1 Chapter 1: Basic Concepts Assembly Language for Intel-Based Computers, 4th edition Kip R. Irvine.

1

Chapter 1: Basic Concepts

Assembly Language for

Intel-Based Computers,4th edition

Kip R. Irvine

Introduction

1.1 Context of assembly language 1.2 Virtual Machine Concept 1.3 Data representation

– 1.3.1 Binary numbers– 1.3.2 Hexadecimal numbers– 1.3.3 Arithmetic

A. Introducing assembly language

1.1 Context of assembly language:

Why study assembly language (ASM)? To learn how high-level language (HLL)

code gets translated into machine language

Assembly language: machine-specific language with a one-to-one correspondence with the machine language for that computer

Machine language: The language a particular processor understands

1.1 Context of assembly language:

ExampleHigh-Level Language: 10 + 20;

Machine language: B8 000A83 C014

MOV AX, 10ADD AX, 20

Assembly Language:

Immediate operand

op code

To learn the computer (i.e., architecture, hardware, operating system…)

by direct access to memory, video controller, sound card, keyboard…

by communicating with the operating system

To speed up applications

direct access to hardware (ex: writing directly to I/O ports instead of doing a system call)

good ASM code is faster and smaller

Avoid restrictions of high level languages

1.1 Context of assembly language:

Why study assembly language (ASM)?

1.1 Context of assembly language:

Problems with assembly language

Provides no structure Is not portable Applications can be very long “Hard” to read and understand Lots of detail required

1.2 Virtual Machine ConceptVirtual Machines

Tanenbaum: Virtual machine concept Programming Language analogy:

– Each computer has a native machine language (language L0) that runs directly on its hardware

– A more human-friendly language is usually constructed above machine language, called Language L1

• Programs written in L1 can run two different ways:• Interpretation – L0 program interprets and executes L1

instructions one by one• Translation – L1 program is completely translated into an L0

program, which then runs on the computer hardware

1.2 Virtual Machine ConceptTranslating Languages

English: Display the sum of A times B plus C.

C++: cout << (A * B + C);

Assembly Language:

mov eax,Amul Badd eax,Ccall WriteInt

Intel Machine Language:

A1 00000000

F7 25 00000004

03 05 00000008

E8 00500000

1.2 Virtual Machine ConceptSpecific Machine Levels

(descriptions of individual levels follow . . . )

1.2 Virtual Machine ConceptHigh-Level Language

Level 5 Application-oriented languages

– C++, Java, Pascal, Visual Basic . . . Programs compile into assembly

language (Level 4)

1.2 Virtual Machine ConceptAssembly Language

Level 4 Instruction mnemonics that have a one-to-

one correspondence to machine language Calls functions written at the operating

system level (Level 3) Programs are translated into machine

language (Level 2)

1.2 Virtual Machine ConceptOperating System

Level 3 Provides services to Level 4 programs Translated and run at the instruction set

architecture level (Level 2)

1.2 Virtual Machine ConceptInstruction Set Architecture

Level 2 Also known as conventional

machine language Executed by Level 1

(microarchitecture) program

1.2 Virtual Machine ConceptMicroarchitecture

Level 1 Interprets conventional machine

instructions (Level 2) Executed by digital hardware

(Level 0)

1.2 Virtual Machine ConceptDigital Logic

Level 0 CPU, constructed from digital logic gates System bus Memory Implemented using bipolar transistors

1.3 Data representation:

Binary Numbers Digital computers store information in binary Binary allows two states

– On yes true 1– Off no false 0

Each digit in a binary number is called a bit– MSB – most significant bit– LSB – least significant bit

0 1 1 0 1 0 0 0MSB

bit

LSB

1.3 Data representation:

Data types byte: 8 bits:

10010101

word: 16 bits or __ bytes:

1110101101100010

double word: __ bits or __ bytes :

11010110111001011101011011001011 quad word: __ bits or __ bytes:

0101111011010101110101001100101111110110110001011101011011001101

2

32 4

64 8

1.3 Data representation:

Range of a unsigned numberQuestion: What is the range of numbers that

can be represented with ?

121~0 1 123~0 2

1215~0 4 )12(~0 n

2 bits?

4 bits?

1 bit?

n bit?

1.3 Data representation:Integer Storage Sizes

Standard sizes:

1.3 Data representation:

Numbering Systems A number system of base n is a system that uses distinct symbols for n digits

A written number is meaningful only with respect to a base

1.3 Data representation:

Numbering Systems

To tell the assembler which base we use:

–Hexadecimal 25 is written as 25h

–Octal 25 is written as 25o or 25q

–Binary 1010 is written as 1010b

–Decimal 1010 is written as 1010 or 1010d

1.3 Data representation:

Number systems (bases) Number systems used

– Binary: The internal representation inside the computer. They may be represented in binary, octal, or hexadecimal.

– Decimal: The system people use. ASCII representations of numbers used for I/O:

– ASCII binary– ASCII octal– ASCII decimal– ASCII hexadecimal

Format Value

ASCII binary "01000001"

ASCII decimal "65"

ASCII hexadecimal "41"

ASCII octal "101"

1.3 Data representation:

Number systems (bases) Example

1.3 Binary numbers:Addition & Multiplication

Binary multiplication * 0 1 0 0 0 1 0 1

Binary addition table + 0 1 0 0 1 1 1 10

1.3 Binary numbers:Addition & Multiplication Examples:

1001100b 1110b + 1101011b * 1101b

10110101b 111101b + 1101101b * 1111b

10110111 10110110

100100010 1110010011

1.3 Hexadecimal numbers:Addition & Multiplication Hex addition and multiplication tables are

largeWe can still do simple calculations by hand B852h 23Ah + 5A65h * 100h

ABCh 2B3h + E F 0 h * 102h

1.3 Arithmetic:

Converting to decimal Binary to Decimal 1010b =1 * 23 + 0 * 22 + 1 * 21 + 0*20

=10

Hex to Decimal ABCh = 10 * 162 + 11 * 161+ 12 * 160

= 10 * 256 + 11 * 16 + 12 * 1 = 2560 + 176 + 12

= 2748

1.3 Arithmetic:

Converting to decimal Octal to Decimal

256o = 2 * 82 + 5 * 81+ 6 * 80 =174

Conclusion:(Number)n

= (multiply each digit by an integer power of n)

1.3 Arithmetic:

Conversion problems

1234 base 5

or (1234)5=_________

111010b = ________ 45677h = ________ 736.4o = ________

1.3 Arithmetic:

Conversion from decimal Decimal integer to Binary

54 = 27 * 2+ 0

27 = 13 * 2 + 1

13 = 6 * 2 +1

1 1 0 1 1 0

remainderquotient

1 = 0 * 2 + 1

3 = 1 * 2 + 1

6 = 3 * 2 + 0

1.3 Arithmetic:

Conversion from decimal Decimal integer to Hex :

Decimal integer to octal

54 = 6 * 8+ 6

6 = 0 * 8 + 6

6 6 o

54= 3 * 16 + 63 = 0 * 16 + 3

3 6 h

Decimal fraction to Binary (Octal, Hex)

1.3 Arithmetic:

Conversion from decimal

(0. 6 8 7 5)102X1 . 3 7 5 0

2X

1 . 0 0 0 0

0 . 7 5 0 02X

1 . 5 0 0 02X

= (0.1011)2

0.1 0 1 1

integer

1.3 Arithmetic:

Conversion from decimal Conclusion

a decimal integer to a base n representation successive division by n accumulation of the remainders

a decimal fraction to a base n representation successive multiplication by n accumulation of the integer digits

1.3 Arithmetic:

Conversion problems

(1234)10=(________)2

(41.6875)10=(________)2

(1234)10=(________)8

(1234)10=(________)16

1.3 Arithmetic: Conversions: hex, octal and binary Conversion between base 2 and base 8 = 23

111010b = 111 010 = 72o Conversion between base 2 and base 16 = 24

111010b = 11 1010 = 3Ah Note: we often write binary numbers in hex:

– binary numbers can be very long– it is easy to convert back and forth

1.3 Arithmetic:

Conversion problems Write ABCh in binary.

Write 101110100010100b in hex.

Write 101110100010100b in octal

1.3 Arithmetic:

Signed numbers Signed numbers: The number of bits must be

fixed. In every case, the highest bit is the sign bit.

0 means +

1 means -

Example:

Unsigned number: 1101b = 13

Signed number: 1101b = ?

1.3 Arithmetic:

Two’s complement 1101 b (signed binary number)

0010 b (reverse all bits)

+ 1 0011 b (unsigned value = 3)

- 3d =

One’s complement

Two’s complement

two’s complement of n = NOT(n) +1

reversible

1.3 Arithmetic:

Practice problems Find the 8 bit two’s complement

representation of -2Ch

Find the 8 bit two’s complement representation of 2Ch

What is decimal representation of the two’s complement number 11110100?

1.3 Arithmetic:

Range of signed numbers Two's complement: Sizes

Signed byte: -128 to +127 = 27 - 1 Signed word: -32,768 to +32,767 = 215 - 1Singed double word: -2,147,483,648 to +2,147,483,647 = 231 - 1

Comments:

• There is one more negative than positive value

• There are (about) half as many positive signed values as unsigned values

1.3 Arithmetic:

Addition & Subtraction (signed) Algorithm to add in signed decimal (sign and magnitude)

If both numbers have the same sign Add the two magnitudes Prefix with the common signelse Subtract the number with smaller magnitude from the number with the larger magnitude Prefix with the sign of the number with the larger magnitude.

1.3 Arithmetic:

Addition & Subtraction (signed)Two's complement:

1) add two numbers including their sign bits

2) discard any carry out of the sign position

Addition

+ 6 0 0 0 0 0 1 1 0+ 13 0 0 0 0 1 1 0 1

+ 19 0 0 0 1 0 0 1 1

Example:

- 19 1 1 1 0 1 1 0 1

- 6 1 1 1 1 1 0 1 0

- 13 1 1 1 1 0 0 1 1

1.3 Arithmetic:

Addition & Subtraction (signed)

Example:

6 0 0 0 0 0 1 1 0- 13

- 7 1 1 1 1 1 0 0 1

Subtraction

1) Take the 2’s complement of the subtrahend (including the sign bit)

2) Add it to the minuend (including the sign bit)

- 6 13 0 0 0 0 1 1 0 1

+ 7 0 0 0 0 0 1 1 1

1 1 1 1 0 0 1 1 1 1 1 1 1 0 1 0

1.3 Arithmetic:

Character Storage Characters are stored as numbers Coding schemes

– ASCII (7 or 8 bit)– EBCDIC (8 bit)– Unicode (16 bit)

ASCII Examples:“A” = 65d = 41h “a” = 97d = 61h“B” = 66d = 42h “b” = 98d = 62h“0” = 48d = 30h “1” = 49d = 31h“ “ = 32d = 20h “*” = 42d = 2Ah

A. Introducing Assembly Language:

MASM Installation

Insert CD in CD-Rom Go to Desktop, select My Computer Select CD-Rom directory (\IRVINE) Double click “index.html” On the new screen, follow the links, install

MASM615, sample program, link library etc.

A. Introducing Assembly Language:

A simple Debug program

MOV AX, 10 ADD AX, 20 SUB AX, 5 MOV [130], AX INT 20

Store 10 into register AX

Terminate program

Assembler: converts programs from assembly language to machine language

Store AX in memory location 130h

A. Introducing Assembly Language:

Running a Debug program Select Start | Programs (| Accessories) |MS-DOS (Command) Prompt Type Debug Type A 100, type the program, press return Type R (registers) to display the registers Type T (trace) as needed to step through program Type G (go) to execute the rest of program Type D 130 131 to display locations 130 and 131 Type Q (quit) to exit Debug

A. Introducing Assembly Language:

Debug register display

Next instruction

Contents of AX (hex) -R

AX=0030 BX=0000 CX=0000 SP=FFEE ...DS=1FA6 ES=1FA6 SS=1FA6 CS=0106 ...1FA6:0106 2D0500 SUB AX, 0005

A. Introducing Assembly Language:

Debug memory display

Contents oflocations 130 and 131.Namely 002Bh

-D 130 1311FA6:0130 2B 00

Note: The individual bytes in binary are reversed

when stored in memory

Example: 234567 would be stored in memory as 674523

A. Introducing Assembly Language:

Debug disassembly -U 100

1FA6:0100 B81000 MOV AX, 00101FA6:0103 052000 ADD AX, 00201FA6:0106 2D0500 SUB AX, 00051FA6:0109 A33001 MOV [0130],AX1FA6:010C CD20 INT 201FA6:010E 1F POP DS

Memory locations

Machinecode

Disassembled assembly code