, ROR R2

Register R0 will become the value of the register R1 rotated
to the right by the number of bits specified by the R2 register.
R1 and R2 are not altered by this operation.

R0 ← R1 >>> R2

Why is there no corresponding Rotate Left operation?

An Add With Carry (ADC, A.1 on page 137) to a zero value provides this service for a single bit.
The designers of the instruction set believe that a Rotate Left by more than one bit would never
be required, thus they have not provided a ROL function.

5.1.6 Rotate Right Extended

This is similar to a Rotate Right by one bit. The extended section of the fact that this function
moves the value of the Carry (C) flag into the most significant bit of the value, and the least
significant bit of the value into the Carry (C) flag. Thus it allows the Carry flag to be propagated
though multi-word values, thereby allowing values larger than 32-bits to be used in calculations.

Rotate Right Extend

MOV

R0, R1 RRX

The register R0 become the same as the value of the regis-
ter R1 rotated though the carry flag by one bit. The most
significant bit of the value becomes the same as the current
Carry flag, while the Carry flag will be the same as the least
significant bit or R1 . The value of R1 will not be changed.

R0 ← C >>> R1 >>> C

5.2 Memory Access Operands: (cid:104)op2 (cid:105)

The memory address used in the memory access instructions may also modified by the barrel
shifter. This provides for more advanced access to memory which is particularly useful when
dealing with more advanced data structures. It allows pre- and post-increment instructions that
update memory pointers as a side effect of the instruction. This makes loops which pass though
memory more efficient. We denote instructions of this type as taking one of its arguments from
(cid:104)op2 (cid:105).

There are three main methods of specifying a memory address ((cid:104)op2 (cid:105)), all of which include an
offset value of some form. This offset can be specified in one of three ways:

Constant Value

An immediate constant value can be provided. If no offset is specified an immediate constant
value of zero is assumed.

Register

The offset can be specified by another register. The value of the register is added to the
address held in another register to form the final address.

5.2. MEMORY ACCESS OPERANDS: (cid:104)OP2 (cid:105)

55

Scaled

The offset is specified by another register which can be scaled by one of the shift operators
used for (cid:104)op1 (cid:105). More specifically by the Logical Shift Left (LSL), Logical Shift Right (LSR),
Arithmetic Shift Right (ASR), ROtate Right (ROR) or Rotate Right Extended (RRX) shift
operators, where the number of bits to shift is specified as a constant value.

5.2.1 Offset Addressing

In offset addressing the memory address is formed by adding (or subtracting) an offset to or from
the value held in a base register.

Zero Offset

LDR

R0, [R1]

Will load the register R0 with the 32-bit word at the memory
address held in the register R1 . In this instruction there is
no offset specified, so an offset of zero is assumed. The value
of R1 is not changed in this instruction.

MAR ← R1
MBR ← M(MAR)
R0 ← MBR

Immediate Offset

LDR

R0, [R1, #4]

Will load the register R0 with the word at the memory ad-
dress calculated by adding the constant value 4 to the mem-
ory address contained in the R1 register. The register R1 is
not changed by this instruction.

MAR ← R1 + IR(value)
MBR ← M(MAR)
R0 ← MBR

Register Offset

LDR

R0, [R1, R2]

Loads the register R0 with the value at the memory address
calculated by adding the value in the register R1 to the value
held in the register R2 . Both R1 and R2 are not altered by
this operation.

MAR ← R1 + R2
MBR ← M(MAR)
R0 ← MBR

56

CHAPTER 5. ADDRESSING MODES

Scaled Register Offset

LDR

R0, [R1, R2, LSL #2]

Will load the register R0 with the 32-bit value at the memory
address calculated by adding the value in the R1 register to
the value obtained by shifting the value in R2 left by 2 bits.
Both registers, R1 and R2 are not effected by this operation.

MAR ← R1 + R2 << IR(value)
MBR ← M(MAR)
R0 ← MBR

This is particularly useful for indexing into a complex data structure. The start of the data
structure is held in a base register, R1 in this case, and the offset to access a particular field within
the structure is then added to the base address. Placing the offset in a register allows it to be
calculated at run time rather than fixed. This allows for looping though a table.

A scaled value can also be used to access a particular item of a table, where the size of the item
is a power of two. For example, to locate item 7 in a table of 32-bit values we need only shift the
index value 6 left by 2 bits (6 × 22) to calculate the value we need to add as an offset to the start
of the table held in a register, R1 in our example. Remember that the computer count from zero,
thus we use an index value of 6 rather than 7. A 32-bit number requires 4 bytes of storage which
is 22, thus we only need a 2-bit left shift.

5.2.2 Pre-Index Addressing

In pre-index addressing the memory address if formed in the same way as for offset addressing.
The address is not only used to access memory, but the base register is also modified to hold
the new value. In the ARM system this is known as a write-back and is denoted by placing a
exclamation mark after at the end of the (cid:104)op2 (cid:105) code.

Pre-Index address can be particularly useful in a loop as it can be used to automatically increment
or decrement a counter or memory pointer.

Immediate Pre-indexed

LDR

R0, [R1, #4]!

Will load the register R0 with the word at the memory ad-
dress calculated by adding the constant value 4 to the mem-
ory address contained in the R1 register. The new memory
address is placed back into the base register, register R1 .

R1 ← R1 + IR(value)

MAR ← R1
MBR ← M(MAR)
R0 ← MBR

5.2. MEMORY ACCESS OPERANDS: (cid:104)OP2 (cid:105)

57

Register Pre-indexed

LDR

R0, [R1, R2]!

Loads the register R0 with the value at the memory address
calculated by adding the value in the register R1 to the value
held in the register R2 . The offset register, R2 , is not altered
by this operation, the register holding the base address, R1 ,
is modified to hold the new address.

R1 ← R1 + R2

MAR ← R1
MBR ← M(MAR)
R0 ← MBR

Scaled Register Pre-indexed

LDR

R0, [R1, R2, LSL #2]!

First calculates the new address by adding the value in the
base address register, R1 , to the value obtained by shifting
the value in the offset register, R2 , left by 2 bits. It will then
load the 32-bit at this address into the destination register,
R0 . The new address is also written back into the base reg-
ister, R1 . The offset register, R2 , will not be effected by this
operation.

R1 ← R1 + R2 << IR(value)

MAR ← R1
MBR ← M(MAR)
R0 ← MBR

5.2.3 Post-Index Addressing

In post-index address the memory address is the base register value. As a side-effect, an offset
is added to or subtracted from the base register value and the result is written back to the base
register.

Post-index addressing uses the value of the base register without modification. It then applies the
modification to the address and writes the new address back into the base register. This can be
used to automatically increment or decrement a memory pointer after it has been used, so it is
pointing to the next location to be used.

As the instruction must preform a write-back we do not need to include an exclamation mark.
Rather we move the closing bracket to include only the base register, as that is the register holding
the memory address we are going to access.

Immediate Post-indexed

LDR

R0, [R1], #4

Will load the register R0 with the word at the memory ad-
dress contained in the base register, R1 . It will then calculate
the new value of R1 by adding the constant value 4 to the
current value of R1 .

MAR ← R1
MBR ← M(MBR)
R0 ← MBR
R1 ← R1 + IR(value)

58

CHAPTER 5. ADDRESSING MODES

Register Post-indexed

LDR

R0, [R1], R2

Loads the register R0 with the value at the memory address
held in the base register, R1 . It will then calculate the new
value for the base register by adding the value in the offset
register, R2 , to the current value of the base register. The
offset register, R2 , is not altered by this operation.

MAR ← R1
MBR ← M(MBR)
R0 ← MBR
R1 ← R1 + R2

Scaled Register Post-indexed

LDR

R0, [R1], R2, LSL #2

First loads the 32-bit value at the memory address contained
in the base register, R1 , into the destination register, R0 . It
will then calculate the new value for the base register by
adding the current value to the value obtained by shifting
the value in the offset register, R2 , left by 2 bits. The offset
register, R2 , will not be effected by this operation.

MAR ← R1
MBR ← M(MBR)
R0 ← MBR
R1 ← R1 + R2 << IR(value)

6 Programs

The only way to learn assembly language programming is through experience. Throughout the rest
of this book each chapter will introduce various aspects of assembly programming. The chapter
will start with a general discussion, then move on to a number of example programs which will
demonstrate the topic under discussion. The chapter will end with a number of programming
problems for you to try.

6.1 Example Programs

Each of the program examples contains several parts:

Title
Purpose

Problem
Algorithm
Source code
Notes

that describes the general problem
statement of purpose that describes the task the program performs
and the memory locations used.
A sample problem complete with data and results.
if the program logic is complex.
for the assembly program.
Explanatry notes that discusses the instructions and methods used
in the program.

Each example is written and assembled as a stand-alone program.

6.1.1 Program Listing Format

The examples in the book are the actual source code used to generate the programs. Sometimes
you may need to use the listing output of the ARM assembler (the .list file), and in any case you
should be aware of the fact that you can generate a listing file. See the section on the ARMulator
environment which follows for details of how to generate a .list listing file.

6.1.2 Guidelines for Examples

We have used the following guidelines in construction of the examples:

1. Standard ARM assembler notation is used, as summarized in Chapter 2.

2. The forms in which data and addresses appear are selected for clarity rather than for con-
sistency. We use hexadecimal numbers for memory addresses, instruction codes, and BCD
data; decimal for numeric constants; binary for logical masks; and ASCII for characters.

3. Frequently used instructions and programming techniques are emphasized.

4. Examples illustrate tasks that microprocessors perform in communication, instrumentation,

computers, business equipment, industrial, and military applications.

59

60

CHAPTER 6. PROGRAMS

Figure 6.1: New Project Dialog

5. Detailed comments are included.

6. Simple and clear structures are emphasised, but programs are written as efficiently as possible
within this guideline. Notes accompanying programs often describe more efficient procedures.

7. Program are written as an independent procedures or subroutines although no assumptions

are made concerning the state of the microprocessor on entry to the procedure.

8. Program end with a SWI &11 (Software Interrupt) instruction. You may prefer to modify

this by replacing the SWI &11 instruction with an endless loop instruction such as:

HERE

BAL HERE

9. Programs use standard ARM assembler directives. We introduced assembler directives con-
ceptually in Chapter 2. When first examining programming examples, you can ignore the
assembler directives if you do not understand them. Assembler directives do not contribute
to program logic, which is what you will be trying to understand initially; but they are a nec-
essary part of every assembly language program, so you will have to learn how to use them
before you write any executable programs.
Including assembler directives in all program
examples will help you become familiar with the functions they perform.

6.2 Trying the examples

To test one of the example programs, first obtain a copy of the source code. The best way of doing
this is to type in the source code presented in this book, as this will help you to understand the
code. Alternatively you can download the source from the web site, although you won’t gain the
same knowledge of the code.

Go to the start menu and call up the “Armulate” program. Next open the source file using the
normal “File | Open” menu option. This will open your program source in a separate window
within the “Armulate” environment.

The next step is to create a new Project within the environment. Select the “Project” menu option,
then “New”. Give your project the same name as the source file that you are using (there is no
need to use a file extension – it will automatically be saved as a .apj file).

Once you have given the file a name, a further dialog will open as shown in the figure 6.1.

Click the “Add” button, and you will again be presented with a file dialog, which will display the
source files in the current directory. Select the relevant source file and “OK” the dialog. You will
be returned to the previous dialog, but you will see now that your source file is included in the
project. “OK” the “Edit Project” dialog, and you will be returned to the Armulate environment,
now with two windows open within it, one for the source code and one for the project.

6.3. TRYING THE EXAMPLES FROM THE COMMAND LINE

61

Figure 6.2: Assembler Options Dialog

We recommend that you always create a .list listing file for each project that you create. Do
this by selecting the “Options” menu with the project window in focus, then the “Assembler” item.
This will open the dialog shown in figure 6.2.

Enter -list [yourfilename].list into the “Other” text box and “OK” the dialog.

You have now created your project and are ready to assemble and debug your code.

Additional information on the Armulator is available via the help menu item.

6.3 Trying the examples from the command line

When developing the example programs, we found the “Armulate” environment too clumsy. We
used the TextPad editor and assembled the programs from the command line. The Armulate
environment provides commands for use from the command line:

1. Assembler

The command line assembler is used to create an object file from the program source code.
During the development of the add program (program 7.3a) we used the command line:

ARMASM -LI -CPU ARM6 -g -list add.list add.s

2. Linker

It is necessary to position the program at a fixed location in memory. This is done using the
linker. In our add example we used the command:

ARMLINK -o add add.o

Which resolves the relative addresses in the add.o file, producing the add load image.

3. Debugger

Finally it is necessary to debug the load image. This can be done in one of two ways, using
a command line debugger or the windows debugger. In either case they require a load image
(add in our example). To use the command line debugger (known as the source debugger)
the following command is used:

ARMSD add

However, the command driven nature of this system is confusing and hard to use for even
the most experienced of developers. Thus we suggest you use the windows based debugger
program:

WINDBG add

Which will provide you with the same debugger you would have seen had you used the
Window based Armulate environment.

62

CHAPTER 6. PROGRAMS

Figure 6.3: TextPad Colour Preferences Dialog

6.3.1 Setting up TextPad

To set up this environment simply download the TextPad editor and the ARM Assembler syntax
file. You can download the editor from the download page of the TextPad web site1.

Download Derek Law’s ARM Assembler Syntax Definition file from the TextPad web site. You
can find this under the Syntax Definition sub-section of the Add-ons section of the Download page.
Unpack the armasm.syn from the arm.zip file into the TextPad Samples directory.

Having installed the Syntax Definitions you should now add a new Document Class to TextPad.
Run TextPad and select the New Document Class. . . wizard from the Configure menu. The wizard
will now take you though the following steps:

1. The Document Class requires a name. We have used the name “ARM Assembler ”.

2. The Class Members, the file name extension to associate with this document class. We

associate all .s and .list files with this class: “*.s,*.list”

3. Syntax Highlighting. The next dialog is where we tell TextPad to use syntax highlighting,
simply check the Enable Syntax Highlighting box. We now need to tell it which syntax
definition file to use. If the armasm.syn file was placed in the Samples directory, this will
appear in the drop down list, and should be selected.

While this will create the new document class, you will almost certainly want to change the colour
settings for this document class. This class uses the different levels of Keyword colouring for
different aspects of the syntax as follows:

Instructions

Keywords 1
Keywords 2 Co-processor and pseudo-instructions
Keywords 3
Shift-addresses and logical directives
Keywords 4 Registers
Keywords 5 Directives
Keywords 6 Arguments and built-in names

You will probably want to set the colour setting for all of these types to the same settings. We
have set all but Keywords 2 to the same colour scheme. To alter the colour setting you should
select the Preferences. . . option from the Configure menu.

In the “Preference” dialog (shown in figure 6.4 on the next page), open the Document Classes
section and then your new document class (ARM Assembler). Now you should select the colors
section. This will now allow you to change the colours for any of the given colour settings.

Finally you may like to consider adding a “File Type Filter” to the “Open File” dialog. This
can be done by selecting the File Type Filter entry in the Preference dialog. Simply click on the

1http://www.textpad.com

6.4. PROGRAM INITIALIZATION

63

Figure 6.4: TextPad File Name Filters Preferences Dialog

New button, add the description (“ARM Assembler (*.s, *.list)”) and wildcard (“*.s;*.list”) details.
Finally click on the OK button.

Note the use of a comma to seperate the wildcards in the description, and the use of a semi-colon
(without spaces) in the wildcard entry.

6.4 Program Initialization

All of the programming examples presented in these notes pay particular attention to the correct
initialization of constants and operands. Often this requires additional instructions that may
appear superfluous, in that they do not contribute directly to the solution of the stated problem.
Nevertheless, correct initialization is important in order to ensure the proper execution of the
program every time.

We want to stress correct initialization; that is why we are going to emphasize this aspect of
problems.

6.5 Special Conditions

For the same reasons that we pay particular attention to special conditions that can cause a pro-
gram to fail. Empty lists and zero indexes are two of the most common circumstances overlooked
in sample problems. It is critically important when using microprocessors that you learn with your
very first program to anticipate unusual circumstances; they frequently cause your program to fail.
You must build in the necessary programming steps to account for these potential problems.

6.6 Problems

Each chapter will now end with a number of programming problems for your to try. They have
been provided to help you understand the ideas presented in the chapter. You should use the
programming examples as guidelines for solving the problems. Don’t forget to run your solutions
on the ARMulator to ensure that they are correct.

The following guidelines will help in solving the problems:

1. Comment each program so that others can understand it. The comments can be brief and
ungrammatical. They should explain the purpose of a section or instruction in the program,
but should not describe the operation of instructions, that description is available in manuals.
For example the following line:

64

CHAPTER 6. PROGRAMS

ADD

R1, R1, #1

could be given the comment “Add one to R1 ” or “Increment R1 ”, both of which provide no
indication as to why the line is there. They tell us what the instruction is doing, but we can
tell that by looking at the instruction itself. We are more interested in why the instruction
is there. A comment such as “Increment loop counter” is much more useful as it explains
why you are adding one to R1 , the loop counter.

You do not have to comment each statement or explain the obvious. You may follow the
format of the examples but provide less detail.

2. Emphasise clarity, simplicity, and good structure in programs. While programs should be
reasonably efficient, do not worry about saving a single byte of program memory or a few
microseconds.

3. Make programs reasonably general. Do not confuse parameters (such as the number of

elements in any array) with fixed constants (such as the code for the letter “C”).

4. Never assume fixed initial values for parameters.

5. Use assembler notation as shown in the examples and defined in Chapter 2.

6. Use symbolic notation for address and data references. Symbolic notation should also be
used even for constants (such as DATA_SELECT instead of 2_00000100). Also use the clearest
possible form for data (such as ’C’ instead of 0x43).

7. Use meaningful names for labels and variables, e.g., SUM or CHECK rather than X or Z.

8. Execute each program with the emulator. There is no other way of ensuring that your
program is correct. We have provided sample data with each problem. Be sure that the
program works for special cases.

7 Data Movement

This chapter contains some very elementary programs. They will introduce some fundamental
features of the ARM. In addition, these programs demonstrate some primitive tasks that are
common to assembly language programs for many different applications.

7.1 Program Examples

7.1.1

16-Bit Data Transfer

Move the contents of one 16-bit variable Value to another 16-bit variable Result.

Sample Problems

Input:
Output:

= C123
Value
Result = C123

Main

TTL
AREA
ENTRY

; 16-Bit data transfer

Ch4ex1 - move16
Program, CODE, READONLY

Program 7.1: move16.s — 16bit data transfer
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16

R1, Value
R1, Result
&11

LDRB
STR
SWI

Result DCW

DCW
ALIGN

&C123

Value

0

END

; Load the value to be moved
; Store it back in a different location

; Value to be moved
; Need to do this because working with 16bit value
; Storage space

This program solves the problem in two simple steps. The first instruction loads data register R1
with the 16-bit value in location Value. The next instruction saves the 16-bit contents of data
register R1 in location Result.

As a reminder of the necessary elements of an assembler program for the ARMulator, notice that
this, and all the other example programs have the following elements. Firstly there must be an
ENTRY directive. This tells the assembler where the first executable instruction is located. Next
there must be at least one AREA directive, at the start of the program, and there may be other
AREA directives to define data storage areas. Finally there must be an END directive, to show where
the code ends. The absence of any of these will cause the assembly to fail with an error.

Another limitation to bear in mind is that ARMulator instructions will only deal with BYTE (8
bits) or WORD (32 bit) data sizes. It is possible to declare HALF-WORD (16 bit) variables by the use

65

66

CHAPTER 7. DATA MOVEMENT

of the DCW directive, but it is necessary to ensure consistency of storage of HALF-WORD by the use
of the ALIGN directive. You can see the use of this in the first worked example.

In addition, under the RISC architecture of the ARM, it is not possible to directly manipulate
data in storage. Even if no actual manipulation of the data is taking place, as in this first example,
it is necessary to use the LDR or LDRB and STR or STRB to move data to a different area of memory.

This version of the LDR instruction moves the 32-bit word contained in memory location Value
into a register and then stores it using the STR instruction at the memory location specified by
Result.

Notice that, by default, every program is allocated a literal pool (a storage area) after the last
executable line. In the case of this, and most of the other programs, we have formalised this by
the use of the AREA
Data1, DATA directive. Instruction on how to find addresses of variables
will be given in the seminars.

7.1.2 One’s Complement

From the bitwise complement of the contents of the 16-bit variable Value.

Sample Problems

Input:
Output:

Value
Result = FFFF3EDC

= C123

Main

TTL
AREA
ENTRY

; Find the one’s compliment (inverse) of a number

Ch4Ex2 - invert
Program, CODE, READONLY

Program 7.2: invert.s — Find the one’s compliment (inverse) of a number
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16

; Load the number to be complimented
; NOT the contents of R1
; Store the result

; Value to be complemented
; Storage for result

R1, Value
R1, R1
R1, Result
&11

Value
DCD
Result DCD

LDR
MVN
STR
SWI

&C123
0

END

This program solves the problem in three steps. The first instruction moves the contents of
location Value into data register R1 . The next instruction MVN takes the logical complement of
data register R1 . Finally, in the third instruction the result of the logical complement is stored in
Value.

Note that any data register may be referenced in any instruction that uses data registers, but note
the use of R15 for the program counter, R14 for the link register and R13 for the stack pointer.
Thus, in the LDR instruction we’ve just illustrated, any of the general purpose registers could have
been used.

The LDR and STR instructions in this program, like those in Program 7.1, demonstrate one of
the ARM’s addressing modes. The data reference to Value as a source operand is an example
of immediate addressing.
In immediate addressing the offset to the address of the data being
referenced (less 8 byes) is contained in the extension word(s) following the operation word of the
instruction. As shown in the assembly listing, the offset to the address corresponding to Value is
found in the extension word for the LDR and STR instructions.

7.1. PROGRAM EXAMPLES

7.1.3

32-Bit Addition

67

Add the contents of the 32-bit variable Value1 to the contents of the 32-bit variable Value2 and
place the result in the 32-bit variable Result.

Sample Problems

Input:

Value1 = 37E3C123
Value2 = 367402AA

Output:

Result = 6E57C3CD

Main

TTL
AREA
ENTRY

; Add two (32-Bit) numbers

Ch4Ex3 - add
Program, CODE, READONLY

Program 7.3a: add.s — Add two numbers
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18

R1, Value1
R2, Value2
R1, R1, R2
R1, Result
&11

Value1 DCD
Value2 DCD
Result DCD

&37E3C123
&367402AA
0

LDR
LDR
ADD
STR
SWI

END

; Load the first number
; Load the second number
; ADD them together into R1 (x = x + y)
; Store the result

; First value to be added
; Second value to be added
; Storage for result

The ADD instruction in this program is an example of a three-operand instruction. Unlike the LDR
instruction, this instruction’s third operand not only represents the instruction’s destination but
may also be used to calculate the result. The format:

DESTINATION ← SOURCE1 operation SOURCE2

is common to many of the instructions.

As with any microprocessor, there are many instruction sequences you can execute which will solve
the same problem. Program 7.3b, for example, is a modification of Program 7.3a and uses offset
addressing instead of immediate addressing.

Main

TTL
AREA
ENTRY

; Add two numbers and store the result

Ch4Ex4 - add2
Program, CODE, READONLY

Program 7.3b: add2.s — Add two numbers and store the result
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19

R0, =Value1
R1, [R0]
R0, R0, #0x4
R2, [R0]
R1, R1, R2
R0, =Result
R1, [R0]
&11

; First value
; Second value
; Space to store result

; Load the address of first value
; Load what is at that address
; Adjust the pointer
; Load what is at the new addr
; ADD together
; Load the storage address
; Store the result
; All done

Value1 DCD
Value2 DCD
Result DCD

&37E3C123
&367402AA
0

LDR
LDR
ADD
LDR
ADD
LDR
STR
SWI

68

20
21

END

CHAPTER 7. DATA MOVEMENT

The ADR pseudo-instruction introduces a new addressing mode — offest addressing, which we
have not used previously. Immediate addressing lets you define a data constant and include that
constant in the instruction’s associated object code. The assembler format identifies immediate
addressing with a # preceding the data constant. The size of the data constant varies depending
on the instruction. Immediate addressing is extremely useful when small data constants must be
referenced.

The ADR pseudo-instruction could be replaced by the use of the instruction LDR together with the
use of the = to indicate that the address of the data should be loaded rather than the data itself.

The second addressing mode — offset addressing — uses immediate addressing to load a pointer
to a memory address into one of the general purpose registers.

Program 7.3b also demonstrates the use of base register plus offset addressing. In this example
we have performed this operation manually on line 10 (ADD R0, R0, #0x4), which increments the
address stored in R0 by 4 bytes or one WORD. There are much simpler and more efficient ways of
doing this, such as pre-index or post-index addressing which we will see in later examples.

Another advantage of this addressing mode is its faster execution time as compared to immediate
addressing. This improvement occurs because the address extension word(s) does not have to be
fetched from memory prior to the actual data reference, after the initial fetch.

A final advantage is the flexibility provided by having R0 hold an address instead of being fixed as
part of the instruction. This flexibility allows the same code to be used for more than one address.
Thus if you wanted to add the values contained in consecutive variables Value3 and Value4, you
could simply change the contents of R0 .

7.1.4 Shift Left One Bit

Shift the contents of the 16-bit variable Value to the left one bit. Store the result back in Result.

Sample Problems

Input:
Output:

= 4242
Value
Result = 8484

(0100 0010 0100 00102)
(1000 0100 1000 01002)

Main

TTL
AREA
ENTRY

; Shift Left one bit

Ch4Ex5 - shiftleft
Program, CODE, READONLY

Program 7.4: shiftleft.s — Shift Left one bit
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16

R1, Value
R1, R1, LSL #0x1
R1, Result
&11

Value
DCD
Result DCD

LDR
MOV
STR
SWI

&4242
0

END

; Load the value to be shifted
; SHIFT LEFT one bit
; Store the result

; Value to be shifted
; Space to store result

The MOV instruction is used to perform a logical shift left. Using the operand format of the MOV
instruction shown in Program 7.4, a data register can be shifted from 1 to 25 bits on either a byte,

7.1. PROGRAM EXAMPLES

69

word or longword basis. Another form of the LSL operation allows a shift counter to be specified
in another data register.

7.1.5 Byte Disassembly

Divide the least significant byte of the 8-bit variable Value into two 4-bit nibbles and store one
nibble in each byte of the 16-bit variable Result. The low-order four bits of the byte will be stored
in the low-order four bits of the least significant byte of Result. The high-order four bits of the
byte will be stored in the low-order four bits of the most significant byte of Result.

Sample Problems

Input:
Output:

Value
Result = 050F

= 5F

Main

TTL
AREA
ENTRY

LDR
LDR
MOV
MOV
AND
ADD

Ch4Ex6 - nibble
Program, CODE, READONLY

; Disassemble a byte into its high and low order nibbles

Program 7.5: nibble.s — Disassemble a byte into its high and low order nibbles
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24

; Load the value to be disassembled
; Load the bitmask
; Copy just the high order nibble into R3
; Now left shift it one byte
; AND the original number with the bitmask
; Add the result of that to
; What we moved into R3
; Store the result

R1, Value
R2, Mask
R3, R1, LSR #0x4
R3, R3, LSL #0x8
R1, R1, R2
R1, R1, R3

; Value to be shifted
; Keep the memory boundaries
; Bitmask = %0000000000001111

; Space to store result

DCB
ALIGN
DCW
ALIGN

R1, Result
&11

Result DCD

STR
SWI

Value

&000F

Mask

&5F

0

END

This is an example of byte manipulation. The ARM allows most instructions which operate on
words also to operate on bytes. Thus, by using the B suffix, all the LDRinstructions in Program 7.5
become LDRB instructions, therefore performing byte operations. The STR instruction must remain,
since we are storing a halfword value. If we were only dealing with a one byte result, we could use
the STRB byte version of the store instruction.

Remember that the MOV instruction performs register-to-register transfers. This use of the MOV
instruction is quite frequent.

Generally, it is more efficient in terms of program memory usage and execution time to minimise
references to memory.

7.1.6 Find Larger of Two Numbers

Find the larger of two 32-bit variables Value1 and Value2. Place the result in the variable Result.
Assume the values are unsigned.

Sample Problems

70

CHAPTER 7. DATA MOVEMENT

Compare Condition Signed Unsigned

greater than or equal
greater than
equal
not equal
less than or equal
less than

BGE
BGT
BEQ
BNE
BLE
BLT

BHS
BHI
BEQ
BNE
BLS
BLO

Table 7.1: Signed/Unsigned Comparisons

Input:

Value1 = 12345678
Value2 = 87654321

a

b

12345678
0ABCDEF1

Output:

Result = 87654321

12345678

Main

TTL
AREA
ENTRY

; Find the larger of two numbers

Ch4Ex7 - bigger
Program, CODE, READONLY

Program 7.6: bigger.s — Find the larger of two numbers
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21

R1, Value1
R2, Value2
R1, R2
Done
R1, R2

Value1 DCD
Value2 DCD
Result DCD

&12345678
&87654321
0

R1, Result
&11

LDR
LDR
CMP
BHI
MOV

STR
SWI

Done

END

; Store the result

; Value to be compared
; Value to be compared
; Space to store result

; Load the first value to be compared
; Load the second value to be compared
; Compare them
; If R1 contains the highest
; otherwise overwrite R1

The Compare instruction, CMP, sets the status register flags as if the destination, R1 , were sub-
tracted from the source R2 . The order of the operands is the same as the operands in the subtract
instruction, SUB.

The conditional transfer instruction BHI transfers control to the statement labeled Done if the
unsigned contents of R2 are greater than or equal to the contents of R1 . Otherwise, the next
instruction (on line 12) is executed. At Done, register R2 will always contain the larger of the two
values.

The BHI instruction is one of several conditional branch instructions. To change the program to
operate on signed numbers, simply change the BHI to BGE (Branch if Greater than or Equal to):

...
CMP
BGE
...

R1, R2
Done

You can use the following table 8.1 to use when performing signed and unsigned comparisons.

Note that the same instructions are used for signal and unsigned addition, subtraction, or com-
parison; however, the comparison operations are different.

7.1. PROGRAM EXAMPLES

71

The conditional branch instructions are an example of program counter relative addressing. In
other words, if the branch condition is satisfied, control will be transfered to an address relative
to the current value of the program counter. Dealing with compares and branches is an important
part of programming. Don’t confuse the sense of the CMP instruction. After a compare, the relation
tested is:

DESTINATION condition SOURCE

For exampe, if the condition is “less than,” then you test for destination less than source. Become
familiar with all of the conditions and their meanings. Unsigned compares are very useful when
comparing two addresses.

7.1.7

64-Bit Adition

Add the contents of two 64-bit variables Value1 and Value2. Store the result in Result.

Sample Problems

Input:

Output:

Value1 = 12A2E640, F2100123
40023F51
Value2 = 001019BF,
32124074
Result = 12B30000,

Main

TTL
AREA
ENTRY

; 64 bit addition

Ch4Ex8 - add64
Program, CODE, READONLY

Program 7.7: add64.s — 64 bit addition
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26

R0, =Value1
R1, [R0]
R2, [R0, #4]
R0, =Value2
R3, [R0]
R4, [R0, #4]
R6, R2, R4
R5, R1, R3
R0, =Result
R5, [R0]

&12A2E640, &F2100123
&001019BF, &40023F51
0

LDR
LDR
LDR
LDR
LDR
LDR
ADDS
ADC
LDR
STR

Value1 DCD
Value2 DCD
Result DCD

R6, [R0, #4]
&11

STR
SWI

END

; Pointer to first value
; Load first part of value1
; Load lower part of value1
; Pointer to second value
; Load upper part of value2
; Load lower part of value2
; Add lower 4 bytes and set carry flag
; Add upper 4 bytes including carry
; Pointer to Result
; Store upper part of result

; Store lower part of result

; Value to be added
; Value to be added
; Space to store result

Here we introduce several important and powerful instructions from the ARM instruction set. As
before, at line 8 we use the LDR instruction which causes register R0 to hold the starting address
of Value1. At line 9 the instruction LDR
R1, [R0] fetches the first 4 bytes (32-bits) of the 64-bit
value, starting at the location pointed to by R0 and places them in the R1 register. Line 10 loads
the second 4 bytes or the lower half of the 64-bit value from the memroy address pointed to by
R0 plus 4 bytes ([R0, #4]. Between them R1 and R2 now hold the first 64-bit value, R1 has
the upper half while R2 has the lower half. Lines 11–13 repeat this process for the second 64-bit
value, reading it into R3 and R4 .

Next, the two low order word s, held in R2 and R4 are added, and the result stored in R6 .

72

CHAPTER 7. DATA MOVEMENT

This is all straightforward, but note now the use of the S suffix to the ADD instruction. This forces
the update of the flags as a result of the ADD operation. In other words, if the result of the addition
results in a carry, the carry flag bit will be set.

Now the ADC (add with carry) instruction is used to add the two high order word s, held in R1 and
R3 , but taking into account any carry resulting from the previous addition.

Finally, the result is stored using the same technique as we used the load the values (lines 16–18).

7.1.8 Table of Factorials

Calculate the factorial of the 8-bit variable Value from a table of factorials DataTable. Store the
result in the 16-bit variable Result. Assume Value has a value between 0 and 7.

Sample Problems

Input:

FTABLE = 0001
= 0001
= 0002
= 0006
= 0018
= 0078
= 02D0
= 13B0
= 05

Value

110)
(0! =
110)
(1! =
210)
(2! =
610)
(3! =
2410)
(4! =
12010)
(5! =
(6! =
72010)
(7! = 504010)

Output:

Result = 0078

(5! =

12010)

Main

LDR
LDR
MOV

ADD
LDR
LDR
STR

TTL
AREA
ENTRY

R0, R0, R1
R2, [R0]
R3, =Result
R2, [R3]

Ch4Ex9 - factorial
Program, CODE, READONLY

R0, =DataTable
R1, Value
R1, R1, LSL #0x2

; Lookup the factorial from a table using the address of the memory location

; Load the address of the lookup table
; Offset of value to be looked up
; Data is declared as 32bit - need
; to quadruple the offset to point at the
; correct memory location
; R0 now contains memory address to store

Program 7.8: factorial.s — Lookup the factorial from a table by using the address of the memory
location
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34

; The address where we want to store the answer
; Store the answer

;0! = 1
;1! = 1
;2! = 2
;3! = 6
;4! = 24
;5! = 120
;6! = 720
;7! = 5040

DCD
DCD
DCD
DCD
DCD
DCD
DCD
DCD
DCB
ALIGN

1
1
2
6
24
120
720
5040
5

; The data table containing the factorials

DataTable, DATA

Result DCW

Value

AREA

&11

SWI

0

END

7.2. PROBLEMS

73

The approach to this table lookup problem, as implemented in this program, demonstrates the
use of offset addressing. The first two LDR instructions, load register R0 with the start address of
the lookup table1, and register R1 contents of Value.

The actual calculation of the entry in the table is determined by the first operand of the R1, R1,
LSL #0x2 instruction. The long word contents of address register R1 are added to the long word
contents of data register R0 to form the effective address used to index the table entry. When R0
is used in this manner, it is referred to as an index register.

7.2 Problems

7.2.1

64-Bit Data Transfer

Move the contents of the 64-bit variable VALUE to the 64-bit variable RESULT.

Sample Problems

Input:

VALUE

Output:

RESULT

3E2A42A1
21F260A0

3E2A42A1
21F260A0

7.2.2

32-Bit Subtraction

Subtract the contents of the 32-bit variable VALUE1 from the contents of the 32-bit variable VALUE2
and store the result back in VALUE1.

Sample Problems

Input:

VALUE1
VALUE2

12343977
56782182

Output:

VALUE1

4443E80B

7.2.3 Shift Right Three Bits

Shift the contents of the 32-bit variable VALUE right three bits. Clear the three most significant
bit postition.

Sample Problems

Input:

VALUE

Test A
415D7834

Test B
9284C15D

Output:

VALUE

082BAF06

1250982B

7.2.4 Halfword Assembly

Combine the low four bits of each of the four consecutive bytes beginning at LIST into one 16-bit
halfword. The value at LIST goes into the most significant nibble of the result. Store the result
in the 32-bit variable RESULT.

1 Note that we are using a LDR instruction as the data table is sufficently far away from the instruction that an

ADR instruction is not valid.

CHAPTER 7. DATA MOVEMENT

74

Sample Problems

Input:

LIST

0C
02
06
09

Output:

RESULT

0000C269

7.2.5 Find Smallest of Three Numbers

The three 32-bit variables VALUE1, VALUE2 and VALUE3, each contain an unsigned number. Store
the smallest of these numbers in the 32-bit variable RESULT.

Sample Problems

Input:

VALUE1
VALUE2
VALUE3

91258465
102C2056
70409254

Output:

RESULT

102C2056

7.2.6 Sum of Squares

Calculate the squares of the contents of word VALUE1 and word VALUE2 then add them together.
Please the result into the word RESULT.

Sample Problems

Input:

VALUE1
VALUE2

00000007
00000032

Output:

RESULT

000009F5

That is 72 + 502 = 49 + 2500 = 2549 (decimal)
or 72 + 322 = 31 + 9C4 = 9F 5 (hexadecimal)

7.2.7 Shift Left n bits

Shift the contents of the word VALUE left. The number of bits to shift is contained in the word
COUNT. Assume that the shift count is less than 32. The low-order bits should be cleared.

Sample Problems

Input:

VALUE
COUNT

Test A
182B
0003

Output:

VALUE C158

Test B
182B
0020

0000

In the first case the value is to be shifted left by three bits, while in the second case the same
value is to be shifted by thirty two bits.

8 Logic

8.1 Program Examples

8.1.1 Find Larger of Two Numbers

Find the larger of two 32-bit variables Value1 and Value2. Place the result in the variable Result.
Assume the values are unsigned.

Sample Problems

Input:

Value1 = 12345678
Value2 = 87654321

a

b

12345678
0ABCDEF1

Output:

Result = 87654321

12345678

Program 8.1: bigger.s — Find the larger of two numbers

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21

; Find the larger of two numbers

TTL
AREA
ENTRY

LDR
LDR
CMP
BHI
MOV

STR
SWI

Main

Done

Ch4Ex7 - bigger
Program, CODE, READONLY

R1, Value1
R2, Value2
R1, R2
Done
R1, R2

R1, Result
&11

; Load the first value to be compared
; Load the second value to be compared
; Compare them
; If R1 contains the highest
; otherwise overwrite R1

; Store the result

Value1 DCD
Value2 DCD
Result DCD

&12345678
&87654321
0

; Value to be compared
; Value to be compared
; Space to store result

END

The Compare instruction, CMP, sets the status register flags as if the destination, R1 , were sub-
tracted from the source R2 . The order of the operands is the same as the operands in the subtract
instruction, SUB.

The conditional transfer instruction BHI transfers control to the statement labelled Done if the
unsigned contents of R2 are greater than or equal to the contents of R1 . Otherwise, the next
instruction (on line 12) is executed. At Done, register R2 will always contain the larger of the two
values.

75

76

CHAPTER 8. LOGIC

Compare Condition Signed Unsigned

greater than or equal
greater than
equal
not equal
less than or equal
less than

BGE
BGT
BEQ
BNE
BLS
BLT

BCC
BHI
BEQ
BNE
BLS
BCS

Table 8.1: Signed/Unsigned Comparisons

The BHI instruction is one of several conditional branch instructions. To change the program to
operate on signed numbers, simply change the BHI to BGE (Branch if Greater than or Equal to):

...
CMP
BGE
...

R1,R2
Done

Table 8.1 gives the branch instructions to use when performing signed and unsigned comparisons.

Note that the same instructions are used for signed and unsigned addition, subtraction, or com-
parison; however, the comparison operations are different.

The conditional branch instructions are an example of program counter relative addressing. In
other words, if the branch condition is satisfied, control will be transferred to an address relative
to the current value of the program counter.

Dealing with compares and branches is an important part of programming. Don’t confuse the
sense of the CMP instruction. After a compare, the relation tested is:

DESTINATION condition SOURCE

For example, if the condition is “less than,” then you test for destination less than source. Become
familiar with all of the conditions and their meanings. Unsigned compares are very useful when
comparing two addresses.

8.1.2

64-Bit Adition

Add the contents of two 64-bit variables Value1 and Value2. Store the result in Result.

Sample Problems

Input:

Output:

Value1 = 12A2E640, F2100123
40023F51
Value2 = 001019BF,
32124074
Result = 12B30000,

TTL
AREA
ENTRY

; 64 bit addition

Ch4Ex8 - add64
Program, CODE, READONLY

Program 8.2: add64.s — 64 bit addition
1
2
3
4
5
6
7
8
9
10
11
12

R0, =Value1
R1, [R0]
R2, [R0, #4]
R0, =Value2
R3, [R0]

LDR
LDR
LDR
LDR
LDR

Main

; Pointer to first value
; Load first part of value1
; Load lower part of value1
; Pointer to second value
; Load upper part of value2

8.1. PROGRAM EXAMPLES

77

13
14
15
16
17
18
19
20
21
22
23
24
25
26

LDR
ADDS
ADC
LDR
STR

STR
SWI

R4, [R0, #4]
R6, R2, R4
R5, R1, R3
R0, =Result
R5, [R0]

R6, [R0, #4]
&11

; Load lower part of value2
; Add lower 4 bytes and set carry flag
; Add upper 4 bytes including carry
; Pointer to Result
; Store upper part of result

; Store lower part of result

Value1 DCD
Value2 DCD
Result DCD

&12A2E640, &F2100123
&001019BF, &40023F51
0

; Value to be added
; Value to be added
; Space to store result

END

Here we introduce several important and powerful instructions from the ARM instruction set. The
first of these is LDMIA, which may appear to be a confusing acronym, but which simply stands for
“Load many increment after”. As before, at line 8 we use the ADR instruction which causes register
R0 to hold the starting address of Value1. At line 9 the instruction LDMIA R0, {R1,R2} fetches
the next two 32-bit values, starting at the location pointed to by R0 and places them in registers
R1 and R2 . This process is repeated for the next two 32-bit values.

Next, the two low order WORDs, held in R2 and R4 are added, and the result stored in R6 .

This is all straightforward, but note now the use of the S suffix to the ADD instruction. This forces
the update of the flags as a result of the ADD operation. In other words, if the result of the addition
results in a carry, the carry flag bit will be set.

Now the ADC (add with carry) instruction is used to add the two high order WORDs, held in R1 and
R3 , but taking into account any carry resulting from the previous addition.

Finally, the STMIA instruction is used to store the result now held in R5 and R6 .

8.1.3 Table of Factorials

Calculate the factorial of the 8-bit variable Value from a table of factorials DataTable. Store the
result in the 16-bit variable Result. Assume Value has a value between 0 and 7.

Sample Problems

Input:

FTABLE = 0001
= 0001
= 0002
= 0006
= 0018
= 0078
= 02D0
= 13B0
= 05

Value

110)
(0! =
110)
(1! =
210)
(2! =
610)
(3! =
2410)
(4! =
12010)
(5! =
(6! =
72010)
(7! = 504010)

Output:

Result = 0078

(5! =

12010)

; Lookup the factorial from a table using the address of the memory location

Program 8.3: factorial.s — Lookup the factorial from a table by using the address of the memory
location
1
2
3
4
5
6
7
8

Ch4Ex9 - factorial
Program, CODE, READONLY

; Load the address of the lookup table

TTL
AREA
ENTRY

R0, =DataTable

Main

LDR

78

9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34

CHAPTER 8. LOGIC

; Offset of value to be looked up
; Data is declared as 32bit - need
; to quadruple the offset to point at the
; correct memory location
; R0 now contains memory address to store

; The address where we want to store the answer
; Store the answer

LDR
MOV

ADD
LDR
LDR
STR

SWI

R1, Value
R1, R1, LSL #0x2

R0, R0, R1
R2, [R0]
R3, =Result
R2, [R3]

&11

AREA

DataTable, DATA

; The data table containing the factorials

;0! = 1
;1! = 1
;2! = 2
;3! = 6
;4! = 24
;5! = 120
;6! = 720
;7! = 5040

DCD
DCD
DCD
DCD
DCD
DCD
DCD
DCD
DCB
ALIGN

1
1
2
6
24
120
720
5040
5

Value

Result DCW

0

END

The approach to this table lookup problem, as implemented in this program, demonstrates the
use of offset addressing. The first two LDR instructions, load register R0 with the start address of
the lookup table1, and register R1 contents of Value.

The actual calculation of the entry in the table is determined by the first operand of the R1, R1,
LSL #0x2 instruction. The long word contents of address register R1 are added to the long word
contents of data register R0 to form the effective address used to index the table entry. When R0
is used in this manner, it is referred to as an index register.

8.2 Problems

8.2.1

64-Bit Data Transfer

Move the contents of the 64-bit variable VALUE to the 64-bit variable RESULT.

Sample Problems

Input:

VALUE

Output:

RESULT

3E2A42A1
21F260A0

3E2A42A1
21F260A0

8.2.2

32-Bit Subtraction

Subtract the contents of the 32-bit variable VALUE1 from the contents of the 32-bit variable VALUE2
and store the result back in VALUE1.

Sample Problems

1 Note that we are using a LDR instruction as the data table is sufficently far away from the instruction that an

ADR instruction is not valid.

8.2. PROBLEMS

Input:

VALUE1
VALUE2

12343977
56782182

Output:

VALUE1

68AC5AF9

8.2.3 Shift Right Three Bits

79

Shift the contents of the 32-bit variable VALUE right three bits. Clear the three most significant
bit position.

Sample Problems

Input:

VALUE

Test A
415D7834

Test B
9284C15D

Output:

VALUE

082BAF06

1250982B

8.2.4 Halfword Assembly

Combine the low four bits of each of the four consecutive bytes beginning at LIST into one 16-bit
halfword. The value at LIST goes into the most significant nibble of the result. Store the result
in the 32-bit variable RESULT.

Sample Problems

Input:

LIST

0C
02
06
09

Output:

RESULT

0000C269

8.2.5 Find Smallest of Three Numbers

The three 32-bit variables VALUE1, VALUE2 and VALUE3, each contain an unsigned number. Store
the smallest of these numbers in the 32-bit variable RESULT.

Sample Problems

Input:

VALUE1
VALUE2
VALUE3

91258465
102C2056
70409254

Output:

RESULT

102C2056

8.2.6 Sum of Squares

Calculate the squares of the contents of word VALUE1 and word VALUE2 then add them together.
Please the result into the word RESULT.

Sample Problems

Input:

VALUE1
VALUE2

00000007
00000032

Output:

RESULT

000009F5

That is 72 + 502 = 49 + 2500 = 2549 (decimal)
or 72 + 322 = 31 + 9C4 = 9F 5 (hexadecimal)

80

CHAPTER 8. LOGIC

8.2.7 Shift Left n bits

Shift the contents of the word VALUE left. The number of bits to shift is contained in the word
COUNT. Assume that the shift count is less than 32. The low-order bits should be cleared.

Sample Problems

Input:

VALUE
COUNT

Test A
182B
0003

Output:

VALUE C158

Test B
182B
0020

0000

In the first case the value is to be shifted left by three bits, while in the second case the same
value is to be shifted by thirty two bits.

9 Program Loops

The program loop is the basic structure that forces the CPU to repeat a sequence of instructions.
Loops have four sections:

1. The initialisation section, which establishes the starting values of counters, pointers, and

other variables.

2. The processing section, where the actual data manipulation occurs. This is the section that

does the work.

3. The loop control section, which updates counters and pointers for the next iteration.

4. The concluding section, that may be needed to analyse and store the results.

The computer performs Sections 1 and 4 only once, while it may perform Sections 2 and 3 many
times. Therefore, the execution time of the loop depends mainly on the execution time of Sections
2 and 3. Those sections should execute as quickly as possible, while the execution times of Sections
1 and 4 have less effect on overall program speed.

There are typically two methods of programming a loop, these are
the “repeat . . . until” loop (Algorithm 9.1a) and the “while” loop
(Algorithm 9.1b). The repeat-until loop results in the computer
always executing the processing section of the loop at least once.
On the other hand, the computer may not execute the processing
section of the while loop at all. The repeat-until loop is more
natural, but the while loop is often more efficient and eliminates
the problem of going through the processing sequence once even where there is no data for it to
handle.

Processing Section
Loop Control Section

Initialisation Section
Repeat

Until task completed
Concluding Section

Algorithm 9.1a

The computer can use the loop structure to process large sets
of data (usually called “arrays”). The simplest way to use one
sequence of instructions to handle an array of data is to have the
program increment a register (usually an index register or stack
pointer) after each iteration. Then the register will contain the
address of the next element in the array when the computer repeats the sequence of instructions.
The computer can then handle arrays of any length with a single program.

Initialisation Section
While task incomplete
Processing Section

Algorithm 9.1b

Repeat

Register indirect addressing is the key to the processing arrays since it allows you to vary the
actual address of the data (the “effective address”) by changing the contents of a register. The
autoincrementing mode is particularly convenient for processing arrays since it automatically up-
dates the register for the next iteration. No additional instruction is necessary. You can even have
an automatic increment by 2 or 4 if the array contains 16-bit or 32-bit data or addresses.

Although our examples show the processing of arrays with autoincrementing (adding 1, 2, or 4 after
each iteration), the procedure is equally valid with autodecrementing (subtracting 1, 2, or 4 before
each iteration). Many programmers find moving backward through an array somewhat awkward
and difficult to follow, but it is more efficient in many situations. The computer obviously does
not know backward from forward. The programmer, however, must remember that the processor

81

82

CHAPTER 9. PROGRAM LOOPS

increments an address register after using it but decrements an address register before using it.
This difference affects initialisation as follows:

1. When moving forward through an array (autoincrementing), start the register pointing to

the lowest address occupied by the array.

2. When moving backward through an array (autodecrementing), start the register pointing

one step (1, 2, or 4) beyond the highest address occupied by the array.

9.1 Program Examples

9.1.1 Sum of Numbers

Algorithm 9.1a

8
9

Pointer ← Table
Sum ← 0

Repeat

10 Count ← M(Length)
11
12
13
14
15
15–16

temp ← M(Pointer)
Sum ← Sum + temp
Pointer ← Pointer + 4
Count ← Count − 1

Until Count = 0

R0
R1

R2

17 M(Result) ← Sum

Program 9.1a: sum1.s — Add a table of 32 bit numbers

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31

*

Add a series of 16 bit numbers by using a table address look-up

Main

Loop

TTL
AREA
ENTRY

LDR
EOR
LDR

LDR
ADD
ADD
SUBS
BNE
STR
SWI

Ch5Ex1
Program, CODE, READONLY

R0, =Table
R1, R1, R1
R2, Length

R3, [R0]
R1, R1, R3
R0, R0, #+4
R2, R2, #0x1
Loop
R1, Result
&11

AREA

Data1, DATA

Table

DCW
DCW
DCW
TablEnd DCD

&2040
&1C22
&0242
0

; load the address of the lookup table
; clear R1 to store sum
; init element count

; get the data
; add it to r1
; increment pointer
; decrement count with zero set
; if zero flag is not set, loop
; otherwise done - store result

; table of values to be added

AREA

Length DCW
Result DCW

Data2, DATA
(TablEnd - Table) / 4
0

; gives the loop count
; storage for result

END

9.1. PROGRAM EXAMPLES

83

Algorithm 9.1b

8
9

Pointer ← Table
Sum ← 0

10 Count ← M(Length)

Repeat

If Count (cid:54)= 0 Then

11–12
15
16
17
18
19
19–20
22
23 M(Result) ← Sum

Until Count = 0

temp ← M(Pointer)
Sum ← Sum + temp
Pointer ← Pointer + 4
Count ← Count − 1

End If

R0
R1
R2

R3

Program 9.1b: sum2.s — Add a table of 32 bit numbers

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35

*
*

Add a table of 32 bit numbers
This example has an empty table, and the program handles this

TTL
AREA
ENTRY

LDR
EOR
LDR
CMP
BEQ

LDR
ADD
ADD
SUBS
BNE

Main

Loop

Done

Ch5Ex2
Program, CODE, READONLY

R0, =Table
R1, R1, R1
R2, Length
R2, #0
Done

R3, [R0]
R1, R1, R3
R0, R0, #+4
R2, R2, #0x1
Loop

STR
SWI

R1, Result
&11

AREA

Data1, DATA

; load the address of the lookup table
; clear R1 to store sum
; init element count
; is count zero?
; Yes => Then we are Done

; get the data that R0 points to
; add it to r1
; increment pointer
; decrement count with zero set
; if zero flag is not set, loop

; store result

Table
TablEnd DCD

0

;Table is empty

AREA

Length DCW
Result DCW

Data2, DATA
(TablEnd - Table) / 4
0

; Calculate table length
; storage for result

END

9.1.2

64-bit Sum of Numbers

Program 9.2: bigsum.s — Add a table of 32-bit numbers into a 64-bit number

File loops/bigsum.s does not exist!

84

CHAPTER 9. PROGRAM LOOPS

9.1.3 Number of negative elements

R0
R1
R2

R3

Algorithm 9.3

Pointer ← Table

8
9 NegCount ← 0

Repeat

temp ← M(Pointer)
If temp (cid:54)> 0 Then

LoopCount ← M(Length)
If LoopCount (cid:54)= 0 Then

10
11–12
13
14
15–16
17
18
19
20
20–21
22
23 M(Result) ← NegCount

Until LoopCount = 0

End If

NegCount ← NegCount + 1

End If
Pointer ← Pointer + 4
LoopCount ← LoopCount − 1

*

Main

Loop

LDR
CMP
BPL
ADD

TTL
AREA
ENTRY

LDR
EOR
LDR
CMP
BEQ

R3, [R0]
R3, #0
Looptest
R1, R1, #1

Ch5Ex3
Program, CODE, READONLY

R0, =Table
R1, R1, R1
R2, Length
R2, #0
Done

Count the number of negative values in a table of 32-bit numbers

; load the address of the lookup table
; clear R1 to store count
; init element count
; is table empty?
; Yes => Skip loop

Program 9.3a: cntneg1.s — Count the number of negative values in a table of 32-bit numbers
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37

; Read data from table
; Is value < 0
; Yes => skip next line
; No => increment -ve number count

; increment pointer
; decrement loop count
; repeat unless count is zero

Data2, DATA
(TablEnd - Table) / 4
0

DCD
DCD
DCD
TablEnd DCD

; gives the loop count
; storage for result

&F1522040
&7F611C22
&80000242
0

R0, R0, #+4
R2, R2, #0x1
Loop

; table of values to be added

Length DCW
Result DCW

R1, Result
&11

; store result

ADD
SUBS
BNE

Data1, DATA

Looptest

STR
SWI

Table

AREA

AREA

Done

END

*

Count the number of negative values in a table of 32-bit numbers

Program 9.3b: cntneg2.s — Count the number of negative values in a table of 32-bit numbers
1
2
3
4
5
6

Ch5Ex4
Program, CODE, READONLY

TTL
AREA
ENTRY

9.1. PROGRAM EXAMPLES

85

7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35

Main

Loop

Done

LDR
EOR
LDR
CMP
BEQ

LDR
MOV
ADDCS
ADD
SUBS
BNE

R0, =Table
R1, R1, R1
R2, Length
R2, #0
Done

R3, [R0]
R3, R3 LSL #1
R1, R1, #1
R0, R0, #+4
R2, R2, #0x1
Loop

STR
SWI

R1, Result
&11

AREA

Data1, DATA

Table

DCD
DCD
DCD
TablEnd DCD

&F1522040
&7F611C22
&80000242
0

; load the address of the lookup table
; clear R1 to store count
; init element count
; is table empty?
; Yes => No need to scan it

; get the data
; Shift top bit into Carry
; Inc NegCount if Carry Set
; Increment pointer
; Decrement Loop Count
; Repeat Loop if Count not zero

; Store result

; table of values to be added

AREA

Length DCW
Result DCW

Data2, DATA
(TablEnd - Table) / 4
0

; Calculate the loop count
; Storage for result

END

9.1.4 Find Maximum Value

R0
R1
R2

R3

Algorithm 9.4

Pointer ← Table

8
9 Max ← 0

10 Count ← Length

temp ← M(Pointer)
If temp (cid:54)≤ Max Then

Repeat

If Count (cid:54)= 0 Then

11–12
13
14
15–16
17
18
19
20
20–21
22
23 M(Result) ← Max

Until Count = 0

End If

Max ← temp

End If
Pointer ← Pointer + 4
Count ← Count − 1

*

TTL
AREA
ENTRY

Find largest unsigned 32 bit value in a table

Ch5Ex5
Program, CODE, READONLY

Program 9.4: largest.s — Find largest unsigned 32 bit value in a table
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15

; load the address of the lookup table
; clear Max
; init loop count
; is Loop Count empty?
; Yes => No table to search!

; Get the data
; Is it Lower or Same as current Max ?

R0, =Table
R1, R1, R1
R2, Length
R2, #0
Done

R3, [R0]
R3, R1

LDR
EOR
LDR
CMP
BEQ

LDR
CMP

Loop

Main

86

CHAPTER 9. PROGRAM LOOPS

16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39

; Yes => skip next line
; No => Make this the current Max

; increment pointer
; decrement loop count
; Repeat if loop count not zero

; store result

; table of values to be tested

Looptest

Done

BLS
MOV

ADD
SUBS
BNE

STR
SWI

Looptest
R1, R3

R0, R0, #+4
R2, R2, #0x1
Loop

R1, Result
&11

AREA

Data1, DATA

Table

DCW
DCW
DCW
DCW
TablEnd DCD

&A152
&7F61
&F123
&8000
0

AREA

Data2, DATA

Length DCW
Result DCW

(TablEnd - Table) / 4
0

; Calculste the loop count
; storage for result

END

9.1.5 Normalise A Binary Number

Algorithm 9.5

Pointer ← Table

8
9 Count ← 0

10 Value ← M(Pointer)

R0
R1
R2

Repeat

Count ← Count + 1
Value ← Value << 1

If Value (cid:54)= 0 Then

11–12
13
14
15
16
17
18 M(Shift) ← Count
19 M(Normal) ← Value

Until Value[31] = 1

End If

*

Main

TTL
AREA
ENTRY

Normalize a Binary Number

Ch5Ex6
Program, CODE, READONLY

Program 9.5a: normal1.s — Normalize a binary number
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22

R0, =Table
R1, R1, R1
R3, [R0]
R3, R1
Done

R1, R1, #1
R3, R3, LSL#0x1
Loop

R1, Shifted
R3, Normal
&11

LDR
EOR
LDR
CMP
BEQ

ADD
MOVS
BPL

STR
STR
SWI

Data1, DATA

AREA

Done

Loop

; load the address of the lookup table
; clear R1 to store shifts
; get the data
; bit is 1
; if table is empty

; increment pointer
; decrement count with zero set
; if negative flag is not set, loop

; otherwise done - store result

9.2. PROBLEMS

87

; table of values to be tested

; storage for shift

; storage for result

23
24
25
26
27
28
29
30
31
32
33
34
35
36
37

Table
*
*
*

DCD
DCD
DCD
DCD

&30001000
&00000001
&00000000
&C1234567

AREA

Result, DATA

Number DCD
Shifted DCB

Table
0

ALIGN

Normal DCD

0

END

Algorithm 9.5b

Value ← M(Number)
Count ← 0
While Value[31] = 0

Value ← Value << 1
Count ← Count + 1

End While
M(Shift) ← Count
M(Normal) ← Value

R0
R1

Program 9.5b: normal2.s — Normalise a binary number

File loops/normal2.s does not exist!

9.2 Problems

9.2.1 Checksum of data

Calculate the checksum of a series of 8-bit numbers. The length of the series is defined by the
variable LENGTH. The label START indicates the start of the table. Store the checksum in the
variable CHECKSUM. The checksum is formed by adding all the numbers in the list, ignoring the
carry over (or overflow).

Note: Checksums are often used to ensure that data has been correctly read. A checksum calcu-
lated when reading the data is compared to a checksum that is stored with the data. If the two
checksums do not agree, the system will usually indicate an error, or automatically read the data
again.

Sample Problem:

Input:

LENGTH
START

Output:

CHECKSUM

(Number of items)
(Start of data table)

(Data Checksum)

00000003
28
55
26

28 + 55 + 26
= 00101000 (28)
+ 01010101 (55)
= 01111101 (7D)
+ 00100110 (26)
= 10100011 (A3)

88

CHAPTER 9. PROGRAM LOOPS

9.2.2 Number of Zero, Positive, and Negative numbers

Determine the number of zero, positive (most significant bit zero, but entire number not zero),
and negative (most significant bit set) elements in a series of signed 32-bit numbers. The length of
the series is defined by the variable LENGTH and the starting series of numbers start with the START
label. Place the number of negative elements in variable NUMNEG, the number of zero elements in
variable NUMZERO and the number of positive elements in variable NUMPOS.

Sample Problem:

Input:

LENGTH

6

(Number of items)

START

(Start of data table — Positive)
(Negative)
(Positive)
(Zero)

76028326
8D489867
21202549
00000000
E605546C (Negative)
(Positive)
00000004

Output:

NUMNEG
NUMZERO
NUMPOS

2
1
3

(2 negative numbers: 8D489867 and E605546C )
(1 zero value)
(3 positive numbers: 76028326, 21202549 and 00000004 )

9.2.3 Find Minimum

Find the smallest element in a series of unsigned bytes. The length of the series is defined by the
variable LENGTH with the series starting at the START label. Store the minimum byte value in the
NUMMIN variable.

Sample Problem:

Input:

LENGTH

5

(Number of items)

START

Output:

NUMMIN

65
79
15
E3
72

15

(Start of data table)

(Smallest of the five)

9.2.4 Count 1 Bits

Determine the number of bits which are set in the 32-bit variable NUM, storing the result in the
NUMBITS variable.

Sample Problem:

Input:
Output:

NUM
NUMBITS

2866B794 = 0011 1000 0110 0110 1011 0111 1001 0100
0F = 15

9.2.5 Find element with most 1 bits

Determine which element in a series of 32-bit numbers has the largest number of bits set. The
length of the series is defined by the LENGTH variable and the series starts with the START label.
Store the value with the most bits set in the NUM variable.

Sample Problem:

9.2. PROBLEMS

89

Input:

LENGTH

5

(Number of items)

START

(0010 0000 0101 1010 0001 0101 1101 0011 — 13)
205A15E3
(0010 0101 0110 1100 1000 0111 0000 0000 — 11)
256C8700
(0010 1001 0101 0100 0110 1000 1111 0010 — 14)
295468F2
29856779
(0010 1001 1000 0101 0110 0111 0111 1001 — 16)
9147592A (1001 0001 0100 0111 0101 1001 0010 1010 — 14)

Output:

NUM

29856779

(Number with most 1-bits)

90

CHAPTER 9. PROGRAM LOOPS

10 Strings

Microprocessors often handle data which represents printed characters rather than numeric quan-
tities. Not only do keyboards, printers, communications devices, displays, and computer terminals
expect or provide character-coded data, but many instruments, test systems, and controllers also
require data in this form. ASCII (American Standard Code for Information Interchange) is the
most commonly used code, but others exist.

We use the standard seven-bit ASCII character codes, as shown in Table 10.1; the character code
occupies the low-order seven bits of the byte, and the most significant bit of the byte holds a 0 or
a parity bit.

10.1 Handling data in ASCII

Here are some principles to remember in handling ASCII-coded data:

• The codes for the numbers and letters form ordered sequences. Since the ASCII codes for
the characters “0” through “9” are 3016 through 3916 you can convert a decimal digit to the
equivalent ASCII characters (and ASCII to decimal) by simple adding the ASCII offset: 3016
= ASCII “0”. Since the codes for letters (4116 through 5A16 and 6116 through 7A16) are in
order, you can alphabetises strings by sorting them according to their numerical values.

• The computer does not distinguish between printing and non-printing characters. Only the

I/0 devices make that distinction.

• An ASCII I/0 device handles data only in ASCII. For example, if you want an ASCII printer
to print the digit “7”, you must send it 3716 as the data; 0716 will ring the bell. Similarly,
if a user presses the “9” key on an ASCII keyboard, the input data will be 3916; 0916 is the
tab key.

• Many ASCII devices do not use the entire character set. For example, devices may ignore

many control characters and may not print lower-case letters.

• Despite the definition of the control characters many devices interpret them differently. For
example they typically uses control characters in a special way to provide features such as
cursor control on a display, and to allow software control of characteristics such as rate of
data transmission, print width, and line length.

• Some widely used ASCII control characters are:

line feed
0A16
LF
carriage return
0D16 CR
0816 BS
backspace
7F16 DEL rub out or delete character

• Each ASCII character occupies eight bits. This allows a large character set but is wasteful
when only a few characters are actually being used. If, for example, the data consists entirely
of decimal numbers, the ASCII format (allowing one digit per byte) requires twice as much

91

92

CHAPTER 10. STRINGS

LSB
0
1
2
3
4
5
6
7
8
9
A
B
C
D
E
F

0

1

NUL DLE
DC1
SOH
STX
DC2
ETX DC3
EOT DC4
ENQ NAK
ACK SYN
ETB
BEL
CAN
BS
EM
HT
SUB
LF
ESC
VT
FS
FF
GS
CR
RS
SO
US
SI

MSB
3
0
1
2
3
4
5
6
7
8
9
:
;
<
=
>
?

2
SP
!
"
#
$
%
&
’
(
)
*
+
,
-
.
/

4
@
A
B
C
D
E
F
G
H
I
J
K
L
M
N
0

5
P
Q
R
S
T
U
V
W
X
Y
Z
[
\
]
^
_

7
p
q
r
s
t
u
v
w
x
y
z
{
|
}
~

6
‘
a
b
c
d
e
f
g
h
i
j
k
l
m
n
o DEL

Control Characters

End of tx

Data link escape
Device control 1
Device control 2
Device control 3
Device control 4

Null
DLE
NUL
Start of heading DC1
SOH
DC2
STX
Start of text
DC3
ETX End of text
DC4
EOT
NAK Negative ack
ENQ Enquiry
SYN
ACK Acknowledge
ETB
Bell, or alarm
BEL
CAN Cancel
Backspace
BS
EM
Horizontal tab
HT
SUB
Line feed
LF
ESC
Vertical tab
VT
FS
Form feed
FF
GS
Carriage return
CR
RS
Shift out
SO
US
Shift in
SI
DEL
Space
SP

End of medium
Substitute
Escape
File separator
Group separator
Record separator
Unit separator
Delete

Synchronous idle
End of tx block

Table 10.1: Hexadecimal ASCII Character Codes

storage, communications capacity, and processing time as the BCD format (allowing two
digits per byte).

The assembler includes a feature to make character-coded data easy to handle, single quotation
marks around a character indicate the character’s ASCII value. For example,

MOV

R3, #’A’

is the same as

MOV

R3, #0x41

The first form is preferable for several reasons. It increases the readability of the instruction, it
also avoids errors that may result from looking up a value in a table. The program does not
depend on ASCII as the character set, since the assembler handles the conversion using whatever
code has been designed for.

10.2 A string of characters

Individual characters on there own are not really all that helpful. As humans we need a string
of characters in order to form meaningful text. In assembly programming it is normal to have
to process one character at a time. However, the assembler does at least allow us to store a
string of byes (characters) in a friendly manner with the DCB directive. For example, line 26 of
program 10.1a is:

DCB

"Hello, World", CR

which will produce the following binary data:

Binary:
Text:

48
H

65
e

6C
l

6C
l

6F
o

2C
,

20
SP

57
W

6F
o

72
r

6C
l

64
d

0D
CR

Use table 10.1 to check that this is correct. In order to make the program just that little bit more
readable, line 5 defines the label CR to have the value for a Carriage Return (0D16).

There are three main methods for handling strings: Fixed Length, Terminated, and Counted. It
is normal for a high level language to support just one method. C/C++ and Java all support the
use of Zero-Terminated strings, while Pascal and Ada use counted strings. Although it is possible

10.2. A STRING OF CHARACTERS

93

to provide your own support for the alternative string type it is seldom done. A good programmer
will use a mix of methods depending of the nature of the strings concerned.

10.2.1 Fixed Length Strings

A fixed length string is where the string is of a predefined and fixed size. For example, in a system
where it is known that all strings are going to be ten characters in length, we can simply reserve
10 bytes for the string.

This has an immediate advantages in that the management of the strings is simple when compared
to the alternative methods. For example we only need one label for an array of strings, and we
can calculate the starting position of the nth string by a simple multiplication.

This advantage is however also a major disadvantage. For example a persons name can be anything
from two characters to any number of characters. Although it would be possible to reserve sufficient
space for the longest of names this amount of memory would be required for all names, including
the two letter ones. This is a significant waist of memory.

It would be possible to reserve just ten characters for each name. When a two letter name appears
it would have to be padded out with spaces in order to make the name ten characters in length.
When a name longer than ten characters appears it would have to be truncated down to just ten
characters thus chopping off part of the name. This requires extra processing and is not entirely
friendly to users who happen to have a long name.

When there is little memory and all the strings are known in advance it may be a good idea to
use fixed length strings. For example, command driven systems tend to use a fixed length strings
for the list of commands.

10.2.2 Terminated Strings

A terminated string is one that can be of any length and uses a special character to mark the end
of the string, this character is known at the sentinel. For example program 10.1a uses the carriage
return as it’s sentinel.

Over the years several different sentinels have been used, these include $ (2616), EOT (End of
Text – 0416), CR (Carriage Return – 0D16), LF (Line Feed – 0A16) and NUL (No character –
0016). Today the most commonly used sentinel is the NUL character, primarily because it is used
by C/C++. The NUL character also has a good feeling about it, as it is represented by the value
0, has no other meaning and it is easier to detected than any other character. This is frequently
referred to as a Null- or Zero-Terminated string or simply as an ASCIIZ string.

The terminated string has the advantage that it can be of any length. Processing the string is
fairly simply, you enter into a loop processing each character at a time until you reach the sentinel.
The disadvantage is that the sentinel character can not appear in the string. This is another reason
why the NUL character is such a good choice for the sentinel.

10.2.3 Counted Strings

A counted string is one in which the first one or two byte holds the length of the string in characters.
Thus a counted string can be of any number of characters up to the largest unsigned number that
can be stored in the first byte/word.

A counted string may appear rather clumsy at first. Having the length of the string as a binary
value has a distinct advantage over the terminated string. It allow the use of the counting instruc-

94

CHAPTER 10. STRINGS

tions that have been included in many instruction sets. This means we can ignore the testing for
a sentinel character and simply decrement our counter, this is a far faster method of working.

To scan through an array of strings we simply point to the first string, and add the length count
to our pointer to obtain the start of the next string. For a terminated string we would have to
scan for the sentinel for each string.

There are two disadvantages with the counted string. The string does have a maximum length,
255 characters or 64K depending on the size of the count value (8- or 16-bit). Although it is
normally felt that 64K should be sufficient for most strings. The second disadvantage is their
perceived complexity. Many people feel that the complexity of the counted string outweighs the
speed advantage.

10.3

International Characters

As computing expands outside of the English speaking world we have to provide support for
languages other than standard American. Many European languages use letters that are not
available in standard ASCII, for example: œ, Œ, ø, Ø, æ, Æ, ł, Ł, ß, ¡, and ¿. This is particularly
important when dealing with names: Ångstrøm, Karlstraße or Łukasiewicz.

The ASCII character set is not even capable of handling English correctly. When we borrow a
word from another language we also use it’s diacritic marks (or accents). For example I would
rather see pâté on a menu rather than pate. ASCII does not provide support for such accents.

To overcome this limitation the international community has produced a new character encoding,
known as Unicode. In Unicode the character code is two bytes long, the first byte indicates which
character set the character comes from, while the second byte indicates the character position
within the character set. The traditional ASCII character set is incorporated into Unicode as
character set zero. In the revised C standard a new data type of wchar was defined to cater for
this new “wide character”.

While Unicode is sufficient to represent the characters from most modern languages, it is not
sufficient to represent all the written languages of the world, ancient and modern. Hence an
extended version, known as Unicode-32 is being developed where the character set is a 23-bit
value (three bytes). Unicode is a subset of Unicode-32, while ASCII is a subset of Unicode.

Although we do not consider Unicode you should be aware of the problem of international character
sets and the solution Unicode provides.

10.4 Program Examples

10.4.1 Length of a String of Characters

TTL

Ch6Ex1 - strlencr

; Find the length of a CR terminated string

Program 10.1a: strlencr.s — Find the length of a Carage Return terminated string
1
2
3
4
5
6
7
8
9
10
11

; Load the address of the lookup table

Program, CODE, READONLY

AREA
ENTRY

R0, =Data1

0x0D

Main

EQU

LDR

CR

10.4. PROGRAM EXAMPLES

95

EOR

R1, R1, R1

; Clear R1 to store count

Loop

Done

Table

LDRB
CMP
BEQ
ADD
BAL

R2, [R0], #1
R2, #CR
Done
R1, R1, #1
Loop

STR
SWI

R1, CharCount
&11

AREA

Data1, DATA

DCB
ALIGN

"Hello, World", CR

AREA

Result, DATA

CharCount

0

DCB

END

; Load the first byte into R2
; Is it the terminator ?
; Yes => Stop loop
; No => Increment count
; Read next char

; Store result

; Storage for count

12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34

; Load the address of the lookup table
; Start count at -1

Main

Loop

LDR
MOV

TTL
AREA
ENTRY

ADD
LDRB
CMP
BNE

R0, =Data1
R1, #-1

; Find the length of a null terminated string

Ch6Ex1 - strlen
Program, CODE, READONLY

R1, R1, #1
R2, [R0], #1
R2, #0
Loop

Program 10.1b: strlen.s — Find the length of a null terminated string
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29

; Increment count
; Load the first byte into R2
; Is it the terminator ?
; No => Next char

R1, CharCount
&11

; Storage for count

"Hello, World", 0

; Store result

Result, DATA

Data1, DATA

DCB
ALIGN

CharCount

STR
SWI

Table

AREA

AREA

DCB

0

END

10.4.2 Find First Non-Blank Character

*

Program 10.2: skipblanks.s — Find first non-blank
1
2
3
4

find the length of a string

Ch6Ex3

TTL

96

CHAPTER 10. STRINGS

5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31

Blank

EQU
AREA
ENTRY

" "
Program, CODE, READONLY

Main

Loop

ADR
MOV

LDRB
CMP
BEQ

SUB
STR
SWI

R0, Data1
R1, #Blank

R2, [R0], #1
R2, R1
Loop

R0, R0, #1
R0, Pointer
&11

AREA

Data1, DATA

Table

DCB
ALIGN

AREA

Pointer DCD

ALIGN

END

"

7

"

Result, DATA
0

;load the address of the lookup table
;store the blank char in R1

;load the first byte into R2
;is it a blank
;if so loop

;otherwise done - adjust pointer
;and store it

;storage for count

10.4.3 Replace Leading Zeros with Blanks

;load the address of the lookup table
;store the zero char in R1
;and the blank char in R3

*

TTL

Loop

Main

Ch6Ex4

LDR
MOV
MOV

Blank
Zero

EQU
EQU
AREA
ENTRY

supress leading zeros in a string

R0, =Data1
R1, #Zero
R3, #Blank

’ ’
’0’
Program, CODE, READONLY

Program 10.3: padzeros.s — Supress leading zeros in a string
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32

R0, R0, #1
R3, [R0]
R0, R0, #1
Loop

R2, [R0], #1
R2, R1
Done

SUB
STRB
ADD
BAL

LDRB
CMP
BNE

"000007000"

Data1, DATA

DCB
ALIGN

;all done

Table

AREA

Done

&11

SWI

;load the first byte into R2
;is it a zero
;if not, done

;otherwise adjust the pointer
;and store it blank char there
;otherwise adjust the pointer
;and loop

10.4. PROGRAM EXAMPLES

97

33
34
35
36
37

AREA

Pointer DCD

Result, DATA
0

;storage for count

ALIGN

END

10.4.4 Add Even Parity to ASCII Chatacters

Program 10.4: setparity.s — Set the parity bit on a series of characters store the amended
string in Result

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53

; Set the parity bit on a series of characters store the amended string in Result

TTL

Ch6Ex5

Main

MainLoop

AREA
ENTRY

LDR
LDR
LDRB
CMP
BEQ

LDRB
MOV
MOV
MOV
MOV

ParLoop

DontAdd

Even
Check

MOVS
BPL
ADD

SUBS
BNE
TST
BEQ
ORR
STRB
BAL
STRB
SUBS
BNE

Program, CODE, READONLY

R0, =Data1
R5, =Pointer
R1, [R0], #1
R1, #0
Done

R2, [R0], #1
R6, R2
R2, R2, LSL #24
R3, #0
R4, #7

R2, R2, LSL #1
DontAdd
R3, R3, #1

R4, R4, #1
ParLoop
R3, #1
Even
R6, R6, #0x80
R6, [R5], #1
Check
R6, [R5], #1
R1, R1, #1
MainLoop

;load the address of the lookup table

;store the string length in R1

;nothing to do if zero length

;load the first byte into R2
;keep a copy of the original char
;shift so that we are dealing with msb
;zero the bit counter
;init the shift counter

;left shift
;if msb is not a one bit, branch
;otherwise add to bit count

;update shift count
;loop if still bits to check
;is the parity even
;if so branch
;otherwise set the parity bit
;and store the amended char

;store the unamended char if even pty
;decrement the character count

Done

SWI

&11

AREA

Data1, DATA

Table

DCB
DCB
DCB
DCB
DCB
DCB
DCB

6
0x31
0x32
0x33
0x34
0x35
0x36

AREA
ALIGN

Result, DATA

;data table starts with byte length of string
;the string

Pointer DCD

0

;storage for parity characters

END

98

CHAPTER 10. STRINGS

10.4.5 Pattern Match

*

TTL

Main

Ch6Ex6

CMP
BEQ

*
Loop

AREA
ENTRY

R3, #0
Same

;load the address of the lookup table

;assume strings not equal - set to -1
;store the first string length in R3
;store the second string length in R4

Program, CODE, READONLY

LDR
LDR
LDR
LDR
LDR
CMP
BNE

compare two counted strings for equality

if we got this far, we now need to check the string char by char

;if they are different lengths,
;they can’t be equal
;test for zero length if both are
;zero length, nothing else to do

R0, =Data1
R1, =Data2
R2, Match
R3, [R0], #4
R4, [R1], #4
R3, R4
Done

Program 10.5a: cstrcmp.s — Compare two counted strings for equality
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47

;character of first string
;character of second string
;are they the same
;if not the strings are different
;use the string length as a counter
;if we got to the end of the count
;the strings are the same
;not done, loop

R5, [R0], #1
R6, [R1], #1
R5, R6
Done
R3, R3, #1
Same

Data2, DATA
3
"CAT"

Data1, DATA
3
"CAT"

R2, #0
R2, Match
&11

;storage for parity characters

LDRB
LDRB
CMP
BNE
SUBS
BEQ

Table2 DCD
DCB

Table1 DCD
DCB

;clear the -1 from match (0 = match)
;store the result

AREA
ALIGN
DCD

Result, DATA

MOV
STR
SWI

Same
Done

&FFFF

Match

AREA

Loop

AREA

B

END

;data table starts with byte length of string
;the string

;data table starts with byte length of string
;the string

; Compare two null terminated strings for equality

Program 10.5b: strcmp.s — Compare null terminated strings for equality assume that we have
no knowledge of the data structure so we must assess the individual strings
1
2
3
4
5
6
7

Program, CODE, READONLY

AREA
ENTRY

Ch6Ex7

TTL

10.4. PROGRAM EXAMPLES

99

Main

Count1

Count2

Next

*
Loop

Same

Done

8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63

LDR
LDR
LDR
MOV
MOV

LDRB
CMP
BEQ
ADD
BAL

LDRB
CMP
BEQ
ADD
BAL

CMP
BNE

CMP
BEQ
LDR
LDR

R0, =Data1
R1, =Data2
R2, Match
R3, #0
R4, #0

R5, [R0], #1
R5, #0
Count2
R3, R3, #1
Count1

R5, [R1], #1
R5, #0
Next
R4, R4, #1
Count2

R3, R4
Done

R3, #0
Same
R0, =Data1
R1, =Data2

;load the address of the lookup table

;assume strings not equal, set to -1
;init register

;load the first byte into R5
;is it the terminator
;if not, Loop
;increment count

;load the first byte into R5
;is it the terminator
;if not, Loop
;increment count

;if they are different lengths,
;they can’t be equal
;test for zero length if both are
;zero length, nothing else to do
;need to reset the lookup table

if we got this far, we now need to check the string char by char

LDRB
LDRB
CMP
BNE
SUBS
BEQ

R5, [R0], #1
R6, [R1], #1
R5, R6
Done
R3, R3, #1
Same

BAL

Loop

R2, #0

R2, Match
&11

MOV

STR
SWI

AREA

;character of first string
;character of second string
;are they the same
;if not the strings are different
;use the string length as a counter
;if we got to the end of the count
;the strings are the same
;not done, loop

;clear the -1 from match (0 = match)

;store the result

Data1, DATA
"Hello, World", 0

;the string

Data2, DATA
"Hello, worl", 0

Result, DATA

;the string

&FFFF

;flag for match

Table1 DCB

ALIGN

AREA

Table2 DCB

Match

AREA
ALIGN
DCD

END

100

10.5 Problems

CHAPTER 10. STRINGS

10.5.1 Length of a Teletypewriter Message

Determine the length of an ASCII message. All characters are 7-bit ASCII with MSB = 0. The
string of characters in which the message is embedded has a starting address which is contained
in the START variable. The message itself starts with an ASCII STX (Start of Text) character
(0216) and ends with ETX (End of Text) character (0316). Save the length of the message, the
number of characters between the STX and the ETX markers (but not including the markers) in
the LENGTH variable.

Sample Problem:

Input:

START

String

Output:

LENGTH

String

(Location of string)

02
47
4F
03

02

(STX — Start Text)
(“G”)
(“O”)
(ETX — End Text)

(“GO”)

10.5.2 Find Last Non-Blank Character

Search a string of ASCII characters for the last non-blank character. Starting address of the string
is contained in the START variable and the string ends with a carriage return character (0D16).
Place the address of the last non-blank character in the POINTER variable.

Sample Problems:

Input:

START

String

Test A

String

37 (“7”)
0D (CR)

Test B

String

41 (“A”)
20 (Space)
48 (“H”)
41 (“A”)
54 (“T”)
20 (Space)
20 (Space)
0D (CR)

Output:

POINTER First Char

Fourth Char

10.5.3 Truncate Decimal String to Integer Form

Edit a string of ASCII decimal characters by replacing all digits to the right of the decimal point
with ASCII blanks (2016). The starting address of the string is contained in the START variable
and the string is assumed to consist entirely of ASCII-coded decimal digits and a possible decimal
point (2E16). The length of the string is stored in the LENGTH variable. If no decimal point appears
in the string, assume that the decimal point is at the far right.

Sample Problems:

101

10.5. PROBLEMS

Input:

START
LENGTH

String

Output:

Test A

String
4

37 (“7”)
2E (“.”)
38 (“8”)
31 (“1”)

37 (“7”)
2E (“.”)
20 (Space)
20 (Space)

Test B
String
3

36 (“6”)
37 (“7”)
31 (“1”)

36 (“6”)
37 (“7”)
31 (“1”)

Note that in the second case (Test B) the output is unchaged, as the number is assumed to be
“671.”.

10.5.4 Check Even Parity and ASCII Characters

Cheek for even parity in a string of ASCII characters. A string’s starting address is contained in
the START variable. The first word of the string is its length which is followed by the string itself.
If the parity of all the characters in the string are correct, clear the PARITY variable; otherwise
place all ones (FFFFFFFF16) into the variable.

Sample Problems:

Input:

START

String

Test A

String

3
B1 (1011 0001)
B2 (1011 0010)
33 (0011 0011)

Test B

String

03
B1 (1011 0001)
B6 (1011 0110)
33 (0011 0011)

Output:

PARITY

00000000 (True)

FFFFFFFF (False)

10.5.5 String Comparison

Compare two strings of ASCII characters to see which is larger (that is, which follows the other in
alphabetical ordering). Both strings have the same length as defined by the LENGTH variable. The
strings’ starting addresses are defined by the START1 and START2 variables. If the string defined
by START1 is greater than or equal to the other string, clear the GREATER variable; otherwise set
the variable to all ones (FFFFFFFF16).

Sample Problems:

Test A

Test B

Test C

Input:

LENGTH
START1
START2

String1

String2

3
String1
String2

43 (“C”)
41 (“A”)
54 (“T”)

42 (“B”)
41 (“A”)
54 (“T”)

3
String1
String2

43 (“C”)
41 (“A”)
54 (“T”)

52 (“C”)
41 (“A”)
54 (“T”)

3
String1
String2

43 (“C”)
41 (“A”)
54 (“T”)

52 (“C”)
55 (“U”)
54 (“T”)

Output:

GREATER

00000000
(CAT > BAT)

00000000
(CAT = CAT)

FFFFFFFF
(CAT < CUT)

102

CHAPTER 10. STRINGS

11 Code Conversion

Code conversion is a continual problem in microcomputer applications. Peripherals provide data
in ASCII, BCD or various special codes. The microcomputer must convert the data into some
standard form for processing. Output devices may require data in ASCII, BCD, seven-segment or
other codes. Therefore, the microcomputer must convert the results to the proper form after it
completes the processing.

There are several ways to approach code conversion:

1. Some conversions can easily be handled by algorithms involving arithmetic or logical func-

tions. The program may, however, have to handle special cases separately.

2. More complex conversions can be handled with lookup tables. The lookup table method
requires little programming and is easy to apply. However the table may occupy a large
amount of memory if the range of input values is large.

3. Hardware is readily available for some conversion tasks. Typical examples are decoders
for BCD to seven-segment conversion and Universal Asynchronous Receiver/Transmitters
(UARTs) for conversion between parallel and serial formats.

In most applications, the program should do as much as possible of the code conversion work.
Most code conversions are easy to program and require little execution time.

11.1 Program Examples

11.1.1 Hexadecimal to ASCII

*

TTL

Main

Ch7Ex1

AREA
ENTRY

Program, CODE, READONLY

convert a single hex digit to its ASCII equivalent

Program 11.1a: nibtohex.s — Convert a single hex digit to its ASCII equivalent
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20

;load the digit
;load the address for the result
;is the number < 10 decimal
;then branch

R0, Digit
R1, =Result
R0, #0xA
Add_0

R0, R0, #"0"
R0, [R1]
&11

;convert to ASCII
;store the result

;add offset for ’A’ to ’F’

R0, R0, #"A"-"0"-0xA

LDR
LDR
CMP
BLT

ADD
STR
SWI

Data1, DATA

Add_0

AREA

ADD

103

104

Digit

CHAPTER 11. CODE CONVERSION

DCD

&0C

;the hex digit

AREA

Result DCD

Data2, DATA
0

;storage for result

END

21
22
23
24
25
26
27

*
*

TTL

Mask

start

Ch7Ex2

MainLoop

0x0000000F

LDR
MOV
MOV

AREA
ENTRY
EQU

Program, CODE, READONLY

R1, Digit
R4, #8
R5, #28

;load the digit
;init counter
;control right shift

now something a little more adventurous - convert a 32 bit
hexadecimal number to an ASCII string and output to the terminal

Program 11.1b: wordtohex.s — Convert a 32 bit hexadecimal number to an ASCII string and
output to the terminal
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37

;copy original word
;right shift the correct number of bits
;reduce the bit shift
;mask out all but the ls nibble
;is the number < 10 decimal
;then branch

R3, R1
R3, R3, LSR R5
R5, R5, #4
R3, R3, #Mask
R3, #0xA
Add_0

;convert to ASCII
;prepare to output
;output to console
;decrement counter

R3, R3, #"0"
R0, R3
&0
R4, R4, #1
MainLoop

;add a CR character
;output it
;all done

R0, #&0D
&0
&11

;add offset for ’A’ to ’F’

Data1, DATA
&DEADBEEF

ADD
MOV
SWI
SUBS
BNE

MOV
MOV
SUB
AND
CMP
BLT

R3, R3, #"A"-"0"-0xA

;the hex word

MOV
SWI
SWI

AREA
DCD

Digit

Add_0

ADD

END

11.1.2 Decimal to Seven-Segment

*

TTL

Ch7Ex3

convert a decimal number to seven segment binary

Program 11.2: nibtoseg.s — Convert a decimal number to seven segment binary
1
2
3
4
5
6
7
8
9
10
11

;load the start address of the table
;clear register for the code
;get the digit to encode

R0, =Data1
R1, R1, R1
R2, Digit

Program, CODE, READONLY

LDR
EOR
LDRB

AREA
ENTRY

Main

11.1. PROGRAM EXAMPLES

105

12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41

CMP
BHI

ADD
LDRB

STR
SWI

AREA
DCB
DCB
DCB
DCB
DCB
DCB
DCB
DCB
DCB
DCB
ALIGN

AREA
DCB
ALIGN

AREA

Done

Table

Digit

Result DCD

END

R2, #9
Done

R0, R0, R2
R1, [R0]

R1, Result
&11

Data1, DATA
&3F
&06
&5B
&4F
&66
&6D
&7D
&07
&7F
&6F

Data2, DATA
&05

Data3, DATA
0

;is it a valid digit?
;clear the result

;advance the pointer
;and get the next byte

;store the result
;all done

;the binary conversions table

;the number to convert

;storage for result

11.1.3 ASCII to Decimal

Program 11.3: dectonib.s — Convert an ASCII numeric character to decimal

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27

*

convert an ASCII numeric character to decimal

TTL

Ch7Ex4

Main

Done

Char

AREA
ENTRY

MOV
LDRB
SUBS
BCC
CMP
BHI
MOV

STR
SWI

AREA
DCB
ALIGN

AREA

Result DCD

END

Program, CODE, READONLY

R1, #-1
R0, Char
R0, R0, #"0"
Done
R0, #9
Done
R1, R0

R1, Result
&11

Data1, DATA
&37

Data2, DATA
0

;set -1 as error flag
;get the character
;convert and check if character is < 0
;if so do nothing
;check if character is > 9
;if so do nothing
;otherwise....

;.....store the decimal no
;all done

;ASCII representation of 7

;storage for result

106

CHAPTER 11. CODE CONVERSION

11.1.4 Binary-Coded Decimal to Binary

*

TTL

Main

Loop

Ch7Ex5

AREA
ENTRY

LDR
MOV
MOV
MOV

Program, CODE, READONLY

convert an unpacked BCD number to binary

R0, =BCDNum
R5, #4
R1, #0
R2, #0

;load address of BCD number
;init counter
;clear result register
;and final register

Program 11.4a: ubcdtohalf.s — Convert an unpacked BCD number to binary
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35

;load digit and incr address
;add the next digit
;decr counter
;if counter != 0, loop

R1, R1, R1
R3, R1
R3, R3, LSL #2
R1, R1, R3

R4, [R0], #1
R1, R1, R4
R5, R5, #1
Loop

;mult by 8 (2 x 4)
;= mult by 10

;store the result
;all done

Data1, DATA
&02,&09,&07,&01

Data2, DATA
0

;an unpacked BCD number

R1, Result
&11

;storage for result

LDRB
ADD
SUBS
BNE

;multiply by 2

ADD
MOV
MOV
ADD

Result DCD

BCDNum DCB

STR
SWI

ALIGN

AREA

AREA

END

*

TTL

Main

Ch7Ex6

AREA
ENTRY

Program, CODE, READONLY

convert an unpacked BCD number to binary using MUL

Program 11.4b: ubcdtohalf2.s — Convert an unpacked BCD number to binary using MUL
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24

;mult by 10
;load digit and incr address
;add the next digit
;decr counter
;if count != 0, loop

;load address of BCD number
;init counter
;clear result register
;and final register
;multiplication constant

R6, R1
R1, R6, R7
R4, [R0], #1
R1, R1, R4
R5, R5, #1
Loop

R0, =BCDNum
R5, #4
R1, #0
R2, #0
R7, #10

;store the result
;all done

MOV
MUL
LDRB
ADD
SUBS
BNE

R1, Result
&11

LDR
MOV
MOV
MOV
MOV

STR
SWI

Loop

11.2. PROBLEMS

107

25
26
27
28
29
30
31
32
33

AREA

BCDNum DCB

Data1, DATA
&02,&09,&07,&01

;an unpacked BCD number

ALIGN

AREA

Result DCD

END

Data2, DATA
0

;storage for result

11.1.5 Binary Number to ASCII String

*

Loop

Main

TTL
AREA
ENTRY

;load the number to process

LDR
ADD
LDRB
MOV
MOV
LDR

Ch7Ex7
Program, CODE, READONLY

store a 16bit binary number as an ASCII string of ’0’s and ’1’s

R0, =String
R0, R0, #16
R6, String
R2, #"1"
R3, #"0"
R1, Number

;load startr address of string
;adjust for length of string
;init counter
;load character ’1’ to register

Program 11.5: halftobin.s — Store a 16bit binary number as an ASCII string of ’0’s and ’1’s
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36

;rotate right with carry
;if carry set branch (LSB was a ’1’ bit)
;otherwise store "0"
;and branch to counter code

R1, R1, ROR #1
Loopend
R3, [R0], #-1
Decr

;decrement counter
;loop while not 0

;a 16 bit binary number number

Data2, DATA
16

Data1, DATA
&31D2

R6, R6, #1
Loop

;storage for result

MOVS
BCS
STRB
BAL

R2, [R0], #-1

;store a "1"

Number DCD

String DCB

SUBS
BNE

Loopend

ALIGN

ALIGN

STRB

AREA

AREA

Decr

&11

SWI

END

11.2 Problems

11.2.1 ASCII to Hexadecimal

Convert the contents of the A_DIGIT variable from an ASCII character to a hexadecimal digit and
store the result in the H_DIGIT variable. Assume that A_DIGIT contains the ASCII representation
of a hexadecimal digit (7 bits with MSB=0).

108

CHAPTER 11. CODE CONVERSION

7
0
0
3F
0
0
06
1
0
5B
2
0
4F
3
66
0
4
6D 0
5
7D 0
6
0
07
7
0
7F
8
0
6F
9
0
A 77
0
7C
B
3A 0
C
D 5E
0
7A 0
E
0
71
F

6
g
0
0
1
1
1
1
1
0
1
1
1
1
0
1
1
1

5
f
1
0
0
0
1
1
1
0
1
1
1
1
1
0
1
1

4
e
1
0
1
0
0
0
1
0
1
0
1
1
1
1
1
1

3
d
1
0
1
1
0
1
1
0
1
1
0
1
1
1
1
0

2
c
1
1
0
1
1
1
1
1
1
1
1
1
0
1
0
0

1
b
1
1
1
1
1
0
0
1
1
1
1
0
0
1
0
0

0
a
1
0
1
1
0
1
1
1
1
1
1
0
1
0
1
1

Figure 11.1: Seven-Segment Display

f

e

a

g

d

b

c

Sample Problems:

Input:

A_DIGIT

Test A
43 (“C”)

Output:

H_DIGIT

0C

Test B
36 (“6”)

06

11.2.2 Seven-Segment to Decimal

Convert the contents of the CODE variable from a seven-segment code to a decimal number and
store the result in the NUMBER variable. If CODE does not contain a valid seven-segment code, set
NUMBER to FF16. Use the seven-segment table given in Figure 11.1 and try to match codes.

Sample Problems:

Input:

CODE

Test A
4F

Output:

NUMBER

03

Test B
28

FF

11.2.3 Decimal to ASCII

Convert the contents of the variable DIGIT from decimal digit to an ASCII character and store
the result in the variable CHAR. If the number in DIGIT is not a decimal digit, set the contents of
CHAR to an ASCII space (2016).

Sample Problems:

Input:

DIGIT

Test A
07

Test B

55

Output:

CHAR

37 (“7”)

20 (Space)

11.2.4 Binary to Binary-Coded-Decimal

Convert the contents of the variable NUMBER to four BCD digits in the STRING variable. The 32-bit
number in NUMBER is unsigned and less than 10,000.

Sample Problem:

11.2. PROBLEMS

109

Input:

NUMBER

1C52

(725010)

Output:

STRING

07
02
05
00

(“7”)
(“2”)
(“5”)
(“0”)

11.2.5 Packed Binary-Coded-Decimal to Binary String

Convert the eight digit packed binary-coded-decimal number in the BCDNUM variable into a 32-bit
number in a NUMBER variable.

Sample Problem:

Input:

BCDNUM

92529679

Output:

NUMBER

0583E40916 (9252967910)

11.2.6 ASCII string to Binary number

Convert the eight ASCII characters in the variable STRING to an 8-bit binary number in the
variable NUMBER. Clear the byte variable ERROR if all the ASCII characters are either ASCII “1” or
ASCII “0”; otherwise set ERROR to all ones (FF16).

Sample Problems:

Input:

STRING

Test A
(“1”)
(“1”)
(“0”)
(“1”)
(“0”)
(“0”)
(“1”)
(“0”)

31
31
30
31
30
30
31
30

Test B
(“1”)
(“1”)
(“0”)
(“1”)
(“0”)
(“7”)
(“1”)
(“0”)

31
31
30
31
30
37
31
30

Output:

NUMBER D2
ERROR

0

(1101 0010)
(No Error )

00
(Valid )
FF (Error )

110

CHAPTER 11. CODE CONVERSION

12 Arithmetic

Much of the arithmetic in some microprocessor applications consists of multiple-word binary or
decimal manipulations. The processor provides for decimal addition and subtraction, but does
not provide for decimal multiplication or division, you must implement these operations with
sequences of instruction.

Most processors provide for both signed and unsigned binary arithmetic. Signed numbers are
represented in two’s complement form. This means that the operations of addition and subtraction
are the same whether the numbers are signed or unsigned.

Multiple-precision binary arithmetic requires simple repetitions of the basic instructions. The
Carry flag transfers information between words. It is set when an addition results in a carry or
a subtraction results in a borrow. Add with Carry and Subtract with Carry use this information
from the previous arithmetic operation.

Decimal arithmetic is a common enough task for microprocessors that most have special instruc-
tions for this purpose. These instructions may either perform decimal operations directly or correct
the results of binary operations to the proper decimal form. Decimal arithmetic is essential in
such applications as point-of-sale terminals, check processors, order entry systems, and banking
terminals.

You can implement decimal multiplication and division as series of additions and subtractions,
respectively. Extra storage must be reserved for results, since a multiplication produces a result
twice as long as the operands. A division contracts the length of the result. Multiplications and
divisions are time-consuming when done in software because of the repeated operations that are
necessary.

12.1 Program Examples

12.1.2

64-Bit Addition

*

64 bit addition

TTL
AREA
ENTRY

64 bit addition
Program, CODE, READONLY

Program 12.2: add64.s — 64 Bit Addition
1
2
3
4
5
6
7
8
9
10
11
12
13
14

R0, =Value1
R1, [R0]
R2, [R0, #4]
R0, =Value2
R3, [R0]
R4, [R0, #4]
R6, R2, R4

LDR
LDR
LDR
LDR
LDR
LDR
ADDS

Main

; Pointer to first value
; Load first part of value1
; Load lower part of value1
; Pointer to second value
; Load upper part of value2
; Load lower part of value2
; Add lower 4 bytes and set carry flag

111

112

CHAPTER 12. ARITHMETIC

15
16
17
18
19
20
21
22
23
24
25

ADC
LDR
STR

STR
SWI

R5, R1, R3
R0, =Result
R5, [R0]

R6, [R0, #4]
&11

; Add upper 4 bytes including carry
; Pointer to Result
; Store upper part of result

; Store lower part of result

Value1 DCD
Value2 DCD
Result DCD
END

&12A2E640, &F2100123
&001019BF, &40023F51
0

; Value to be added
; Value to be added
; Space to store result

12.1.3 Decimal Addition

*

EQU

Mask

Loop

Main

0x0000000F

TTL
AREA
ENTRY

LDR
LDR
LDR
LDRB
ADD
MOV

Ch8Ex3
Program, CODE, READONLY

add two packed BCD numbers to give a packed BCD result

R0, =Result
R1, BCDNum1
R2, BCDNum2
R8, Length
R0, R0, #3
R5, #0

;address for storage
;load the first BCD number
;and the second
;init counter
;adjust for offset
;carry

Program 12.3: addbcd.s — Add two packed BCD numbers to give a packed BCD result
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48

;copy what is left in the data register
;and the other number
;mask out everything except low order nibble
;mask out everything except low order nibble
;shift the original number one nibble
;shift the original number one nibble
;add the digits
;and the carry
;is it over 10?
;if not, reset the carry to 0
;otherwise set the carry
;and subtract 10

;copy what is left in the data register
;and the other number
;mask out everything except low order nibble
;mask out everything except low order nibble
;shift the original number one nibble
;shift the original number one nibble
;add the digits
;and the carry
;is it over 10?
;if not, reset the carry to 0
;otherwise set the carry
;and subtract 10

R3, R1
R4, R2
R3, R3, #Mask
R4, R4, #Mask
R1, R1, LSR #4
R2, R2, LSR #4
R7, R3, R4
R7, R7, R5
R7, #0xA
RCarry2
R5, #1
R7, R7, #0xA
Loopend

R3, R1
R4, R2
R3, R3, #Mask
R4, R4, #Mask
R1, R1, LSR #4
R2, R2, LSR #4
R6, R3, R4
R6, R6, R5
R6, #0xA
RCarry1
R5, #1
R6, R6, #0xA
Next

MOV
MOV
AND
AND
MOV
MOV
ADD
ADD
CMP
BLT
MOV
SUB
B

MOV
MOV
AND
AND
MOV
MOV
ADD
ADD
CMP
BLT
MOV
SUB
B

;carry reset to 0

RCarry1

R5, #0

Next

MOV

12.1. PROGRAM EXAMPLES

113

49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70

RCarry2

Loopend

MOV

R5, #0

;carry reset to 0

MOV
ORR
STRB
SUBS
BNE
SWI

AREA

Length DCB

ALIGN

R7, R7, LSL #4
R6, R6, R7
R6, [R0], #-1
R8, R8, #1
Loop
&11

Data1, DATA
&04

;shift the second digit processed to the left
;and OR in the first digit to the ls nibble
;store the byte, and decrement address
;decrement loop counter
;loop while > 0

BCDNum1 DCB

&36, &70, &19, &85

;an 8 digit packed BCD number

AREA

BCDNum2 DCB

Data2, DATA
&12, &66, &34, &59

;another 8 digit packed BCD number

AREA

Result DCD

Data3, DATA
0

;storage for result

END

12.1.4 Multiplication

16-Bit

*

*

Main

TTL
AREA
ENTRY

LDR
LDR
MUL
MUL
STR

16 bit binary multiplication

Ch8Ex1
Program, CODE, READONLY

R0, Number1
R1, Number2
R0, R1, R0
R0, R0, R1
R0, Result

Program 12.4a: mul16.s — 16 bit binary multiplication
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25

Data1, DATA
&706F
&0161

Data2, DATA
0

Number1 DCD
Number2 DCD

Result DCD

;all done

ALIGN

ALIGN

AREA

AREA

&11

SWI

END

;a 16 bit binary number
;another

;storage for result

;load first number
;and second
;x:= y * x
;won’t work - not allowed

32-Bit

Program 12.4b: mul32.s — Multiply two 32 bit number to give a 64 bit result (corrupts R0 and
R1)

114

CHAPTER 12. ARITHMETIC

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39

*
*

multiply two 32 bit number to give a 64 bit result
(corrupts R0 and R1)

Ch8Ex4
Program, CODE, READONLY

R0, Number1
R1, Number2
R6, =Result
R5, R0, LSR #16
R3, R1, LSR #16
R0, R0, R5, LSL #16
R1, R1, R3, LSL #16
R2, R0, R1
R0, R3, R0
R1, R5, R1
R3, R5, R3
R0, R1, R0
R3, R3, #&10000
R2, R2, R0, LSL #16
R3, R3, R0, LSR #16

R2, [R6]
R6, R6, #4
R3, [R6]
&11

Data1, DATA
&12345678
&ABCDEF01

Data2, DATA
0

;load first number
;and second
;load the address of result
;top half of R0
;top half of R1
;bottom half of R0
;bottom half of R1
;partial result
;partial result
;partial result
;partial result
;add middle parts
;add in any carry from above
;LSB 32 bits
;MSB 32 bits

;store LSB
;increment pointer
;store MSB
;all done

;a 16 bit binary number
;another

;storage for result

Main

TTL
AREA
ENTRY

LDR
LDR
LDR
MOV
MOV
BIC
BIC
MUL
MUL
MUL
MUL
ADDS
ADDCS
ADDS
ADC

STR
ADD
STR
SWI

AREA

Number1 DCD
Number2 DCD

ALIGN

AREA

Result DCD

ALIGN

END

12.1.5

32-Bit Binary Divide

*
*
*

TTL
AREA
ENTRY

Ch8Ex2
Program, CODE, READONLY

divide a 32 bit binary no by a 16 bit binary no
store the quotient and remainder
there is no ’DIV’ instruction in ARM!

Program 12.5: divide.s — Divide a 32 bit binary no by a 16 bit binary no store the quotient and
remainder there is no ’DIV’ instruction in ARM!
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20

;is the divisor less than the dividend?
;if so, finished
;add one to quotient
;take away the number you first thought of
;and loop

;load first number
;and second
;clear register for quotient

R1, #0
Err
R0, R1
Done
R3, R3, #1
R0, R0, R1
Loop

R0, Number1
R1, Number2
R3, #0

CMP
BEQ
CMP
BLT
ADD
SUB
B

;test for divide by 0

LDR
LDR
MOV

Main

Loop

12.2. PROBLEMS

115

21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38

Err

Done

MOV

STR
STR
SWI

AREA

Number1 DCD
Number2 DCD

ALIGN

R3, #0xFFFFFFFF

;error flag (-1)

R0, Remain
R3, Quotient
&11

Data1, DATA
&0075CBB1
&0141

;store the remainder
;and the quotient
;all done

;a 16 bit binary number
;another

AREA
Quotient DCD
Remain DCD

Data2, DATA
0
0

;storage for result
;storage for remainder

ALIGN

END

12.2 Problems

12.2.1 Multiple precision Binary subtraction

Subtract one multiple-word number from another. The length in words of both numbers is in the
LENGTH variable. The numbers themselves are stored (most significant bits First) in the variables
NUM1 and NUM2 respectively. Subtract the number in NUM2 from the one in NUM1. Store the
difference in NUM1.

Sample Problem:

Input:

LENGTH

3

(Number of words in each number )

NUM1

NUM2

2F5B8568
84C32546
706C9567

14DF4098
85B81095
A3BC1284

(First number is 2F5B856884C32546706C956716)

(The second number is 14DF409885B81095A3BC128416)

Output:

NUM1

1A7C44CF (Difference is 1A7C44CFFF0B14B0CCB082E316)
FF0B14B0
CCB082E3

That is,

−

2F5B856884C32546706C9567
14DF409885B81095A3BC1284
1A7C44CFFF0B14B0CCB082E3

12.2.2 Decimal Subtraction

Subtract one packed decimal (BCD) number from another. The length in bytes of both numbers
is in the byte variable LENGTH. The numbers themselves are in the variables NUM1 and NUM2 re-
spectively. Subtract the number contained in NUM2 from the one contained in NUM1. Store the
difference in NUM1.

Sample Problem:

CHAPTER 12. ARITHMETIC

(Number of bytes in each number )

(The first number is 36701985783410)

(The second number is 12663459326910)

(Difference is 24038526456510)

116

Input:

LENGTH

NUM1

NUM2

Output:

NUM1

4

36
70
19
85
78
34

12
66
34
59
32
69

24
03
85
26
45
65

That is,

367019857834
− 126634593269
240385264565

12.2.3

32-Bit by 32-Bit Multiply

Multiply the 32-bit value in the NUM1 variable by the value in the NUM2 variable. Use the MULU
instruction and place the result in the 64-bit variable PROD1.

Sample Problem:

Input:

NUM1

NUM2

Output:

PROD1

PROD2

0024
68AC

0328
1088

(The first number is 2468AC16)

(The second number is 328108810)

0000
72EC (MULU product is 72ECB8C25B6016)
B8C2
5B60

0000
72EC (Shift product is 72ECB8C25B6016)
B8C2
5B60

13 Tables and Lists

Tables and lists are two of the basic data structures used with all computers. We have already seen
tables used to perform code conversions and arithmetic. Tables may also be used to identify or
respond to commands and instructions, provide access to files or records, define the meaning of keys
or switches, and choose among alternate programs. Lists are usually less structured than tables.
Lists may record tasks that the processor must perform, messages or data that the processor must
record, or conditions that have changed or should be monitored.

13.1 Program Examples

13.1.1 Add Entry to List

*
*

Main

Loop

TTL
AREA
ENTRY

LDR
LDR
LDR
LDR
LDR

Ch9Ex1
Program, CODE, READONLY

R0, List
R1, NewItem
R3, [R0]
R2, [R0], #4
R4, [R0], #4

examine a table for a match - store a new entry at
the end if no match found

;load the start address of the list
;load the new item
;copy the list counter
;init counter and increment pointer

Program 13.1a: insert.s — Examine a table for a match - store a new entry at the end if no
match found
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34

;adjust the pointer
;increment the number of items
;and store it back
;store the new item at the end of the list

;does the item match the list?
;found it - finished
;no - get the next item
;get the next item
;and loop

Data1, DATA
&4
&5376
&7615
&138A
&21DC
&20

R1, R4
Done
R2, R2, #1
R4, [R0], #4
Loop

R0, R0, #4
R3, R3, #1
R3, Start
R1, [R0]

;length of list
;items

AREA
DCD
DCD
DCD
DCD
DCD
%

;reserve 20 bytes of storage

CMP
BEQ
SUBS
LDR
BNE

SUB
ADD
STR
STR

Start

Store

Done

SWI

&11

117

118

CHAPTER 13. TABLES AND LISTS

35
36
37
38
39
40

AREA

NewItem DCD
DCD
List

Data2, DATA
&16FA
Start

END

*
*

Main

Loop

TTL
AREA
ENTRY

CMP
BEQ
SUBS
LDR
BNE

LDR
LDR
LDR
LDR
CMP
BEQ
LDR

Ch9Ex2
Program, CODE, READONLY

R0, List
R1, NewItem
R3, [R0]
R2, [R0], #4
R3, #0
Insert
R4, [R0], #4

examine a table for a match - store a new entry if no match found
extends Ch9Ex1

;load the start address of the list
;load the new item
;copy the list counter
;init counter and increment pointer
;it’s an empty list
;so store it
;not empty - move to 1st item

Program 13.1b: insert2.s — Examine a table for a match - store a new entry if no match found
extends insert.s
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42

;does the item match the list?
;found it - finished
;no - get the next item
;get the next item
;and loop

;adjust the pointer
;incr list count
;and store it
;store new item at the end

Data1, DATA
&4
&5376
&7615
&138A
&21DC
&20

R1, R4
Done
R2, R2, #1
R4, [R0], #4
Loop

SUB
Insert ADD
STR
STR

R0, R0, #4
R3, R3, #1
R3, Start
R1, [R0]

Data2, DATA
&16FA
Start

;length of list
;items

AREA
DCD
DCD
DCD
DCD
DCD
%

;reserve 20 bytes of storage

NewItem DCD
DCD
List

;all done

Start

Store

AREA

Done

SWI

&11

END

13.1.2 Check an Ordered List

*

examine an ordered table for a match

Program 13.2: search.s — Examine an ordered table for a match
1
2
3
4
5
6
7

Ch9Ex3
Program, CODE, READONLY

TTL
AREA
ENTRY

Main

13.1. PROGRAM EXAMPLES

119

;load the address past the list
;adjust pointer to point at last element of list
;load the item to test
;init counter by reading index from list
;are there zero items
;zero items in list - error condition

;does the item match the list?
;found it - finished
;if the one to test is higher, it’s not in the list
;no - decr counter
;get the next item
;and loop
;if we get to here, it’s not there either
;flag it as missing

;store the index (either index or -1)
;all done

;length of list
;items

8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40

Loop

LDR
SUB
LDR
LDR
CMP
BEQ
LDR

CMP
BEQ
BHI
SUBS
LDR
BNE

R0, =NewItem
R0, R0, #4
R1, NewItem
R3, Start
R3, #0
Missing
R4, [R0], #-4

R1, R4
Done
Missing
R3, R3, #1
R4, [R0], #-4
Loop

Missing MOV

R3, #0xFFFFFFFF

Done

Start

STR
SWI

AREA
DCD
DCD
DCD
DCD
DCD

AREA

NewItem DCD
DCW
Index
DCD
List

END

R3, Index
&11

Data1, DATA
&4
&0000138A
&000A21DC
&001F5376
&09018613

Data2, DATA
&001F5376
0
Start

13.1.3 Remove an Element from a Queue

*

Main

LDR
STR
CMP
BEQ

TTL
AREA
ENTRY

remove the first element of a queue

R0, Queue
R1, Pointer
R0, #0
Done

Ch9Ex4
Program, CODE, READONLY

Program 13.3: head.s — Remove the first element of a queue
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26

AREA
Queue
DCD
Pointer DCD

Data1, DATA
Item1
0

R1, [R0]
R1, Queue

;pointer
;data

Item2
30, 20

* each item consists of a pointer to the next item, and some data
Item1

;load the head of the queue
;and save it in ’Pointer’
;is it NULL?
;if so, nothing to do

;space for new entries

DCD
DCB

LDR
STR

DArea

Done

SWI

&11

20

%

;pointer to the start of the queue
;space to save the pointer

;otherwise get the ptr to next
;and make it the start of the queue

CHAPTER 13. TABLES AND LISTS

120

27
28
29
30
31
32
33
34

Item2

Item3

DCD
DCB

DCD
DCB

END

Item3
30, 0xFF

0
30,&87,&65

;pointer
;data

;pointer (NULL)
;data

13.1.4 Sort a List

Program 13.4: sort.s — Sort a list of values – simple bubble sort

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41

*

sort a list of values - simple bubble sort

TTL
AREA
ENTRY

Ch9Ex5
Program, CODE, READONLY

Main

Sort

Next

NoSwitch

LDR
MOV
LDRB
MOV

ADD
MOV
ADD

LDRB
LDRB
CMP
BCC

STRB
STRB
ADD
SUB

CMP
BHI
CMP
BNE

R6, List
R0, #0
R0, [R6]
R8, R6

R7, R6, R0
R1, #0
R8, R8, #1

R2, [R7], #-1
R3, [R7]
R2, R3
NoSwitch

R2, [R7], #1
R3, [R7]
R1, R1, #1
R7, R7, #1

R7, R8
Next
R1, #0
Sort

;pointer to start of list
;clear register
;get the length of list
;make a copy of start of list

;get address of last element
;zero flag for changes
;move 1 byte up the list each
;iteration
;load the first byte
;and the second
;compare them
;branch if r2 less than r3

;otherwise swap the bytes
;like this
;flag that changes made
;decrement address to check

;have we checked enough bytes?
;if not, do inner loop
;did we mke changes
;if so check again - outer loop

Done

SWI

&11

;all done

Start

List

AREA
DCB
DCB

AREA
DCD

END

Data1, DATA
6
&2A, &5B, &60, &3F, &D1, &19

Data2, DATA
Start

13.2. PROBLEMS

121

13.1.5 Using an Ordered Jump Table

13.2 Problems

13.2.1 Remove Entry from List

Remove the value in the variable ITEM at a list if the value is present. The address of the list is
in the LIST variable. The first entry in the list is the number (in words) of elements remaining in
the list. Move entries below the one removed up one position and reduce the length of the list by
one.

Sample Problems:

Test A

Test B

Input

Output

Input

Output

ITEM
LIST

Table

D010257
Table

00000004
2946C121
2054A346
05723A64
12576C20

No change
since item
not in list

D0102596
Table

0003
00000004
C1212546
C1212546
D0102596
3A64422B
3A64422B 6C20432E
6C20432E

—

13.2.2 Add Entry to Ordered List

Insert the value in the variable ITEM into an ordered list if it is not already there. The address
of the list is in the LIST variable. The first entry in the list is the list’s length in words. The list
itself consists of unsigned binary numbers in increasing order. Place the new entry in the correct
position in the list, adjust the element below it down, and increase the length of the list by one.

Sample Problems

Test A

Test B

Input

Output

Input

Output

ITEM
LIST

Table

7A35B310
Table

0005
00000004
09250037
09250037
29567322
29567322
7A35B310
A356A101
E235C203 A356A101
E235C203

—

7A35B310
Table

00000005
09250037
29567322
7A35B310
A356A101
E235C203

No change
since ITEM
already in
list.

13.2.3 Add Element to Queue

Add the value in the variable ITEM to a queue. The address of the first element in the queue is
in the variable QUEUE. Each element in the queue contains a structure of two items (value and
next) where next is either the address of the next element in the queue or zero if there is no next
element. The new element is placed at the end (tail) of the queue; the new element’s address will
be in the element that was at the end of the queue. The next entry of the new element will contain
zero to indicate that it is now the end of the queue.

Sample Problem:

122

CHAPTER 13. TABLES AND LISTS

Input

Output

Value

Next

Value

Next

ITEM
QUEUE
item1
item2
item3

23854760
00000001
00000123
00123456
—

item1
item2
00000000
—

23854760
00000001
00000123
00123456
23854760

item1
item2
item3
00000000

13.2.4

4-Byte Sort

Sort a list of 4-byte entries into descending order. The first three bytes in each entry are an unsigned
key with the first byte being the most significant. The fourth byte is additional information and
should not be used to determine the sort order, but should be moved along with its key. The
number of entries in the list is defined by the word variable LENGTH. The list itself begins at
location LIST.

Sample Problem:

Input

Output

LENGTH
LIST

00000004
414243 07
4A4B4C 13
4A4B41 37
444B41 3F

(“ABC”)
(“JKL”)
(“JKA”)
(“DKA”)

4A4B4C 13
4A4B41 37
444B41 3F
414243 07

(“JKL”)
(“JKA”)
(“DKA”)
(“ABC”)

13.2.5 Using a Jump Table with a Key

Using the value in the variable INDEX as a key to a jump table (TABLE). Each entry in the jump
table contains a 32-bit identifier followed by a 32-bit address to which the program should transfer
control if the key is equal to that identifier.

Sample Problem:

INDEX

00000010

TABLE

00000001
Proc1
00000010
Proc2
0000001E Proc3

Proc1 NOP
Proc2 NOP
Proc3 NOP

Control should be transferred to Proc2, the second entry in the table.

14 Subroutines

None of the examples that we have shown thus far is a typical program that would stand by itself.
Most real programs perform a series of tasks, many of which may be used a number of times or
be common to other programs.

The standard method of producing programs which can be used in this manner is to write subrou-
tines that perform particular tasks. The resulting sequences of instructions can be written once,
tested once, and then used repeatedly.

There are special instructions for transferring control to subroutines and restoring control to the
main program. We often refer to the special instruction that transfers control to a subroutine as
Call, Jump, or Brach to a Subroutine. The special instruction that restores control to the main
program is usually called Return.

In the ARM the Branch-and-Link instruction (BL) is used to Branch to a Subroutine. This saves the
current value of the program counter (PC or R15 ) in the Link Register (LR or R14 ) before placing
the starting address of the subroutine in the program counter. The ARM does not have a standard
Return from Subroutine instruction like other processors, rather the programmer should copy the
value in the Link Register into the Program Counter in order to return to the instruction after
the Branch-and-Link instruction. Thus, to return from a subroutine you should the instruction:

Should the subroutine wish to call another subroutine it will have to save the value of the Link
Register before calling the nested subroutine.

MOV

PC, LR

14.1 Types of Subroutines

Sometimes a subroutine must have special characteristics.

Relocatable

The code can be placed anywhere in memory. You can use such a subroutine easily, regardless
of other programs or the arrangement of the memory. A relocating loader is necessary to
place the program in memory properly; the loader will start the program after other programs
and will add the starting address or relocation constant to all addresses in the program.

Position Independent

The code does not require a relocating loader — all program addresses are expressed relative
to the program counter’s current value. Data addresses are held in-registers at all times. We
will discuss the writing of position independent code later in this chapter.

Reentrant

The subroutine can be interrupted and called by the interrupting program, giving the correct
results for both the interrupting and interrupted programs. Reentrant subroutines are re-
quired for good for event based systems such as a multitasking operating system (Windows
It is not difficult to make a subroutine
or Unix) and embedded real time environments.

123

124

CHAPTER 14. SUBROUTINES

reentrant. The only requirement is that the subroutine uses just registers and the stack for
its data storage, and the subroutine is self contained in that it does not use any value defined
outside of the routine (global values).

Recursive

The subroutine can call itself. Such a subroutine clearly must also be reentrant.

14.2 Subroutine Documentation

Most programs consist of a main program and several subroutines. This is useful as you can
use known pre-written routines when available and you can debug and test the other subroutines
properly and remember their exact effects on registers and memory locations.

You should provide sufficient documentation such that users need not examine the subroutine’s
internal structure. Among necessary specifications are:

• A description of the purpose of the subroutine

• A list of input and output parameters

• Registers and memory locations used

• A sample case, perhaps including a sample calling sequence

The subroutine will be easy to use if you follow these guidelines.

14.3 Parameter Passing Techniques

In order to be really useful, a subroutine must be general. For example, a subroutine that can
perform only a specialized task, such as looking for a particular letter in an input string of fixed
length, will not be very useful. If, on the other hand, the subroutine can look for any letter, in
strings of any length, it will be far more helpful.

In order to provide subroutines with this flexibility, it is necessary to provide them with the ability
to receive various kinds of information. We call data or addresses that we provide the subroutine
parameters. An important part of writing subroutines is providing for transferring the parameters
to the subroutine. This process is called Parameter Passing.

There are three general approaches to passing parameters:

1. Place the parameters in registers.

2. Place the parameters in a block of memory.

3. Transfer the parameters and results on the hardware stack.

The registers often provide a fast, convenient way of passing parameters and returning results.
The limitations of this method are that it cannot be expanded beyond the number of available
registers; it often results in unforeseen side effects; and it lacks generality.

The trade-off here is between fast execution time and a more general approach. Such a trade-
off is common in computer applications at all levels. General approaches are easy to learn and
consistent; they can be automated through the use of macros. On the other hand, approaches
that take advantage of the specific features of a particular task require less time and memory. The
choice of one approach over the other depends on your application, but you should take the general
approach (saving programming time and simplifying documentation and maintenance) unless time
or memory constraints force you to do otherwise.

14.3. PARAMETER PASSING TECHNIQUES

125

14.3.1 Passing Parameters In Registers

The first and simplest method of passing parameters to a subroutine is via the registers. After
calling a subroutine, the calling program can load memory addresses, counters, and other data
into registers. For example, suppose a subroutine operates on two data buffers of equal length.
The subroutine might specify that the length of the two data buffers be in the register R0 while
the staring address of the two data buffer are in the registers R1 and R2 . The calling program
would then call the subroutine as follows:

MOV
LDR
LDR
BL

R0, #BufferLen
R1, =BufferA
R2, =BufferB
Subr

; Length of Buffer in R0
; Buffer A beginning address in R1
; Buffer B beginning address in R2
; Call subroutine

Using this method of parameter passing, the subroutine can simply assume that the parameters
are there. Results can also be returned in registers, or the addresses of locations for results can
be passed as parameters via the registers. Of course, this technique is limited by the number of
registers available.

Processor features such as register indirect addressing, indexed addressing, and the ability to use
any register as a stack pointer allow far more powerful and general ways of passing parameters.

14.3.2 Passing Parameters In A Parameter Block

Parameters that are to be passed to a subroutine can also be placed into memory in a parameter
block. The location of this parameter block can be passed to the subroutine via a register.

LDR
BL

R0, =Params
Subr

; R0 Points to Parameter Block
; Call the subroutine

If you place the parameter block immediately after the subroutine call the address of the pa-
rameter block is automatically place into the Link Register by the Branch and Link instruction.
The subroutine must modify the return address in the Link Register in addition to fetching the
parameters. Using this technique, our example would be modified as follows:

BL
DCD
DCD
DCD

Subr
BufferLen
BufferA
BufferB

;Buffer Length
;Buffer A starting address
;Buffer B starting address

The subroutine saves’ prior contents of CPU registers, then loads parameters and adjusts the
return address as follows:

Subr

LDR
LDR
LDR

R0, [LR], #4
R1, [LR], #4
R2, [LR], #4

; Read BuufferLen
; Read address of Buffer A
; Read address of Buffer B
; LR points to next instruction

The addressing mode [LR], #4 will read the value at the address pointed to by the Link Register
and then move the register on by four bytes. Thus at the end of this sequence the value of LR has
been updated to point to the next instruction after the parameter block.

This parameter passing technique has the advantage of being easy to read. It has, however, the
disadvantage of requiring parameters to be fixed when the program is written. Passing the address
of the parameter block in via a register allows the papa meters to be changed as the program is
running.

126

CHAPTER 14. SUBROUTINES

14.3.3 Passing Parameters On The Stack

Another common method of passing parameters to a subroutine is to push the parameters onto
the stack. Using this parameter passing technique, the subroutine call illustrated above would
occur as follows:

MOV
STR
LDR
STR
LDR
STR
BL

R0, #BufferLen
R0, [SP, #-4]!
R0, =BufferA
R0, [SP, #-4]!
R0, =BufferA
R0, [SP, #-4]!
Subr

; Read Buffer Length
; Save on the stack
; Read Address of Buffer A
; Save on the stack
; Read Address of Buffer B
; Save on the stack

The subroutine must begin by loading parameters into CPU registers as follows:

Subr

STMIA
LDR
LDR
LDR

...

LDMIA
MOV

R12, {R0, R1, R2, R12, R14}
R0, [R12, #0]
R1, [R12, #4]
R2, [R12, #8]

; save working registers to stack
; Buffer Length in D0
; Buffer A starting address
; Buffer B starting address

R12, {R0, R1, R2, R12, R14}
PC, LR

; Recover working registers
; Return to caller

; Main function of subroutine

In this approach, all parameters are passed and results are returned on the stack.

The stack grows downward (toward lower addresses). This occurs because elements are pushed
onto the stack using the pre-decrement address mode. The use of the pre-decrement mode causes
the stack pointer to always contain the address of the last occupied location, rather than the
next empty one as on some other microprocessors. This implies that you must initialise the stack
pointer to a value higher than the largest address in the stack area.

When passing parameters on the stack, the programmer must implement this approach as follows:

1. Decrement the system stack pointer to make room for parameters on the system stack, and
store them using offsets from the stack pointer, or simply push the parameters on the stack.

2. Access the parameters by means of offsets from the system stack pointer.

3. Store the results on the stack by means of offsets from the systems stack pointer.

4. Clean up the stack before or after returning from the subroutine, so that the parameters are

removed and the results are handled appropriately.

14.4 Types Of Parameters

Regardless of our approach to passing parameters, we can specify the parameters in a variety of
ways. For example, we can:

pass-by-value

Where the actual values are placed in the parameter list. The name comes from the fact
that it is only the value of the parameter that is passed into the subroutine rather than the
parameter itself. This is the method used by most high level programming languages.

pass-by-reference

The address of the parameters are placed in the parameter list. The subroutine can access
the value directly rather than a copy of the parameter. This is much more dangerous as the
subroutine can change a value you don’t want it to.

pass-by-name

Rather than passing either the value or a reference to the value a string containing the name

14.5. PROGRAM EXAMPLES

127

of the parameter is passed. This is used by very high level languages or scripting languages.
This is very flexible but rather time consuming as we need to look up the value associated
with the variable name every time we wish to access the variable.

14.5 Program Examples

*

Main

LDR
LDR
LDR
LDR

TTL
AREA
ENTRY

initiate a simple stack

R1, Value1
R2, Value2
R3, Value3
R4, Value4

Ch10Ex1
Program, CODE, READONLY

Program 14.1a: init1.s — Initiate a simple stack
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29

Stack1, DATA
0xFFFF
0xDDDD
0xAAAA
0x3333

Value1 DCD
Value2 DCD
Value3 DCD
Value4 DCD

R7, =Data2
R7, {R1 - R4}

Data2, DATA
40

Stack
StackEnd

LDR
STMFD

;all done

AREA
%

AREA

&11

SWI

DCD

0

END

;put some data into registers

;load the top of stack
;push the data onto the stack

;reserve 40 bytes of memory for the stack

*

Main

LDR
LDR
LDR
LDR

TTL
AREA
ENTRY

initiate a simple stack

R1, Value1
R2, Value2
R3, Value3
R4, Value4

Ch10Ex2
Program, CODE, READONLY

Program 14.1b: init2.s — Initiate a simple stack
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23

Stack1, DATA
0xFFFF
0xDDDD
0xAAAA
0x3333

Value1 DCD
Value2 DCD
Value3 DCD
Value4 DCD

R7, =Data2
R7, {R1 - R4}

LDR
STMDB

;all done

AREA

&11

SWI

;put some data into registers

128

CHAPTER 14. SUBROUTINES

24
25
26
27
28
29

AREA
%

Stack
StackEnd

Data2, DATA
40

0

DCD

END

;reserve 40 bytes of memory for the stack

*

EQU

0x9000

LDR
LDR
LDR
LDR

TTL
AREA
ENTRY

StackStart
Main

initiate a simple stack

R1, Value1
R2, Value2
R3, Value3
R4, Value4

Ch10Ex3
Program, CODE, READONLY

Program 14.1c: init3.s — Initiate a simple stack
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29

Data1, DATA
0xFFFF
0xDDDD
0xAAAA
0x3333

Value1 DCD
Value2 DCD
Value3 DCD
Value4 DCD

Data2, DATA
StackStart
0

R7, =StackStart
R7, {R1 - R4}

Stack1 DCD

LDR
STMDB

;all done

AREA
^

AREA

&11

SWI

END

;put some data into registers

;Top of stack = 9000
;push R1-R4 onto stack

;some data to put on stack

;reserve 40 bytes of memory for the stack

*

EQU

0x9000

LDR
LDR
LDR
LDR

TTL
AREA
ENTRY

StackStart
start

initiate a simple stack

R1, Value1
R2, Value2
R3, Value3
R4, Value4

Ch10Ex4
Program, CODE, READONLY

Program 14.1d: init3a.s — Initiate a simple stack
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25

Data1, DATA
0xFFFF
0xDDDD
0xAAAA
0x3333

Value1 DCD
Value2 DCD
Value3 DCD
Value4 DCD

R7, =StackStart
R7, {R1 - R4}

LDR
STMDB

;all done

Data2, DATA

AREA

AREA

&11

SWI

;put some data into registers

;Top of stack = 9000
;push R1-R4 onto stack

14.5. PROGRAM EXAMPLES

129

26
27
28
29

^

Stack1 DCD

StackStart
0

END

;reserve 40 bytes of memory for the stack

*
*

EQU

*
*
*

0x9000

TTL
AREA
ENTRY

StackStart
Main

LDRB
BL
STRB
SWI
SWI

Ch10Ex4
Program, CODE, READONLY

R0, HDigit
Hexdigit
R0, AChar
&0
&11

=========================
Hexdigit subroutine
=========================

a simple subroutine example
program passes a variable to the routine in a register

Purpose
Hexdigit subroutine converts a Hex digit to an ASCII character

;variable stored to register
;branch/link
;store the result of the subroutine
;output to console
;all done

Program 14.1e: byreg.s — A simple subroutine example program passes a variable to the routine
in a register
1
2
3
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5
6
7
8
9
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15
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29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49

Final Condition
R0 contains ASCII character in the range ’0’ ... ’9’ or ’A’ ... ’F’

Initial Condition
R0 contains a value in the range 00 ... 0F

R0, #0xA
Addz
R0, R0, #"A" - "0" - 0xA

;digit to convert
;storage for ASCII character

Sample case
Initial condition
Final condition

;convert to ASCII
;return from subroutine

;is it > 9
;if not skip the next

Registers changed
R0 only

Data1, DATA
6
0

R0 = 6
R0 = 36 (’6’)

R0, R0, #"0"
PC, LR

HDigit DCB
DCB
AChar

;adjust for A .. F

*
*
*
*
*
*
*
*
*
*
*
*
*
*
*

CMP
BLE
ADD

Hexdigit

ADD
MOV

AREA

Addz

END

Program 14.1f: bystack.s — A more complex subroutine example program passes variables to the
routine using the stack
1
2
3
4
5

a more complex subroutine example
program passes variables to the routine using the stack

Ch10Ex5
Program, CODE, READONLY

TTL
AREA

*
*

130

CHAPTER 14. SUBROUTINES

6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72

ENTRY

StackStart
Mask

EQU
EQU

0x9000
0x0000000F

;declare where top of stack will be
;bit mask for masking out lower nibble

Main

*
*
*

*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*

LDR
LDR
LDR
STR
STR
BL
SWI

R7, =StackStart
R0, Number
R1, =String
R1, [R7], #-4
R0, [R7], #-4
Binhex
&11

=========================
Binhex subroutine
=========================

;Top of stack = 9000
;Load number to register
;load address of string
;and store it
;and store number to stack
;branch/link
;all done

Purpose
Binhex subroutine converts a 16 bit value to an ASCII string

Initial Condition
First parameter on the stack is the value
Second parameter is the address of the string

Final Condition
the HEX string occupies 4 bytes beginning with
the address passed as the second parameter

Registers changed
No registers are affected

Sample case
Initial condition

Final condition

top of stack : 4CD0
Address of string
The string ’4’’C’’D’’0’ in ASCII
occupies memory

Binhex

Loop

MOV
STMDA
MOV
ADD
LDR
LDR
ADD

MOV
AND
BL
STRB
MOV
SUBS
BNE

R8, R7
R7, {R0-R6,R14}
R1, #4
R7, R7, #4
R2, [R7], #4
R4, [R7]
R4, R4, #4

R0, R2
R0, R0, #Mask
Hexdigit
R0, [R4], #-1
R2, R2, LSR #4
R1, R1, #1
Loop

;save stack pointer for later
;push contents of R0 to R6, and LR onto the stack
;init counter
;adjust pointer
;get the number
;get the address of the string
;move past the end of where the string is to be stored

;copy the number
;get the low nibble
;convert to ASCII
;store it
;shift to next nibble
;decr counter
;loop while still elements left

LDMDA
MOV

R8, {R0-R6,R14}
PC, LR

;restore the registers
;return from subroutine

=========================
Hexdigit subroutine
=========================

Purpose
Hexdigit subroutine converts a Hex digit to an ASCII character

Initial Condition

*
*
*

*
*
*
*

14.5. PROGRAM EXAMPLES

131

73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98

*
*
*
*
*
*
*
*
*
*
*

R0 contains a value in the range 00 ... 0F

Final Condition
R0 contains ASCII character in the range ’0’ ... ’9’ or ’A’ ... ’F’

Registers changed
R0 only

Sample case
Initial condition
Final condition

R0 = 6
R0 = 36 (’6’)

Hexdigit

Addz

CMP
BLE
ADD

ADD
MOV

R0, #0xA
Addz
R0, R0, #"A" - "0" - 0xA

;is the number 0 ... 9?
;if so, branch

;adjust for A ... F

R0, R0, #"0"
PC, LR

;change to ASCII
;return from subroutine

AREA

Number DCD
String DCB

Data1, DATA
&4CD0
4, 0

END

;number to convert
;counted string for result

;branch/link
;address of parameter 1
;address of parameter 2

*

SWI

&11

*
*
*

Main

;all done

BL
DCD
DCD

TTL
AREA
ENTRY

Add64
Value1
Value2

a 64 bit addition subroutine

Ch10Ex6
Program, CODE, READONLY

=========================
Add64 subroutine
=========================

Program 14.1g: add64.s — A 64 bit addition subroutine
1
2
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5
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13
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18
19
20
21
22
23
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25
26
27
28
29
30
31
32
33
34
35
36
37
38

Initial Condition
The two parameter values are passed immediately
following the subroutine call

Sample case
Initial condition
para 1 =
para 2 =

Purpose
Add two 64 bit values

= &0420147AEB529CB8
= &3020EB8520473118

Registers changed
R0 and R1 only

Final condition
R0 = &34410000

*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*

Final Condition
The sum of the two values is returned in R0 and R1

132

CHAPTER 14. SUBROUTINES

39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61

*

R1 = &0B99CDD0

Add64

STMIA
MOV
SUB
LDR
LDR
LDR
LDR

ADDS
BCC
ADD

ADD
LDMIA
MOV

AREA

Next

Value1 DCD
Value2 DCD
END

R12, {R2, R3, R14}
R7, R12
R7, R7, #4
R3, [R7], #-4
R2, [R7], #-4
R1, [R7], #-4
R0, [R7], #-4

;save registers to stack
;copy stack pointer
;adjust to point at LSB of 2nd value
;load successive bytes

R1, R1, R3
Next
R0, R0, #1

;add LS bytes & set carry flag
;branch if carry bit not set
;otherwise add the carry

R0, R0, R2
R12, {R2, R3, R14}
PC, LR

;add MS bytes
;pop from stack
;and return

Data1, DATA
&0420147A, &EB529CB8
&3020EB85, &20473118

;number1 to add
;number2 to add

*

SWI

&11

*
*
*

Main

;all done

LDR
BL
STR

TTL
AREA
ENTRY

R0, Number
Factor
R0, FNum

a subroutine to find the factorial of a number

Ch10Ex6
Program, CODE, READONLY

;get number
;branch/link
;store the factorial

=========================
Factor subroutine
=========================

Program 14.1h: factorial.s — A subroutine to find the factorial of a number
1
2
3
4
5
6
7
8
9
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11
12
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15
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29
30
31
32
33
34
35
36
37
38
39
40
41

Purpose
Recursively find the factorial of a number

;push to stack
;push the return address
;subtract 1 from number

Initial Condition
R0 contains the number to factorial

Sample case
Initial condition
Number = 5

Final Condition
R0 = factorial of number

R0, [R12], #4
R14, [R12], #4
R0, R0, #1

Final condition
FNum = 120 = 0x78

Registers changed
R0 and R1 only

*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*

STR
STR
SUBS

Factor

14.6. PROBLEMS

133

42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60

BNE

MOV
SUB
B

F_Cont

;not finished

R0, #1
R12, R12, #4
Return

;Factorial == 1
;adjust stack pointer
;done

BL

Factor

;if not done, call again

F_Cont

Return

LDR
LDR
MUL
MOV

AREA

Number DCD
DCD
FNum
END

R14, [R12], #-4
R1, [R12], #-4
R0, R1, R0
PC, LR

Data1, DATA
5
0

;return address
;load to R1 (can’t do MUL R0, R0, xxx)
;multiply the result
;and return

;number
;factorial

14.6 Problems

Write both a calling program for the sample problem and at least one properly documented
subroutine for each problem.

14.6.1 ASCII Hex to Binary

Write a subroutine to convert the least significant eight bits in register R0 from the ASCII repre-
sentation of a hexadecimal digit to the 4-bit binary representation of the digit. Place the result
back into R0 .

Sample Problems:

Input:
Output:

R0
R0

Test A
43 ‘C’
0C

Test B
36 ‘6’
06

14.6.2 ASCII Hex String to Binary Word

Write a subroutine that takes the address of a string of eight ASCII characters in R0 . It should
convert the hexadecimal string into a 32-bit binary number, which it return is R0 .

Sample Problem:

Input:

R0
STRING

Output:

R0

String
42 ‘B’
32 ‘2’
46 ‘F’
30 ‘0’
0000B2F0

14.6.3 Test for Alphabetic Character

Write a subroutine that checks the character in register R0 to see if it is alphabetic (upper- or
lower-case). It should set the Zero flag if the character is alphabetic, and reset the flag if it is not.

CHAPTER 14. SUBROUTINES

134

Sample Problems:

Input:
Output:

R0
Z

Test A
47 ‘G’
FF

Test B
36 ‘6’
00

Test C
6A ‘j’
FF

14.6.4 Scan to Next Non-alphabetic

Write a subroutien that takes the address of the start of a text string in register R1 and returns
the address of the first non-alphabetic character in the string in register R1 . You should consider
using the isalpha subroutine you have just define.

Sample Problems:

Input:

R1
String

Output:

R1

Test A

String
43 ‘C’
61 ‘a’
74 ‘t’
0D CR
String + 4
(CR)

B

6100
32 ‘2’
50 ‘P’
49 ‘I’
0D CR
String + 0
(2)

14.6.5 Check Even Parity

Write a subroutine that takes the address of a counted string in the register R0 . It should check
for an even number of set bits in each character of the string. If all the bytes have an even parity
then it should set the Z-flag, if one or more bytes have an odd parity it should clear the Z-flag.

Sample Problems:

Input:

R0
String

Output:

Z

Test A
String
03
47
AF
18
00

Test B
String
03
47
AF
19
FF

Note that 1916 is 0001 10012 which has three 1 bits and is thus has an odd parity.

14.6.6 Check the Checksum of a String

Write a subroutine to calculate the 8-bit checksum of the counted string pointed to by the register
R0 and compares the calculated checksum with the 8-bit checksum at the end of the string. It
should set the Z-flag if the checksums are equal, and reset the flag if they are not.

Sample Problems:

Input:

R0
String

Output:

Z

Test A

Test B

String
03
41
42
43
C6

Set

(Length)
(‘A’)
(‘B’)
(‘C’)
(Checksum)

String
03
61
62
63
C6

Clear

(Length)
(‘a’)
(‘b’)
(‘c’)
(Checksum should be 26</em>)

14.6. PROBLEMS

135

14.6.7 Compare Two Counted Strings

Write a subroutine to compare two ASCII strings. The first byte in each string is its length.
Return the result in the condition codes; i.e., the N-flag will be set if the first string is lexically
less than (prior to) the second, the Z-flag will be set if the strings are equal, no flags are set if the
second is prior to the first. Note that “ABCD” is lexically greater than “ABC”.

136

CHAPTER 14. SUBROUTINES

A ARM Instruction Definitions

This appendix describes every ARM instruction, in terms of:

Syntax

This is a description of the way the instruction should be written in the assembler.
A number of shorthand descriptions are used. The most common of these are:

Condition Codes

(cid:104)cc(cid:105)
(cid:104)op1 (cid:105) Data Movement Addressing Modes
(cid:104)op2 (cid:105) Memory Addressing Modes
(cid:104)S (cid:105)

Set Flags bit

(see section 4.1.2 on page 42)
(see section 5.1 on page 51)
(see section 5.2 on page 54)
(see section 4.1.1 on page 41)

Additional notations will be described as they are introduced.

Operation

A Register Transfer Language (RTL) / pseudo-code description of what the in-
struction does. This will use the same shorthand notations used for the Syntax.
For details of the register transfer language, see section 3.5 on page 30.

Description Written description of what the instruction does. This will interpret the formal

description given in the operation part.

Exceptions This gives details of which exceptions can occur during the instruction. Prefetch

Abort is not listed because it can occur for any instruction.

Usage

Suggestions and other information relating to how an instruction can be used
effectively.

Condition Codes

Indicates what happens to the CPU Condition Code Flags if the set flags option
where to be set.

Notes

Contain any additional explanation that we can not fit into the previous categories.

Appendix B provides a summary of the more common instructions in a more compact manner,
using the operation section only.

ADC

Add with Carry

Syntax

ADC(cid:104)cc(cid:105)(cid:104)S (cid:105) Rd , Rn, (cid:104)op1 (cid:105)

Operation

(cid:104)cc(cid:105):

Rd ← Rn + (cid:104)op1 (cid:105) + C

(cid:104)cc(cid:105)(cid:104)S (cid:105): CPSR ← ALU(flags)

Description The ADC (Add with Carry) instruction adds the value of (cid:104)op1 (cid:105) and the Carry
flag to the value of Rn and stores the result in Rd . The condition code flags are
optionally updated, based on the result.

Usage

ADC is used to synthesize multi-word addition. If register pairs R0 , R1 and R2 , R3
hold 64-bit values (where R0 and R2 hold the least significant words) the following
instructions leave the 64-bit sum in R4 , R5 :

137

138

A.2 Add (ADD)

ADDS
ADC

R4,R0,R2
R5,R1,R3

If the second instruction is changed from:

ADC

R5,R1,R3

to:

ADCS

R5,R1,R3

the resulting values of the flags indicate:

N The 64-bit addition produced a negative result.

C An unsigned overflow occurred.

V A signed overflow occurred.

Z

The most significant 32 bits are all zero.

The following instruction produces a single-bit Rotate Left with Extend operation
(33-bit rotate through the Carry flag) on R0 :

ADCS

R0,R0,R0

See Data-processing operands - Rotate right with extend for information on how
to perform a similar rotation to the right.

Condition Codes

The N and Z flags are set according to the result of the addition, and the C and
V flags are set according to whether the addition generated a carry (unsigned
overflow) and a signed overflow, respectively.

Notes

If the destination register (Rd ) is the program counter (PC or R15 ), then the
current status register (CPSR) is restored from the saved status register (SPSR).
This form of the instruction is unpredictable if executed in User mode or System
mode, because these modes do not have an SPSR.

ADD

Add

Syntax

ADD(cid:104)cc(cid:105)(cid:104)S (cid:105) Rd , Rn, (cid:104)op1 (cid:105)

Operation

(cid:104)cc(cid:105):

Rd ← Rn + (cid:104)op1 (cid:105)
(cid:104)cc(cid:105)(cid:104)S (cid:105): CPSR ← ALU(flags)

Description Adds the value of (cid:104)op1 (cid:105) to the value of register Rn, and stores the result in the
destination register Rd . The condition code flags are optionally updated, based
on the result.

Usage

The ADD instruction is used to add two values together to produce a third.

To increment a register value in Rx use:

ADD

Rx, Rx, #1

Constant multiplication of Rx by 2n + 1 into Rd can be performed with:

ADD

Rd, Rx, Rx, LSL #n

To form a PC-relative address use:

ADD

Rs, PC, #offset

A.3 Bitwise AND (AND)

139

where the (cid:104)offset(cid:105) must be the difference between the required address and the
address held in the PC, where the PC is the address of the ADD instruction itself
plus 8 bytes.

Condition Codes

The N and Z flags are set according to the result of the addition, and the C and
V flags are set according to whether the addition generated a carry (unsigned
overflow) and a signed overflow, respectively.

Notes

If the destination register (Rd ) is the program counter (PC or R15 ), then the
current status register (CPSR) is restored from the saved status register (SPSR).
This form of the instruction is unpredictable if executed in User mode or System
mode, because these modes do not have an SPSR.

AND

Bitwise AND

Syntax

AND(cid:104)cc(cid:105)(cid:104)S (cid:105) Rd , Rn, (cid:104)op1 (cid:105)

Operation

(cid:104)cc(cid:105):

Rd ← Rn ∧ (cid:104)op1 (cid:105)
(cid:104)cc(cid:105)(cid:104)S (cid:105): CPSR ← ALU(flags)

Description The AND instruction performs a bitwise AND of the value of register Rn with the
value of (cid:104)op1 (cid:105), and stores the result in the destination register Rd . The condition
code flags are optionally updated, based on the result.

Usage

AND is most useful for extracting a field from a register, by and ing the register
with a mask value that has 1 s in the field to be extracted, and 0 s elsewhere.

Condition Codes

The N and Z flags are set according to the result of the operation, and the C flag
is set to the carry output generated by (cid:104)op1 (cid:105) (see 5.1 on page 51), the V flag is
unaffected.

Notes

If the destination register (Rd ) is the program counter (PC or R15 ), then the
current status register (CPSR) is restored from the saved status register (SPSR).
This form of the instruction is unpredictable if executed in User mode or System
mode, because these modes do not have an SPSR.

B, BL

Branch, Branch and Link

Syntax

B(cid:104)L(cid:105)(cid:104)cc(cid:105) (cid:104)offset(cid:105)

Operation

(cid:104)cc(cid:105)(cid:104)L(cid:105): LR ← PC + 8

(cid:104)cc(cid:105): PC ← PC + (cid:104)offset(cid:105)

Description The B (Branch) and BL (Branch and Link) instructions cause a branch to a target
address, and provide both conditional and unconditional changes to program flow.

The BL (Branch and Link) instruction stores a return address in the link register
(LR or R14 ).

The (cid:104)offset(cid:105) specifies the target address of the branch. The address of the next
instruction is calculated by adding the offset to the program counter (PC) which
contains the address of the branch instruction plus 8.

The branch instructions can specify a branch of approximately ±32MB.

140

Usage

A.6 Compare Negative (CMN)

The BL instruction is used to perform a subroutine call. The return from subrou-
tine is achieved by copying the LR to the PC. Typically, this is done by one of the
following methods:

• Executing a MOV PC,R14 instruction.

• Storing a group of registers and R14 to the stack on subroutine entry, using

an instruction of the form:

STMFD R13!,{(cid:104)registers(cid:105),R14}

and then restoring the register values and returning with an instruction of
the form:

LDMFD R13!,{(cid:104)registers(cid:105),PC}

Condition Codes

The condition codes are not effected by this instruction.

Notes

Branching backwards past location zero and forwards over the end of the 32-bit
address space is unpredictable.

BIC

Bit Clear

Syntax

BIC(cid:104)cc(cid:105)(cid:104)S (cid:105) Rd , Rn, (cid:104)op1 (cid:105)

Operation

(cid:104)cc(cid:105):

Rd ← Rn ∧ (cid:104)op1 (cid:105)
(cid:104)cc(cid:105)(cid:104)S (cid:105): CPSR ← ALU(flags)

Description The BIC (Bit Clear) instruction performs a bitwise AND of the value of register Rn
with the complement of the value of (cid:104)op1 (cid:105), and stores the result in the destination
register Rd . The condition code flags are optionally updated, based on the result.

Usage

The instruction can be used to clear selected bits in a register. For each bit, BIC
with 1 clears the bit, and BIC with 0 leaves it unchanged.

Condition Codes

The N and Z flags are set according to the result of the operation, and the C flag
is set to the carry output generated by (cid:104)op1 (cid:105) (see 5.1 on page 51), the V flag is
unaffected.

Notes

If the destination register (Rd ) is the program counter (PC or R15 ), then the
current status register (CPSR) is restored from the saved status register (SPSR).
This form of the instruction is unpredictable if executed in User mode or System
mode, because these modes do not have an SPSR.

CMN

Compare Negative

Syntax

CMN(cid:104)cc(cid:105) Rn, (cid:104)op1 (cid:105)

Operation

(cid:104)cc(cid:105): ALU(0) ← Rn + (cid:104)op1 (cid:105)
(cid:104)cc(cid:105): CSPR ← ALU(Flags)

Description The CMN (Compare Negative) instruction compares a register value with the neg-
ative of another arithmetic value. The condition flags are updated, based on the
result of adding the second arithmetic value to the register value, so that subse-
quent instructions can be conditionally executed.

Usage

The instruction performs a comparison by adding the value of (cid:104)op1 (cid:105) to the value
of register Rn, and updates the condition code flags (based on the result). This is

A.7 Compare (CMP)

141

almost equivalent to subtracting the negative of the second operand from the first
operand, and setting the flags on the result. The difference is that the flag values
generated can differ when the second operand is 0 or 0x80000000.

For example, this instruction always leaves the C flag set:

CMP

Rn, #0

while this instruction always leaves the C flag clear:

CMN

Rn, #0

CMP

Compare

Syntax

CMP(cid:104)cc(cid:105) Rn, (cid:104)op1 (cid:105)

Operation

(cid:104)cc(cid:105): ALU(0) ← Rn − (cid:104)op1 (cid:105)
(cid:104)cc(cid:105): CSPR ← ALU(Flags)

Description The CMP (Compare) instruction compares a register value with another arithmetic
value. The condition flags are updated, based on the result of subtracting (cid:104)op1 (cid:105)
from Rn, so that subsequent instructions can be conditionally executed.

Condition Codes

The N and Z flags are set according to the result of the subtraction, and the C and
V flags are set according to whether the subtraction generated a borrow (unsinged
underflow) and a signed overflow, respectively.

EOR

Exclusive OR

Syntax

EOR(cid:104)cc(cid:105)(cid:104)S (cid:105) Rd , Rn, (cid:104)op1 (cid:105)

Operation

(cid:104)cc(cid:105):

Rd ← Rn ⊕ (cid:104)op1 (cid:105)
(cid:104)cc(cid:105)(cid:104)S (cid:105): CPSR ← ALU(Flags)

Description The EOR (Exclusive OR) instruction performs a bitwise Exclusive-OR of the value
of register Rn with the value of (cid:104)op1 (cid:105), and stores the result in the destination
register Rd . The condition code flags are optionally updated, based on the result.

Usage

EOR can be used to invert selected bits in a register. For each bit, a 1 inverts that
bit, and a 0 leaves it unchanged.

Condition Codes

The N and Z flags are set according to the result of the operation, and the C flag
is set to the carry output bit generated by the shifter. The V flag is unaffected.

Notes

If the destination register (Rd ) is the program counter (PC or R15 ), then the
current status register (CPSR) is restored from the saved status register (SPSR).
This form of the instruction is unpredictable if executed in User mode or System
mode, because these modes do not have an SPSR.

LDM

Load Multiple

Syntax

LDM(cid:104)cc(cid:105)(cid:104)mode(cid:105) Rn(cid:104)! (cid:105), {(cid:104)registers(cid:105)}

where (cid:104)mode(cid:105) is one of:

Increment Before
IB
DB Decrement Before

Increment After
IA
DA Decrement After

142

Operation

if (cid:104)cc(cid:105)

A.10 Load Register (LDR)

IA: MAR ← Rn
IB: MAR ← Rn + 4
DA: MAR ← Rn − (#(cid:104)registers(cid:105) × 4) + 4
DB: MAR ← Rn − (#(cid:104)registers(cid:105) × 4)

for each register Ri in (cid:104)registers(cid:105)

IB: MAR ← MAR + 4
DB: MAR ← MAR − 4
Ri ← M(MAR)
IA: MAR ← MAR + 4
DA: MAR ← MAR − 4

(cid:104)! (cid:105):
end if

Rn ← MAR

Description The LDM (Load Multiple) instruction is useful for block loads, stack operations and
procedure exit sequences. It loads a subset, or possibly all, of the general-purpose
registers from sequential memory locations.

The register Rn points to the memory local to load the values from. Each of the
registers listed in (cid:104)registers(cid:105) is loaded in turn, reading each value from the next
memory address as directed by (cid:104)mode(cid:105).

The base register writeback option ((cid:104)! (cid:105)) causes the base register to be modified to
hold the address of the final valued loaded.

The register are loaded in sequence, the lowest-numbered register from the low-
est memory address, through to the highest-numbered register from the highest
memory address.

If the PC (R15 ) is specified in the register list, the instruction causes a branch to
the address loaded into the PC.

Exceptions Data Abort — This exception is generated if the address is not word aligned.

Condition Codes

The condition codes are not effected by this instruction.

Notes

The base register Rn should not be specified in (cid:104)registers(cid:105) when the base register
writeback ((cid:104)! (cid:105)) is used.

LDR

Load Register

Syntax

LDR(cid:104)cc(cid:105) Rd , (cid:104)op2 (cid:105)

Operation

(cid:104)cc(cid:105): Rd ← M((cid:104)op2 (cid:105))

Description The LDR (Load Register) instruction loads a word from the memory address cal-

culated by (cid:104)op2 (cid:105) and writes it to register Rd .

If the PC is specified as register Rd , the instruction loads a data word which it
treats as an address, then branches to that address.

Exceptions Data Abort — This exception is generated if the memory address produced by

(cid:104)op2 (cid:105) is not word aligned.

Usage

Using the PC as the base register allows PC-relative addressing, which facilitates
position-independent code. Combined with a suitable addressing mode, LDR allows
32-bit memory data to be loaded into a general-purpose register where its value

A.11 Load Register Byte (LDRB)

143

can be manipulated. If the destination register is the PC, this instruction loads a
32-bit address from memory and branches to that address.

To synthesise a Branch with Link, precede the LDR instruction with MOV LR, PC.

Condition Codes

The condition codes are not effected by this instruction.

Notes

If (cid:104)op2 (cid:105) specifies an address that is not word-aligned, the instruction will load a
32-bit value, but the value will be unpredictable. The LDRB (load byte) instruction
should be used.

If (cid:104)op2 (cid:105) specifies base register writeback (!), and the same register is specified for
Rd and Rn, the results are unpredictable.

If the PC (R15 ) is specified for Rd , the value must be word aligned otherwise the
result is unpredictable.

LDRB

Load Register Byte

Syntax

LDR(cid:104)cc(cid:105)B Rd , (cid:104)op2 (cid:105)

Operation

(cid:104)cc(cid:105): Rd (7:0) ← M((cid:104)op2 (cid:105))
(cid:104)cc(cid:105): Rd (31:8) ← 0

Description The LDRB (Load Register Byte) instruction loads a byte from the memory address
calculated by (cid:104)op2 (cid:105), zero-extends the byte to a 32-bit word, and writes the word
to register Rd .

Exceptions Data Abort — This exception is generated if the memory address produced by

(cid:104)op2 (cid:105) is not word aligned.

Usage

LDRB allows 8-bit memory data to be loaded into a general-purpose register where
it can be manipulated.

Using the PC as the base register allows PC-relative addressing, to facilitate
position-independent code.

Condition Codes

The condition codes are not effected by this instruction.

Notes

If the PC (R15 ) is specified for Rd , the result is unpredictable.

If (cid:104)op2 (cid:105) specifies base register writeback (!), and the same register is specified for
Rd and Rn, the results are unpredictable.

MLA

Multiply Accumulate

Syntax

MLA(cid:104)cc(cid:105)(cid:104)S (cid:105) Rd , Rm, Rs, Rn

Operation

(cid:104)cc(cid:105): ALU ← (Rm × Rs)(0:31)
Rd ← ALU + Rn
(cid:104)cc(cid:105):
(cid:104)cc(cid:105)(cid:104)S (cid:105): CPSR ← ALU(flags)

Description The MLA (Multiply Accumulate) multiplies signed or unsigned operands to pro-
duce a 32-bit result, which is then added to a third operand, and written to the
destination register. The condition code flags are optionally updated, based on
the result.

144

A.14 Move to Register from Status (MRS)

As it only produces the lower 32-bits of the 64-bit product, it gives the same
answer for multiplication of both signed and unsigned numbers. That is, it does
not take the sign into account.

Condition Codes

The N and Z flags are set according to the result of the operation. The V flag is
unaffected. Unfortunately the C flag will have a random value, and should not be
used after this instruction.

Notes

Specifying the same register for Rd and Rm has unpredictable results.

MOV

Move

Syntax

MOV(cid:104)cc(cid:105)(cid:104)S (cid:105) Rd , (cid:104)op1 (cid:105)

Operation

(cid:104)cc(cid:105):

Rd ← (cid:104)op1 (cid:105)

(cid:104)cc(cid:105)(cid:104)S (cid:105): CPSR ← ALU(Flags)

Description The MOV (Move) instruction moves the value of (cid:104)op1 (cid:105) to the destination register

Rd . The condition code flags are optionally updated, based on the result.

Usage

MOV is used to:

• Move a value from one register to another.

• Put a constant value into a register.

• Perform a shift without any other arithmetic or logical operation. A left shift

by n can be used to multiply by 2n.

• When the PC is the destination of the instruction, a branch occurs. The

instruction:

MOV

PC, LR

can therefore be used to return from a subroutine (see instructions B, and
BL on page 139 and chapter 14).

Condition Codes

The N and Z flags are set according to the value moved (post-shift if a shift is
specified), and the C flag is set to the carry output bit generated by the shifter
(see 5.1 on page 51). The V flag is unaffected.

Notes

If the destination register (Rd ) is the program counter (PC or R15 ), then the
current status register (CPSR) is restored from the saved status register (SPSR).
This form of the instruction is unpredictable if executed in User mode or System
mode, because these modes do not have an SPSR.

MRS

Move to Register from Status

Syntax

MRS(cid:104)cc(cid:105) Rd , CPSR
MRS(cid:104)cc(cid:105) Rd , SPSR

Operation

(cid:104)cc(cid:105)(cid:104)CPSR(cid:105): Rd ← CPSR
(cid:104)cc(cid:105)(cid:104)SPSR(cid:105): Rd ← SPSR

Description The MRS (Move to Register from Status) instruction moves the value of the sta-
tus register, either the current status (CPSR) or the saved status (SPSR) into a
destination register Rd . The value can then be examined, or manipulated, with
normal data-processing instructions.

A.15 Move to Status from Register (MSR)

145

Usage

This is commonly used for three purposes:

• As part of a read/modify/write sequence for updating a status (see MSR below

for a discussion).

• When an exception occurs and there is a possibility of a nested exception of
the same type occurring, the saved process status (SPSR) of the exception
mode is in danger of being corrupted. The value can be copied out and saved
before the nested exception can occur, and later restored in preparation for
the exception return.

• When swapping process, the current state of the process being swapped out
needs to be saved, including the current flags, and processor mode. Similarly
the state of the process being swapped in needs to be restored.

Condition Codes

The condition codes are not effected by this instruction, just copied into the des-
tination register.

Notes

The user mode does not have a saved process status register (SPSR) so attempting
to access the SPSR when in User mode or System mode is unpredictable.

MSR

Move to Status from Register

Syntax

MSR(cid:104)cc(cid:105) CPSR_(cid:104)fields(cid:105), (cid:104)arg(cid:105)
MSR(cid:104)cc(cid:105) SPSR_(cid:104)fields(cid:105), (cid:104)arg(cid:105)

where (cid:104)arg(cid:105) can be either an Immediate value or a register (see section 5.1.1 on
page 51). (cid:104)fields(cid:105) is one or more of the following field specifiers:

c Control field
x Extension field
Status field
s
flag field
f

bits 0–7

bits 24–31

Operation

(cid:104)cc(cid:105)(cid:104)CPSR(cid:105): CPSR((cid:104)fields(cid:105)) ← (cid:104)arg(cid:105)
(cid:104)cc(cid:105)(cid:104)SPSR(cid:105): SPSR((cid:104)fields(cid:105)) ← (cid:104)arg(cid:105)

Description The MSR (Move to Status from Register) instruction copies the value of the source
register (Rs) or immediate constant to the CPSR or the SPSR of the current mode.

Which parts of the status register are to be modified are specified via the (cid:104)fields(cid:105)
field.

Usage

This instruction is used to update the value of the condition code flags, interrupt
enables, or the processor mode.

The value of a status register should normally be updated by moving the reg-
ister to a general-purpose register (using the MRS instruction on the preceding
page), modifying the relevant bits of the general-purpose register, and restoring
the updated general-purpose register value back into the status (using the MSR
instruction). For example, a good way to switch to Supervisor mode from another
privileged mode is:

MRS
BIC
ORR
MSR

R0, CPSR
R0, R0, #0x1F
R0, R0, #0x13
CPSR_c, R0

; Read CPSR
; Remove current mode (lower 5 bits)
; substitute Supervisor mode
; Write modified register back

146

A.16 Multiply (MUL)

You should only write to those fields that they can potentially change. The MSR
instruction in the above code can only change the control field, the other bits/field
are unchanged since they were read from the CPSR by the first instruction. So it
writes to CPSR_c, not CPSR_fsxc or some other combination of fields.

Although the state or extension fields, it is a good idea to write to these fields
when writing the whole register. Rather than just writing to the flags and control
fields (CPSR_fc) you shoud write to all four fields (CPSR_fsxc).

Note that due to a bug in the ARM Software Development Tookit, version 2.50
and earlier, you can only write to the flags and control fields.

The immediate form of this instruction can be used to set any of the fields of a
status register, but you must take care to adhere to the read-modify-write tech-
nique. The immediate form of the instruction is equivalent to reading the register
concerned, replacing all the bits in the relevant field by the corresponding bits of
the immediate constant and writing the result back to the status register. The
immediate form must therefore only be used when the intention is to modify all
the bits in the specified fields. Failure to observe this rule might result in code
which has unanticipated side-effects on future versions of the architecture.

As an exception to this rule, it is legitimate to use the immediate form of the
instruction to modify the flags byte despite the fact that bits 24–26 of the status
register have not been allocated at present. For example, this instruction can be
used to set all four flags:

MSR

CPSR_f, #0xF0000000

Condition Codes

The current state will be effected by the value written into the current process
status register (CPSR). If one of the saved process status registers (SPSR) is used
the current status does not change.

Notes

Any writes into the control, extension, and status fields (lower 24 bits) of the
CPSR when in User mode will be ignored, so that User mode programs cannot
change to a privileged mode.

As there is no saved process status register in User or System mode, any attempt
to write into the SPSR while in User or System mode will be unpredictable.

MUL

Multiply

Syntax

MUL(cid:104)cc(cid:105)(cid:104)S (cid:105) Rd , Rm, Rs

Operation

(cid:104)cc(cid:105):

Rd ← (Rm × Rs)(0:31)(cid:104)arg(cid:105)

(cid:104)cc(cid:105)(cid:104)S (cid:105): CPSR ← ALU(flags)

Description The MUL (Multiply) instruction is used to multiply signed or unsigned variables to
produce a 32-bit result. The condition code flags are optionally updated, based
on the result.

Condition Codes

The N and Z flags are set according to the result of the multiplication. While
the C flag should be left unchanged, a bug in the design means that the C flag is
unpredictable after a MULS instruction.

Notes

Specifying the same register for Rd and Rm has unpredictable results.

As the MUL instruction produces only the lower 32 bits of the 64-bit product, it
gives the same answer for multiplication of both signed and unsigned numbers.

A.17 Move Negative (MVN)

147

MVN

Move Negative

Syntax

MVN(cid:104)cc(cid:105)(cid:104)S (cid:105) Rd , (cid:104)op1 (cid:105)

Operation

(cid:104)cc(cid:105):

Rd ← (cid:104)op1 (cid:105)

(cid:104)cc(cid:105)(cid:104)S (cid:105): CPSR ← ALU(Flags)

Description The MVN (Move Negative) instruction moves the logical one’s complement of the
value of (cid:104)op1 (cid:105) to the destination register Rd . The condition code flags are option-
ally updated, based on the result.

Usage

MVN is used to:

• Write a negative value into a register.

• Form a bit mask.

• Take the one’s complement of a value.

Condition Codes

The N and Z flags are set according to the result of the operation, and the C flag
is set to the carry output bit generated by the shifter (see 5.1 on page 51). The V
flag is unaffected.

Notes

If the destination register (Rd ) is the program counter (PC or R15 ), then the
current status register (CPSR) is restored from the saved status register (SPSR).
This form of the instruction is unpredictable if executed in User mode or System
mode, because these modes do not have an SPSR.

ORR

Bitwise OR

Syntax

ORR(cid:104)cc(cid:105)(cid:104)S (cid:105) Rd , Rn, (cid:104)op1 (cid:105)

Operation

(cid:104)cc(cid:105):

Rd ← Rn ∨ (cid:104)op1 (cid:105)
(cid:104)cc(cid:105)(cid:104)S (cid:105): CPSR ← ALU(Flags)

Description The ORR (Logical OR) instruction performs a bitwise (inclusive) OR of the value
of register Rn with the value of (cid:104)op1 (cid:105), and stores the result in the destination
register Rd . The condition code flags are optionally updated, based on the result.

Usage

ORR can be used to set selected bits in a register. For each bit, OR with 1 sets the
bit, and OR with 0 leaves it unchanged.

Condition Codes

The N and Z flags are set according to the result of the operation, and the C flag
is set to the carry output bit generated by the shifter (see 5.1 on page 51). The V
flag is unaffected.

Notes

If the destination register (Rd ) is the program counter (PC or R15 ), then the
current status register (CPSR) is restored from the saved status register (SPSR).
This form of the instruction is unpredictable if executed in User mode or System
mode, because these modes do not have an SPSR.

RSB

Reverse Subtract

Syntax

RSB(cid:104)cc(cid:105)(cid:104)S (cid:105) Rd , Rn, (cid:104)op1 (cid:105)

Operation

(cid:104)cc(cid:105):

Rd ← (cid:104)op1 (cid:105) − Rn
(cid:104)cc(cid:105)(cid:104)S (cid:105): CPSR ← ALU(Flags)

148

A.20 Reverse Subtract with Carry (RSC)

Description The RSB (Reverse Subtract) instruction subtracts the value of register Rn from the
value of (cid:104)op1 (cid:105), and stores the result in the destination register Rd . The condition
code flags are optionally updated, based on the result.

Usage

The following instruction stores the negation (two’s complement) of Rx in Rd :

RSB

Rd , Rx, #0

Constant multiplication (of Rx) by 2n − 1 (into Rd ) can be performed with:

RSB

Rd , Rx, Rx, LSL #n

Condition Codes

Notes

The N and Z flags are set according to the result of the subtraction, and the C and
V flags are set according to whether the subtraction generated a borrow (unsigned
underflow) and a signed overflow, respectively.

In other
The C flag is set if no borrow occurs, and clear if no borrow occurs.
words, the C flag is used as a borrow (not borrow) flag. This inversion of the
borrow condition is usually compensated for by subsequent instructions. The SBC
(subtract with carry) and RSC (Reverse Subtract with Carry) instructions use
the C flag as a borrow operand, performing a normal subtraction if C is set and
subtracting one more than usual if C is clear. The HS (unsigned higher or same)
and LO (unsigned lower) conditions are equivalent to CS (carry set) and CC (carry
clear) respectively.

If the destination register (Rd ) is the program counter (PC or R15 ), then the
current status register (CPSR) is restored from the saved status register (SPSR).
This form of the instruction is unpredictable if executed in User mode or System
mode, because these modes do not have an SPSR.

RSC

Reverse Subtract with Carry

Syntax

RSC(cid:104)cc(cid:105)(cid:104)S (cid:105) Rd , Rn, (cid:104)op1 (cid:105)

Operation

(cid:104)cc(cid:105):

Rd ← (cid:104)op1 (cid:105) − Rn − C

(cid:104)cc(cid:105)(cid:104)S (cid:105): CPSR ← ALU(Flags)

Description The RSC (Reverse Subtract with Carry) instruction subtracts the value of register
Rn and the value of Carry flag from the value of (cid:104)op1 (cid:105), and stores the result in the
destination register Rd . The condition code flags are optionally updated, based
on the result.

Usage

To negate the 64-bit value in R0 , R1 , use the following sequence (R0 holds the
least significant word) which stores the result in R2 , R3 :

RSBS
RSC

R2, R0, #0
R3, R1, #0

Condition Codes

Notes

The N and Z flags are set according to the result of the subtraction, and the C and
V flags are set according to whether the subtraction generated a borrow (unsigned
underflow) and a signed overflow, respectively.

The C flag is set if no borrow occurs, and clear if no borrow occurs.
In other
words, the C flag is used as a borrow (not borrow) flag. This inversion of the
borrow condition is usually compensated for by subsequent instructions. The SBC
(subtract with carry) instructions use the C flag as a borrow operand, performing
a normal subtraction if C is set and subtracting one more than usual if C is

A.21 Subtract with Carry (SBC)

149

clear. The HS (unsigned higher or same) and LO (unsigned lower) conditions are
equivalent to CS (carry set) and CC (carry clear) respectively.

If the destination register (Rd ) is the program counter (PC or R15 ), then the
current status register (CPSR) is restored from the saved status register (SPSR).
This form of the instruction is unpredictable if executed in User mode or System
mode, because these modes do not have an SPSR.

SBC

Subtract with Carry

Syntax

SBC(cid:104)cc(cid:105)(cid:104)S (cid:105) Rd , Rn, (cid:104)op1 (cid:105)

Operation

(cid:104)cc(cid:105):

Rd ← Rn − (cid:104)op1 (cid:105) − (C)

(cid:104)cc(cid:105)(cid:104)S (cid:105): CPSR ← ALU(Flags)

Description The SBC (Subtract with Carry) instruction is used to synthesise multi-word sub-
traction. SBC subtracts the value of (cid:104)op1 (cid:105) and the value of Carry flag (Not Carry)
from the value of register Rn, and stores the result in the destination register Rd .
The condition code flags are optionally updated, based on the result.

Usage

If register pairs R0 , R1 and R2 , R3 hold 64-bit values (R0 and R2 hold the least
significant words), the following instructions leave the 64-bit difference in R4 , R5 :

Condition Codes

SUBS
SBC

R4,R0,R2
R5,R1,R3

Notes

The N and Z flags are set according to the result of the subtraction, and the C and
V flags are set according to whether the subtraction generated a borrow (unsigned
underflow) and a signed overflow, respectively.

The C flag is set if a borrow occurs, and if clear no borrow occurs.
In other
words, the C flag is used as a borrow (not borrow) flag. This inversion of the
borrow condition is usually compensated for by subsequent instructions. The RSC
(reverse subtract with carry) instructions use the C flag as a borrow operand,
performing a normal subtraction if C is set and subtracting one more than usual
if C is clear. The HS (unsigned higher or same) and LO (unsigned lower) conditions
are equivalent to CS (carry set) and CC (carry clear) respectively.

If the destination register (Rd ) is the program counter (PC or R15 ), then the
current status register (CPSR) is restored from the saved status register (SPSR).
This form of the instruction is unpredictable if executed in User mode or System
mode, because these modes do not have an SPSR.

STM

Store Multiple

Syntax

STM(cid:104)cc(cid:105)(cid:104)mode(cid:105) Rn(cid:104)! (cid:105), {(cid:104)registers(cid:105)}

where (cid:104)mode(cid:105) is one of:

Increment Before
IB
DB Decrement Before

Increment After
IA
DA Decrement After

Operation

150

A.23 Store Register (STR)

if (cid:104)cc(cid:105)

IA: MAR ← Rn
IB: MAR ← Rn + 4
DA: MAR ← Rn − (#(cid:104)registers(cid:105) × 4) + 4
DB: MAR ← Rn − (#(cid:104)registers(cid:105) × 4)

for each register Ri in (cid:104)registers(cid:105)

IB:
DB:

IA:
DA:

MAR ← MAR + 4
MAR ← MAR − 4

M(MAR) ← Ri

MAR ← MAR + 4
MAR ← MAR − 4

(cid:104)! (cid:105):
end if

Rn ← MAR

Description The STM (Store Multiple) instruction stores a subset (or possibly all) of the general-

purpose registers to sequential memory locations.

The register Rn specifies the base register used to store the registers. Each register
given in Rregisters is stored in turn, storing each register in the next memory
address as directed by (cid:104)mode(cid:105).

If the base register writeback option ((cid:104)! (cid:105)) is specified, the base register (Rn) is
modified with the new base address.

(cid:104)registers(cid:105) is a list of registers, separated by commas and specifies the set of
registers to be stored. The registers are stored in sequence, the lowest-numbered
register to the lowest memory address, through to the highest-numbered register
to the highest memory address.

If R15 (PC) is specified in (cid:104)registers(cid:105), the value stored is unknown.

Exceptions Data Abort — This exception is generated if the address (Rn) is not word aligned,

or is not accessible in the current processor mode.

Usage

STM is useful as a block store instruction (combined with LDM it allows efficient
block copy) and for stack operations. A single STM used in the sequence of a
procedure can push the return address and general-purpose register values on to
the stack, updating the stack pointer in the process. (See chapter 14 for further
discussion.)

Condition Codes

The condition codes are not effected by this instruction.

Notes

The base register Rn should not be specified in (cid:104)registers(cid:105) when base register
writeback ((cid:104)! (cid:105)) is used.

R15 (PC) should not be used as the base register (Rn).

STR

Store Register

Syntax

STR(cid:104)cc(cid:105) Rd , (cid:104)op2 (cid:105)

Operation

(cid:104)cc(cid:105): M((cid:104)op2 (cid:105)) ← Rd

Description The STR (Store Register) instruction stores a word from register Rd to the memory

address calculated by (cid:104)op2 (cid:105).

A.24 Store Register Byte (STRB)

151

Exceptions Data Abort — This exception is generated if the address ((cid:104)op2 (cid:105)) is not word

aligned, or is not accessible in the current processor mode.

Usage

Combined with a suitable addressing mode, STR stores 32-bit data from a general-
purpose register into memory. Using the PC as the base register allows PC-relative
addressing, which facilitates position-independent code.

Condition Codes

The condition codes are not effected by this instruction.

Notes

Using the PC as the source register (Rd ) will cause an unknown value to be written.

If (cid:104)op2 (cid:105) specifies base register writeback (!), and the same register is specified for
Rd and Rn, the results are unpredictable.

STRB

Store Register Byte

Syntax

STR(cid:104)cc(cid:105)B Rd , (cid:104)op2 (cid:105)

Operation

(cid:104)cc(cid:105): M((cid:104)op2 (cid:105)) ← Rd (7:0)

Description The STRB (Store Register Byte) instruction stores a byte from the least significant

byte of register Rd to the memory address calculated by (cid:104)op2 (cid:105).

Exceptions Data Abort — This exception is generated if the address ((cid:104)op2 (cid:105)) is not accessible

in the current processor mode.

Usage

Combined with a suitable addressing mode, STRB writes the least significant byte
of a general-purpose register to memory. Using the PC as the base register allows
PC-relative addressing, which facilitates position-independent code.

Condition Codes

The condition codes are not effected by this instruction.

Notes

Specifying the PC as the source register (Rd ) is unpredictable.

If (cid:104)op2 (cid:105) specifies base register writeback (!), and the same register is specified for
Rd and Rn, the results are unpredictable.

SUB

Subtract

Syntax

SUB(cid:104)cc(cid:105)(cid:104)S (cid:105) Rd , Rn, (cid:104)op1 (cid:105)

Operation

(cid:104)cc(cid:105):

Rd ← Rn - (cid:104)op1 (cid:105)
(cid:104)cc(cid:105)(cid:104)S (cid:105): CPSR ← ALU(Flags)

Description Subtracts the value of (cid:104)op1 (cid:105) from the value of register Rn, and stores the result
in the destination register Rd . The condition code flags are optionally updated,
based on the result.

Usage

SUB is used to subtract one value from another to produce a third. To decrement
a register value (in Rx) use:

SUBS

Rx, Rx, #1

SUBS is useful for decrementing a loop counter, as the branch instruction can test
the (Z) flag for the appropriate termination condition, without the need for a
compare instruction:

CMP

Rx, #0 BNE Loop

152

A.26 Software Interrupt (SWI)

This both decrements the loop counter in Rx and branches back to the start of
the loop (the BEQ instruction) if it has not reached zero.

Condition Codes

Notes

The N and Z flags are set according to the result of the subtraction, and the C and
V flags are set according to whether the subtraction generated a borrow (unsigned
underflow) and a signed overflow, respectively.

The C flag is set if no borrow occurs, and if clear a borrow does occur. In other
words, the C flag is used as a borrow (not borrow) flag. This inversion of the
borrow condition is usually compensated for by subsequent instructions. The SBC
(Subtract with Carry) and RSC (Reverse Subtract with Carry) instructions use
the C flag as a borrow operand, performing a normal subtraction if C is set and
subtracting one more than usual if C is clear. The HS (unsigned higher or same)
and LO (unsigned lower) conditions are equivalent to CS (carry set) and CC (carry
clear) respectively.

If the destination register (Rd ) is the program counter (PC or R15 ), then the
current status register (CPSR) is restored from the saved status register (SPSR).
This form of the instruction is unpredictable if executed in User mode or System
mode, because these modes do not have an SPSR.

SWI

Software Interrupt

Syntax

SWI(cid:104)cc(cid:105) (cid:104)value(cid:105)

Operation

SPSR_svc ← CPSR

R14_svc ← PC + 8

(cid:104)cc(cid:105):
(cid:104)cc(cid:105):
(cid:104)cc(cid:105): CPSR(mode) ← Supervisor
(cid:104)cc(cid:105):
(cid:104)cc(cid:105):

PC ← 0x00000008

CPSR(I) ← 1 (Disable Interrupts)

Description Causes a SWI exception (see 3.4 on page 29).

Exceptions

Software interrupt

Usage

The SWI instruction is used as an operating system service call. The method used
to select which operating system service is required is specified by the operating
system, and the SWI exception handler for the operating system determines and
provides the requested service.

One of two methods are used: First, the (cid:104)value(cid:105) can specify which service is
required, and any parameters needed by the selected service are passed in general-
purpose registers. Alternatively, the (cid:104)value(cid:105) is ignored, and the register R0 is used
to select which service is required, any parameters are passed in other registers.

Condition Codes

The flags will be effected by the operation of the software interrupt.
It is not
possible to say how they will be effected. The status of the condition code flags is
unknown after a software interrupt is unknown, unless specified by the operating
system.

A.27 Swap (SWP)

SWP

Swap

Syntax

SWP(cid:104)cc(cid:105) Rd , Rm, [Rn]

Operation

(cid:104)cc(cid:105): ALU(0) ← M(Rn)
(cid:104)cc(cid:105): M(Rn) ← Rm
(cid:104)cc(cid:105):

Rd ← ALU(0)

153

Description SWP (swap) will swap a word between registers and memory. It loads a word from
the memory address given by the value of register Rn. The value of register Rm is
then stored in the memory address given by the value of Rn, and the original value
is loaded into register Rd . If the same register is specified for Rd and Rm, this
instruction swaps the value of the register and the value at the memory address.

Exceptions Data Abort — This exception is generated if the instruction attempts to access
a part of memory which has been reserved for privileged mode access while
the system is in user mode.

If a data abort occurs during the load phase, the store access does not occur.

Usage

The SWP instruction can be used to implement semaphores. Where one process
can send a message to another processes running on the same processor, or on a
separate linked processor.

Condition Codes

The condition codes are not effected by this instruction.

Notes

If the address contained in Rn is non word-aligned the system will attempt an
access, but the effect is unpredictable.

If the PC is specified as the destination (Rd ), address (Rn) or the value (Rm), the
result is also unpredictable.

If the same register is specified as Rn and Rm, or Rn and Rd , the result is unpre-
dictable.

SWPB

Swap Byte

Syntax

SWP(cid:104)cc(cid:105)B Rd , Rm, [Rn]

Operation

MBR ← M(Rn)(0:7)

(cid:104)cc(cid:105):
(cid:104)cc(cid:105): Rd (0:7) ← MBR
(cid:104)cc(cid:105): Rd (8:31) ← 0
(cid:104)cc(cid:105): M(Rn) ← Rm(7:0)

Description The SWPB (swap byte) instruction swaps a byte between registers and memory.
It loads a byte from the memory address given by the value of register Rn. The
value of the least significant byte of register Rm is stored to the memory address
given by Rn, the original value is zero-extended into a 32-bit word, and the word
is written to register Rd . If the same register is specified for Rd and Rm, this
instruction swaps the value of the least significant byte of the register and the
byte value at the memory address.

Exceptions Data Abort — This exception is generated if the instruction attempts to access
a part of memory which has been reserved for privileged mode access while
the system is in user mode.

If a data abort occurs during the load phase, the store access does not occur.

154

Usage

The SWPB instruction can be used to implement semaphores, in a similar manner
to that shown for the SWP instruction.

A.30 Test (TST)

Condition Codes

The condition codes are not effected by this instruction.

Notes

If the PC is specified for Rd , Rn, or Rm, the result is unpredictable.

If the same register is specified as Rn and Rm, or Rn and Rd , the result also
unpredictable.

TEQ

Test Equivalence

Syntax

TEQ(cid:104)cc(cid:105) Rn, (cid:104)op1 (cid:105)

Operation

(cid:104)cc(cid:105): ALU ← Rn ⊕ (cid:104)op1 (cid:105)
(cid:104)cc(cid:105): CPSR ← ALU(flags)

Description The TEQ (Test Equivalence) instruction compares a register value (Rn)with another
arithmetic value ((cid:104)op1 (cid:105)). The condition flags are updated, based on the result of
logically exclusive-ORing the two values, so that subsequent instructions can be
conditionally executed.

Usage

The TEQ instruction is used to test if two values are equal, without affecting the
V flag, as CMP (compaire) does. The C flag is also unaffected in many cases.
TEQ is also useful for testing whether two values have the same sign. After the
comparison, the N flag is the logical Exclusive OR of the sign bits of the two
operands.

Notes

If the PC is specified for Rd , Rn, or Rm, the result is unpredictable.

If the same register is specified as Rn and Rm, or Rn and Rd , the result also
unpredictable.

TST

Test

Syntax

TST(cid:104)cc(cid:105) Rn, (cid:104)op1 (cid:105)

Operation

(cid:104)cc(cid:105): ALU ← Rn ∧ (cid:104)op1 (cid:105)
(cid:104)cc(cid:105): CPSR ← ALU(flags)

Description The TST (Test) instruction compares a register value (Rn with another arithmetic
value ((cid:104)op1 (cid:105)). The condition flags are updated, based on the result of logically
ANDing the two values, so that subsequent instructions can be conditionally ex-
ecuted.

Usage

TST is used to determine whether a particular subset of register bits includes at
least one set bit. A very common use for TST is to test whether a single bit is set
or clear.

Condition Codes

The N and Z flags are set according to the result of the operation, and the C flag
is set to the carry output bit generated by the shifter (see 5.1 on page 51). The V
flag is unaffected.

B ARM Instruction Summary

cc: Condition Codes

Generic

Unsigned

Signed

CS Carry Set
CC Carry Clear
EQ Equal (Zero Set)
NE Not Equal (Zero Clear)
VS Overflow Set
VC Overflow Clear

Higer Than

HI
HS Higer or Same
LO Lower Than
LS

Lower Than or Same

GT Greater Than
GE Greater Than or Equal
LT
LE
MI Minus (Negative)
PL

Less Than
Less Than or Equal

Plus (Positive)

ARM Instructions

Add with Carry
Add
Bitwise AND
Branch
Branch and Link

ADC(cid:104)cc(cid:105)(cid:104)S (cid:105) Rd , Rn, (cid:104)op1 (cid:105)
ADD(cid:104)cc(cid:105)(cid:104)S (cid:105) Rd , Rn, (cid:104)op1 (cid:105)
AND(cid:104)cc(cid:105)(cid:104)S (cid:105) Rd , Rn, (cid:104)op1 (cid:105)
B(cid:104)cc(cid:105)
BL(cid:104)cc(cid:105)

(cid:104)offset(cid:105)
(cid:104)offset(cid:105)

Compare
Exclusive OR
Load Register
Load Register Byte

Move
Move Negative
Bitwise OR
Subtract with Carry
Store Register
Store Register Byte
Subtract
Software Interrupt
Swap

Rn, (cid:104)op1 (cid:105)

CMP(cid:104)cc(cid:105)
EOR(cid:104)cc(cid:105)(cid:104)S (cid:105) Rd , Rn, (cid:104)op1 (cid:105)
LDR(cid:104)cc(cid:105)
LDR(cid:104)cc(cid:105)B

Rd , (cid:104)op2 (cid:105)
Rd , (cid:104)op2 (cid:105)

MOV(cid:104)cc(cid:105)(cid:104)S (cid:105) Rd , (cid:104)op1 (cid:105)
MVN(cid:104)cc(cid:105)(cid:104)S (cid:105) Rd , (cid:104)op1 (cid:105)
ORR(cid:104)cc(cid:105)(cid:104)S (cid:105) Rd , Rn, (cid:104)op1 (cid:105)
SBC(cid:104)cc(cid:105)(cid:104)S (cid:105) Rd , Rn, (cid:104)op1 (cid:105)
STR(cid:104)cc(cid:105)
Rd , (cid:104)op2 (cid:105)
STR(cid:104)cc(cid:105)(cid:104)S (cid:105) Rd , (cid:104)op2 (cid:105)
SUB(cid:104)cc(cid:105)(cid:104)S (cid:105) Rd , Rn, (cid:104)op1 (cid:105)
SWI(cid:104)cc(cid:105)
SWP(cid:104)cc(cid:105)

(cid:104)value(cid:105)
Rd , Rm, [Rn]

Swap Byte

SWP(cid:104)cc(cid:105)B

Rd , Rm, [Rn]

155

← Rn + (cid:104)op1 (cid:105) + CPSR(C)
(cid:104)cc(cid:105): Rd
← Rn + (cid:104)op1 (cid:105)
(cid:104)cc(cid:105): Rd
← Rn & (cid:104)op1 (cid:105)
(cid:104)cc(cid:105): Rd
← PC + (cid:104)offset(cid:105)
(cid:104)cc(cid:105): PC
← PC + 8
(cid:104)cc(cid:105): LR
← PC + (cid:104)offset(cid:105)
(cid:104)cc(cid:105): PC
← (Rn - (cid:104)op1 (cid:105))
(cid:104)cc(cid:105): CSPR
← Rn ⊕ (cid:104)op1 (cid:105)
(cid:104)cc(cid:105): Rd
(cid:104)cc(cid:105): Rd
← M((cid:104)op2 (cid:105))
(cid:104)cc(cid:105): Rd (7:0) ← M((cid:104)op2 (cid:105))
(cid:104)cc(cid:105): Rd (31:8) ← 0
(cid:104)cc(cid:105): Rd
(cid:104)cc(cid:105): Rd
(cid:104)cc(cid:105): Rd
(cid:104)cc(cid:105): Rd
(cid:104)cc(cid:105): M((cid:104)op2 (cid:105)) ← Rd
(cid:104)cc(cid:105): M((cid:104)op2 (cid:105)) ← Rd (7:0)
(cid:104)cc(cid:105): Rd

← (cid:104)op1 (cid:105)
← (cid:104)op1 (cid:105)
← Rn | (cid:104)op1 (cid:105)
← Rn - (cid:104)op1 (cid:105) - CPSR(C)

← Rn - (cid:104)op1 (cid:105)

(cid:104)cc(cid:105): Rd
(cid:104)cc(cid:105): M(Rn)
(cid:104)cc(cid:105): Rd (7:0) ← M(Rn)(7:0)
(cid:104)cc(cid:105): M(Rn)(7:0) ← Rm(7:0)

← M(Rn)
← Rm

156

APPENDIX B. ARM INSTRUCTION SUMMARY

op1: Data Access

Immediate

Register

#(cid:104)value(cid:105)

Rm

(cid:104)op1 (cid:105) ← IR(value)

(cid:104)op1 (cid:105) ← Rm

Logical Shift Left Immediate
Logical Shift Left Register

Rm, LSL #(cid:104)value(cid:105)
Rm, LSL Rs

(cid:104)op1 (cid:105) ← Rm << IR(value)
(cid:104)op1 (cid:105) ← Rm << Rs(7:0)

Logical Shift Right Immediate
Logical Shift Right Register

Rm, LSR #(cid:104)value(cid:105)
Rm, LSR Rs

(cid:104)op1 (cid:105) ← Rm >> IR(value)
(cid:104)op1 (cid:105) ← Rm >> Rs(7:0)

Arithmetic Shift Right Immediate
Arithmetic Shift Right Register

Rm, ASR #(cid:104)value(cid:105)
Rm, ASR Rs

(cid:104)op1 (cid:105) ← Rm +>> IR(value)
(cid:104)op1 (cid:105) ← Rm +>> Rs(7:0)

Rotate Right Immediate
Rotate Right Register

Rm, ROR #(cid:104)value(cid:105)
Rm, ROR Rs

(cid:104)op1 (cid:105) ← Rm >>> (cid:104)value(cid:105)
(cid:104)op1 (cid:105) ← Rm >>> Rs(4:0)

Rotate Right with Extend

Rm, RRX

(cid:104)op1 (cid:105) ← C >>> Rm >>> C

op2: Memory Access

Immediate Offset
Register Offset
Scaled Register Offset

[Rn, #±(cid:104)value(cid:105)]
[Rn, Rm]
[(cid:104)Rn(cid:105), Rm, (cid:104)shift(cid:105) #(cid:104)value(cid:105)]

(cid:104)op2 (cid:105) ← Rn + IR(value)
(cid:104)op2 (cid:105) ← Rn + Rm
(cid:104)op2 (cid:105) ← Rn + (Rm shift IR(value))

Immediate Pre-indexed

[Rn, #±(cid:104)value(cid:105)]!

Register Pre-indexed

[Rn, Rm]!

Scaled Register Pre-indexed

[Rn, Rm, (cid:104)shift(cid:105) #(cid:104)value(cid:105)]!

(cid:104)op2 (cid:105) ← Rn + IR(value)
Rn ← (cid:104)op2 (cid:105)
(cid:104)op2 (cid:105) ← Rn + Rm
Rn ← (cid:104)op2 (cid:105)
(cid:104)op2 (cid:105) ← Rn + (Rm shift IR(value))
Rn ← (cid:104)op2 (cid:105)

Immediate Post-indexed

[Rn], #±(cid:104)value(cid:105)

Register Post-indexed

[Rn], Rm

Scaled Register Post-indexed

[Rn], Rm, (cid:104)shift(cid:105) #(cid:104)value(cid:105)

(cid:104)op2 (cid:105) ← Rn
Rn ← Rn + IR(value)
(cid:104)op2 (cid:105) ← Rn
Rn ← Rn + Rm
(cid:104)op2 (cid:105) ← Rn
Rn ← Rn + Rm shift IR(value)

Where (cid:104)shift(cid:105) is one of: LSL, LSR, ASR, ROR or RRX and has the same effect as for (cid:104)op1 (cid:105)

Index

Addressing Mode, 51–58

Arithmetic Shift Right (ASR), 52
Immediate, 51
Logical Shift Left (LSL), 51
Logical Shift Right (LSR), 52
Offset, 55
Post-Index, 57
Pre-Index, 56
Register, 51
Rotate Right (ROR), 53
Rotate Right Extended (RRX), 54

Arithmetic and Logic Unit (ALU), 31–33

Barrel Shifter, 33
Booth Multiplier, 32

Bit Field, 30

Characters

ASCII, 91
International, 94
Unicode, 94

Complex Instruction Set Computer (CISC), 33, 41
Condition Codes, 28–29, 42–43

Carry Flag, 28
Mnemonics, 42
Negative Flag, 28
Overflow Flag, 29
Zero Flag, 28

Conditional Execution, 30

Exceptions, 29–30

Data Abort, 29
Fast Interrupt, 29
Interrupt, 29
Prefetch Abort, 29
Reset, 29
Software Interrupt, 29
Undefined, 29

Fetch/Execute Cycle, see Instruction Pipeline

Instruction Pipeline, 33–37

Execute, 36
Instruction Decode, 35
Instruction Fetch, 34
Operand Fetch, 36
Operand Store, 36

Instructions, 137–154
ADC, 43, 137

ADD, 43, 138
AND, 47, 139
B, BL, 45, 139
BAL, see B, BL
BIC, 47, 140
BLAL, see B, BL
CDP, 49
CMN, 45, 140
CMP, 44, 141
EOR, 47, 141
LDC, 49
LDM, 46, 141
LDRB, 46, 143
LDR, 46, 142
MCR, 49
MLA, 43, 143
MOV, 41, 144
MRC, 49
MRS, 48, 144
MSR, 48, 145
MUL, 43, 146
MVN, 47, 147
ORR, 47, 147
RSB, 43, 147
RSC, 43, 148
SBC, 43, 149
STC, 49
STM, 46, 149
STRB, 46, 151
STR, 46, 150
SUB, 43, 151
SWI, 48, 152
SWPB, 48, 153
SWP, 48, 153
TEQ, 45, 154
TST, 45, 154

Programs

add.s, 67
add2.s, 67–68
add64.s, 71, 76–77, 111–112, 131–132
addbcd.s, 112–113
bigger.s, 70, 75
bigsum.s, 83
byreg.s, 129
bystack.s, 129–131
cntneg1.s, 84
cntneg2.s, 84–85
cstrcmp.s, 98

157

158

INDEX

Rotate Right Extended (RRX), 54

Strings, 92–94

Counted, 93
Fixed Length, 93
Terminated, 93

dectonib.s, 105
divide.s, 114–115
factorial.s, 72–73, 77–78, 132–133
halftobin.s, 107
head.s, 119–120
init1.s, 127
init2.s, 127–128
init3.s, 128
init3a.s, 128–129
insert.s, 117–118
insert2.s, 118
invert.s, 66
largest.s, 85–86
move16.s, 65
mul16.s, 113
mul32.s, 113–114
nibble.s, 69
nibtohex.s, 103–104
nibtoseg.s, 104–105
normal1.s, 86–87
normal2.s, 87
padzeros.s, 96–97
search.s, 118–119
setparity.s, 97–98
shiftleft.s, 68
skipblanks.s, 95–96
sort.s, 120
strcmp.s, 98–99
strlen.s, 95
strlencr.s, 94–95
sum1.s, 82
sum2.s, 83
ubcdtohalf.s, 106
ubcdtohalf2.s, 106–107
wordtohex.s, 104

Reduced Instruction Set Computer (RISC), 23, 34,

39

Register Transfer Language, 30–33

Arithmetic and Logic Unit, 31–33

Barrel Shifter, 33
Booth Multiplier, 32

Bit Field, 30
Guard, 30
Memory Access, 31
Named Field, 30

Registers, 25–28

General Purpose (R0 –R12 ), 25
Instruction Register (IR), 33, 35–36
Link Register (LR/R14 ), 27, 45
Program Counter (PC/R15 ), 27
Stack Pointer (SP/R13 ), 26
Status Register (CPSR/SPSR), 28

Shift

Arithmetic Shift Right (ASR), 52
Logical Shift Left (LSL), 51
Logical Shift Right (LSR), 52
Rotate Right (ROR), 53