kaizen cso002 l1

CX323L1

COMPUTER SYSTEM ORGANIZATION

COMPUTER SYSTEM ORGANIZATION

Prelim Midterm Finals

Kaizen 2006 - 2007

TERMS

PRELIM

Week 1 – Microprocessor

Week 2 – Basic Structure of hardware and Software

Week 3 – Register

Week 4 – System Operation

Week 5 - Editor

Microprocessors

Topics:

1.Microprocessor

2.History of Microprocessors

3.Types of Microprocessors

4.Number System

* Decimal

* Binary

* Octal

* Hexadecimal

Objectives

•To define what is a microprocessor.• To discuss the history of microprocessor.• Identify the types of microprocessor• To recall numbers systems.

Microprocessors

The brain or engine of the PC is the processor (sometimes called microprocessor), or central processing unit (CPU). The CPU performs the system's calculating and processing.

Brief History• Intel is generally credited with creating the first microprocessor in 1971 with the introduction of a chip called the 4004.

• All PC-compatible systems use either Intel processors or Intel-compatible processors from a handful of competitors

Example: AMD or Cyrix.

•Late 1970s the two most popular processors for PCs were not from Intel (although one was a clone of an Intel processor). Personal computers of that time primarily used the Z-80 by Zilog and the 6502 by MOS Technologies.

• Z-80 was noted for being an improved and less expensive clone of the Intel 8080 processor, similar to the way companies today such as AMD, Cyrix, IDT, and Rise Technologies have cloned Intel's Pentium

processors.

Types of Microprocessor

Processors can broadly be divided into the categories of: CISC, RISC, hybrid, and special purpose.

Complex Instruction Set Computers (CISC)

have a large instruction set, with hardware support for a wide variety of operations. In scientific, engineering, and mathematical operations with hand coded assembly language (and some business applications with hand coded assembly language), CISC processors usually perform the most work in the shortest time.

Reduced Instruction Set Computers (RISC)

have a small, compact instruction set. In most business applications and in programs created by compilers from high level language source, RISC processors usually perform the most work in the shortest time.

Hybrid processors are some combination of CISC and RISC approaches, attempting to balance the advantages of each approach.

Special purpose processors are optimized to perform specific functions. Digital signal processors and various kinds of co-processors are the most common kinds of special purpose processors.

Function of Microprocessor

• To coordinate the functioning of the various components of the system by generating and responding to control signals and to perform various arithmetic and logic functions.

Number Systems

Decimal, Binary, Octal and Hexadecimal Numeric Systems.

Conversions among the different systems

Decimal Numbering systems

Base: 10

Digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9

Example: 523410

103 = 1000 102 = 100 101 = 10 100 = 1 5 2 3 4

5,234 = 5 x 1000 + 2 x 100 + 3 x 10 + 4 x 1

Binary Numbering systems

• Base: 2

• Digits: 0, 1

• binary number: 1101012

positional powers of 2: 25 24 23 22 21 20

decimal positional value: 32 16 8 4 2 1

binary number: 1 1 0 1 0 1

Binary to Decimal Conversion

• To convert to base 10, add all the values where a one digit occurs.

Ex: 1101012


decimal positional value: 32 16 8 4 2 1

binary number: 1 1 0 1 0 1

32 + 16 + 4 + 1 = 5310

Ex: 1010112


decimal positional value: binary number:

Ex: 1010112


decimal positional value: 32 16 8 4 2 1binary number: 1 0 1 0 1 1

32 + 8 + 2 + 1 = 4310

The Division Method. Divide by 2 until you reach zero, and then collect the remainders in reverse. Ex 1: 5610 = 1110002

2 ) 56 Rem:2 ) 28 02 ) 14 02 ) 7 02 ) 3 12 ) 1 1 0 1

Ex 2: 3510 =

2 ) Rem:2 ) 2 ) 2 ) 2 ) 2 )

Answer: 3510 = 2

Character Representation

Ex: Find the binary ASCII and decimal ASCII values for the ‘&’ character.

Rightmost Leftmost Three BitsFour Bits 000 001 010 011 100 101 110 1110000 NUL DLE Space 0 @ P ` p0001 SOH DC1 ! 1 A Q a q0010 STX DC2 " 2 B R b r0011 ETX DC3 # 3 C S c s0100 EOT DC4 $ 4 D T d t0101 ENQ NAK % 5 E U e u0110 ACK SYN && 6 F V f v0111 BEL ETB ' 7 G W g w1000 BS CAN ( 8 H X h x1001 HT EM ) 9 I Y I y1010 LF SUB * : J Z j z1011 VT ESC + ; K [ k {1100 FF FS , < L \ l |1101 CR GS - = M ] m } 1110 SO RS . > N ^ n ~1111 SI US / ? O _ o DEL

Character Representation ASCII Table

From the chart:

‘&’ = 0100110 (binary ASCII value)

Convert the binary value to decimal:

01001102 = 32 + 4 + 2 = 3810

Therefore:

‘&’ = 38 (decimal ASCII value)

Octal Numbering systems

• Base: 8• Digits: 0, 1, 2, 3, 4, 5, 6, 7• Octal number: 12468

powers of : 84 83 82 81 80

decimal value: 4096 512 64 8 1Octal number: 1 2 4 6

Octal to Decimal Conversion

• To convert to base 10, beginning with the rightmost digit multiply each nth digit by 8(n-1), and add all of the results together.

Ex: 12468

positional powers of 8: 83 82 81 80

decimal positional value: 512 64 8 1Octal number: 1 2 4 6

512 + 128 + 32 + 6 = 67810

Ex: 103528

positional powers of 8: 84 83 82 81 80

decimal positional value: Octal number:

The Division Method. Divide by 8 until you reach zero, and then collect the remainders in reverse. Ex 1: 433010 = 103528

8 ) 4330 Rem:8 ) 541 28 ) 67 58 ) 8 38 ) 1 0

0 1

Ex 2: 81010 =

8 ) 810 Rem:8 ) 8 ) 8 )

Answer: 81010 = 8

Hexadecimal Numbering systems

• Base: 16• Digits: 0, 1, 2, 3, 4, 5, 6, 7,8,9,A,B,C,D,E,F

• Hexadecimal number: 1F416

powers of : 164 163 162 161 160

decimal value: 65536 4096 256 16 1

Hexadecimal number: 1 F 4

Four-bit Group Decimal Digit Hexadecimal Digit 0000 0 0 0001 1 1 0010 2 2 0011 3 3 0100 4 4 0101 5 5 0110 6 6 0111 7 7 1000 8 8 1001 9 9 1010 10 A 1011 11 B 1100 12 C 1101 13 D 1110 14 E 1111 15 F

Hexa to Decimal Conversion• To convert to base 10, beginning with the

rightmost digit multiply each nth digit by 16(n-1), and add all of the results together.

Ex: 1F416


decimal positional value: 4096 256 16 1Hexadecimal number: 1 F 4

256 + 240 + 4 = 50010

Ex: 7E16


decimal positional value: Hexa number:

• The Division Method. Divide by 16 until you reach zero, and then collect the remainders in reverse.

Ex 1: 12610 = 7E16

16) 126 Rem:

16) 7 14=E

0 7

Ex 2: 81010 =

16 ) 810 Rem:16 ) 16 )

Answer: 81010 = 16

Binary to Octal Conversion

• Since the maximum value represented in 3 bit is equal to:

23 – 1 = 7

• i.e. using 3 bits we can represent values from 0 –7

which are the digits of the Octal numbering system.

• Thus, three binary digits can be converted to one octal digit and visa versa.

Three-bit Group Decimal Digit Octal Digit 000 0 0 001 1 1 010 2 2 011 3 3 100 4 4 101 5 5 110 6 6 111 7 7

Ex :Convert 7428 = 2

7 = 1114 = 1002 = 010

7428 = 111 100 0102

Ex :Convert 101001102 = 8

110 = 6100 = 4010 = 2 ( pad empty digits with 0)

101001102 = 2468

Four-bit Group Decimal Digit Hexadecimal Digit 0000 0 0 0001 1 1 0010 2 2 0011 3 3 0100 4 4 0101 5 5 0110 6 6 0111 7 7 1000 8 8 1001 9 9 1010 10 A 1011 11 B 1100 12 C 1101 13 D 1110 14 E 1111 15 F

Ex :Convert 3D916 = 2

3 = 0011 D = 1101 9 = 1001

3D916 = 0011 1101 10012

Ex :Convert 101001102 = 16

0110 = 61010 = A

101001102 = A616

To convert between Octal to Hexadecimal numbering systems and visa versa convert from one system to binary first then convert from binary to the new numbering system

Ex :Convert E8A16 = 8

1110 1000 10102

111 010 001 010 (group by 3 bits)

7 2 1 2

E8A16 = 72178


111 101 0102 (group by 4 bits)

0001 1110 1010

1 E A

7528 = 1EA16


111 101 0102 (group by 4 bits)

0001 1110 1010

1 E A

7528 = 1EA16

Basic Structure of hardware and Software

Topics:

• Internal Hardware Features• External Hardware Features• Program Organization• Memory Organization• Basics of Assembly Programming• SK Editor• Using Debug

Objectives

• To familiarize with the basic structure of hardware and software

• To discuss programming model and memory organization

• To discuss assembly programming using debug.

Basic Structure of Hardware and Software

Hardware – is the physical part of the computer system

Software - A program or set of instructions that controls the operation of a computer.

Distinguished from the actual hardware of the computer.

Internal Hardware Features

1. Processor

2. Memory

3. Registers

External Hardware Features

Computer’s Input/Output

example:

1. Keyboard

2. Monitor

3. Disk

4. CD-ROM

Program Organization

• Software consists of the operating system and various programs and data files stored on disk.

• To execute (or run) a program:

1. the system copies it from external device into internal memory.

2. the processor executes the program instructions.

3. The registers handle the requested arithmetic, data movement, and addressing.

Memory OrganizationBit – the fundamental building block of

computer storage

- off means 0, on means 1

Bytes – consists of eight bits for data and one bit for parity

0 0 0 0 0 0 0 0 1

Data bits parity

According to the rule of parity, the number of bits that are on in each byte must always be odd.

Related Bytes

• A program can treat a group of one or more related bytes as a unit of data, such as time or distance. A group of bytes that defines a particular value is commonly known as a data item or field.

• Word – a 2-byte (16-bit) data item • Doubleword – A 4-byte (32-bit) data item• Quadword – An 8-byte (64-bit) data item

• Paragraph – a 16-byte (128 bit) area

• Kilobyte (KB) – The number 210 equals, 1,024, which happens to be the value K, for kilobyte.

• Megabyte (MB) – The number 220 equals 1,048,576, or 1 megabyte.

Basics of Assembly Programming

Programming Language

A programming language is a notation for expressing instructions to be carried out by the computer. It is a medium of communication between the human and the machine, and often between one human being and another.

Classification of programming Languages

• High-Level Language – A high-level language is an English-like language that provides for more natural expression of algorithms.

Examples: BASIC, Pascal, COBOL, & Prolog

• Mid-level Language– It is a computer language which combines low-level efficiency with high-level language structure, abstraction and portability.

Examples: C & MODULO-2

• Low-level Language – A low-level language is considered to be machine-oriented since it is closely follows the built in instruction set of the underlying computer.

Examples: Assembly, machine languages.

Assembly Language

• It is a low-level language that provides a kind of shorthand notation called mnemonic (memory aid) for the actual machine language instruction. It is a much easier to remember SUB AX<BX than 0010101111000011 which is the actual instruction code that the machine uses.

SideKick Editor

• It is a text editor used in coding the source code. SideKick or SK editor can display the ASCII code for coding assembly language program.

Debug

• BUG - It is a computer terminology for error or mistake in a program or computer system and the term debugging is a methodical process of finding and reducing the number of bugs or defects in a computer program.

• DEBUGGER – It is a program tool that provides an environment for testing load modules. Load modules are executable programs

Registers

Topics:

1. Register

2. General Purpose Registers

3. Pointer Registers

4. Segment Registers

5. Index Registers

6. Flag Registers

Objectives

1. To define what is a register.

2. Identify the types of registers.

3. Identify the function of each register.

Register Set

Registers are fast memory, almost always connected to circuitry that allows various arithmetic, logical, control, and other manipulations, as well as possibly setting internal flags.

General-Purpose Registers

AX - Accumulator Register mostly used for calculations and for input/output

BX - Base Register Only register that can be used as an index

AH AL

AX

EAX

BH BL

BX

EBX

• CX - Count Register used for the loop instruction

• DX - Data Register input/output and used by multiply and divide

CH CL

CX

ECX

DH DL

DX

EDX

Pointer Registers

IP - Instruction Pointer 16-bit number that points to the offset of the next instruction

SP - Stack Pointer 16-bit number that points to the offset that the stack is using

BP - Base Pointer used to pass data to and from the stack

Segment Registers

CS - Code Segment 16-bit number that points to the active code-segment

DS - Data Segment 16-bit number that points to the active data-segment

SS - Stack Segment 16-bit number that points to the active stack-segment

ES - Extra Segment 16-bit number that points to the active extra-segment

Index Registers

SI - Source Index used by string operations as source

DI - Destination Index used by string operations as destination

Flag Registers

Abr. Name bit nº Description

OF Overflow Flag

11 indicates an overflow when set

DF Direction Flag

10 used for string operations to check direction

IF Interrupt Flag

9 if set, interrupt are enabled, else disabled

TF Trap Flag 8 if set, CPU can work in single step mode

SF Sign Flag 7 if set, resulting number of calculation is negative

ZF Zero Flag 6 if set, resulting number of calculation is zero

AF Auxiliary Carry

4 some sort of second carry flag

PF Parity Flag 2 indicates even or odd parity

CF Carry Flag 0 contains the left-most bit after calculations

System Operation

Topics:

• Disk Operating System

• Debug

• Loading of Debug from DOS

• Loading SK through DOS

Objectives

• To recall common DOS commands.

• To discuss the procedures in starting and quitting Debug.

• To familiarize with the Debug commands.

• To know the rules in using debug commands.

System Operation

Common Disk Operating System Commands

1. CLS - clear 5. DEL - delete

2. CD – change dir 6. REN - rename

3. DIR - directory 7. MD – make directory

4. COPY - copy 8. RD – remove directory

DEBUG

Debug – is a software that is classified as debugger which is used for testing and debugging executable programs.

- It displays all program code and data in hexadecimal format and any data that you enter into memory must also in hex form.

Advantages of Debug

1. It is free.

2. It is universally available.

3. It is simple to use.

4. It requires relatively little memory.

Debug Commands

1. Q (Quit)

Finishes the Debug session and exits back to DOS environment

Ex: -Q

2. H (Hexarithmetic)

Shows the sum and difference of two 4-bit hexadecimal numbers, coded as H <hex value> <hex value>

Ex: - 000C 0008

3. R (Register)Allows you to display all registers and their values.

It also shows the next instruction and permits you to change the value of a particular register.

Ex: -R -R CX

4. E (Enter)Enables you to key in data or machine instructions

into memory beginning at a specific location address.

Ex: -E 0200

5. D (Display or Dump)Displays the content of a portion memory in hex

and ASCII forms starting with the given address.Ex: D 0200

6. A (Assemble)Allows you to create program in mnemonic or

symbolic code. It also translates this assembly source statements that you create into machine code.

Ex: -A 0100

7. T (Trace)

Runs the program in a single-step mode. It also displays the new values of the registers and the next instructions to be executed.

Ex: -T

8. G (Go)

Runs the program as a whole in memory and displays the output.

Ex: -G

9. U (Unassemble)Lists all the instructions contained in the program

beginning at the given address. You can also specify the last address location.

Ex: -U 0100 -U 0100 0109

10.N (Name)Gives a name to your program, coded as N

<path><filename>. The base name of the filename must be eight characters long and the extension name is .COM

Ex: N A:SAMPLE>COM

11.W (Write)

Saves the program onto the disk storage.

Ex: -W

Rules of Debug Commands

•It is not case sensitive

•It assumes that all numbers given are in hexadecimal format.

•You can enter a space only when it is needed to separate parameters of a particular command.

•You should specify segments and offsets with a colon, in the form <segment>:<offset>.

Starting Debug

1. Key in the word debug and press enter. A hyphen (-) appears on the screen, that is the DEBUG prompt. DEBUG is now ready to accept commands.

C:\>debug

-

Quitting and Closing Debug

To close Debug and go back to DOS environment: type Q and press ENTER key.

C:\>debug

-q

C:\>

Loading SK Through DOS

• From the C:\> prompt type SK then the SideKick main menu will appear. From the menu select Notepad or press F2.

C:\>sk

Basic Assembly Instructions Used in Debug

1. MOV (Move data)

It copies and transfers data between two registers, or between an immediate data to a register.

Format: MOV <register>,<register>

MOV <register>, <immediate data>

Example: MOV AX,BX

MOV CX, 5083

MOV CL,DL

MOV BL,33

2. ADD (Add Data)

It is used to get the sum of two registers or a register and an immediate data, and stores the result to the left most register.

Format: ADD <register>,<register>

ADD <register>,<immediate data>

Example: ADD CX,BX

ADD AX,0308

ADD AL,BL

ADD DH, 95

3. SUB (Subtract Data)

It is used to get the difference of two registers or a register and an immediate data, and stores the result to the left most register.

Format: SUB <register>,<register>

SUB <register>,<immediate data>

Example: SUB CX,BX

SUB AX,0308

SUB AL,BL

SUB CL,95

3. MUL (Multiply Data)

It is used to get the product of the given register and AX register, and stores the result to AX register. If the product is greater than 16 bits, the overflow is stored in DX register.

Format: MUL <register>

Example: MUL CX

5. DIV (Divide Data)

It is used to divide the value of a given register and AX register, and stores the quotient to AX and the remainder to DX registers respectively.

Format: DIV <register>

Example: DIV BX

6. INC (Increment by one)It is used to increase the value of the register by

one (1).

Format: INC <register>

Example: INC AXINC CH

7. DEC (Decrement by one)The opposite of INC, instead of increasing, it

decreases the value of the register by one (1)

Format: DEC <register>

Example: DEC AXDEC CH

8. LOOP (Loop Until Complete)It controls the execution of a program segment in

a specified number of times. The CX register should contain a count value before starting the loop and automatically decrements by one (1). If CX is not equal to zero (0), it transfers to its operand address which points to the start of the loop; otherwise it drops through to the next instruction.

Format: LOOP <offset address>

Example: LOOP 0108

Sample Debug Programs

Displaying a smiley character on the screen.

-A 0100

1358:0100 mov ah,02 ;request display character

1358:0102 mov dl,01 ;character to display

1358:0104 int 21 ;call interrupt service

1358:0106 int 20 ;end

13587:008

-G

Program terminated normally

-

Using control characters (0A – Line Feed and 0D – Carriage Return) in a screen display.

-A 0100

1358:0100 MOV AH, 02 ;request display character

1358:0102 MOV DL,54 ;character ‘T’

1358:0104 INT 21 ;call interrupt service

1358:0106 MOV DL, 0A ;line feed character


1358:010A MOV DL,0D ;carriage return character

1358:010C INT 21 ;call interrupt service

1358:010E MOV DL,49 ;character ‘I’


1358:0112 MOV DL,0A ;line feed character


1358:0106 MOV DL,0D ;carriage return character


1358:011A MOV DL,50 ;character ‘P’

1358:011C INT 21 ;call interrupt service

1358:011E INT 20 ;end

Displaying the same character (lowercase ‘z’) thirty times using loop on the screen.

-A 0100

1358:0100 MOV CX,001E ;SETTING THE NUMBER OF LOOOPS

1358:0103 MOV AH,2 ;REQUEST DISPLAY CHARACTER

1358:0105 MOV DL,7A ;CHARCTER TO DISPLAY

1358:0107 INT 21 ;CALL INTERRUPT SERVICE

1358:0109 LOOP 0107 ;GO TO OFFSET 0107

1358:010B INT 20 ;END

Editor

Topics:

1. Editor

2. Types of language translator

3. Advantages of assembly language

Objectives

• To identify the different editors used in assembly language.

• Differentiate the different types of language translator.

• Enumerate the advantages of assembly language.

Editor

Text Editors used in Turbo Assembler

1. Edit

2. Sidekick

3. Pilot

Turbo Assembler

Types of language Translator

1. Compiler – Is a translator program that transforms an entire source program composed of high/mid-level language statements into an object program consisting of machine language executable code.

High/Mid-level language

Source Program COMPILERMachine language

Instruction

2. Interpreter – processes the source program on a statement-by-statement basis. It translates high/mid-level language statements and then executes it one statement at a time.

2. Assembler – is a language translator that transforms assembly language to machine language instruction.

High/Mid-level language

Source Program INTERPRETERMachine language

Instruction

Assembly language

Source Program ASSEMBLERMachine language

Instruction

Assembly Language

It is a low-level language that provides a kind of shorthand notation called

mnemonic(memory aid) for the actual machine language instruction. It is much easier to remember SUB, AX, BX than 00101111000011 which is the actual instruction code that machine uses…

Advantages of Assembly language

•It provides more control over handling particular hardware requirements.

• It generates smaller, more compact executable modules.

• It results in faster execution.

Tasm – Turbo Assembler

T-link t - Turbo Assembler Linker

Linker is a computer program for high, mid, and low levels of programming language that completes the process by converting the object code into executable machine code.

STRUCTURE OF AN ASSEMBLY LANGUAGE PROGRAM

TITLE MYPROG.ASMDOSSEG. MODEL SMALL. STACK 0100h. DATA

X DB “Hello everyone! $”. CODE

MOV AX, @DATAMOV DS, AXMOV AH, 09hMOV DX, OFFSET XMOV AX, 4C00hINT 21h

END

INSTRUCTIONS FOR INITIALIZING THE PROGRAM

TITLE MYPROG.ASM

DOSSEG

. MODEL SMALL

. STACK 0100h

. DATA

. CODE

MOV AX, @DATA

MOV DS, AX

INSTRUCTIONS FOR CONCLUDING THE PROGRAM

MOV AX, 4C00h

INT 21h

END

EXAMPLES OF ASSEMBLY LANGUAGE PROGRAMS

• Displaying a character (uppercase ‘A’) on screen.

TITLE MYPROG.ASM ; name of the programDOSSEG. MODEL SMALL. STACK 0100h. DATA. CODE

MOV AX, @DATAMOV DS, AXMOV AH, 02h ; request display

characterMOV DL, ‘A’ ; character to

displayMOV AX, 4C00h ; endINT 21h

END

Displaying character (lowercase ‘b’) twice on screen.


MOV AX, @DATAMOV DS, AXMOV AH, 02h ; request display characterMOV DL, ‘b’ ; character to displayINT 21h ; call interrupt serviceINT 21h ; call interrupt serviceMOV AX, 4C00h ; endINT 21h

END

• Using control characters (0A – Line Feed and 0D – Carriage return) in a screen display.


MOV AX, @DATAMOV DS, AXMOV AH, 02h ; request display characterMOV DL, ‘T’ ; character to displayINT 21h ; call interrupt serviceMOV DL, 0Ah ; line feed characterINT 21h ; call interrupt serviceMOV DL, 0DH ; carriage return characterINT 21h ; call interrupt serviceMOV DL, ‘I’ ; character ‘I’INT 21h ; call interrupt serviceMOV DL, 0Ah ; line feed characterINT 21h ; call interrupt serviceMOV DL, 0Dh ;carriage return characterINT 21h ; call interrupt serviceMOV DL, ‘P’ ; character ‘P’INT 21h ; call interrupt serviceMOV AX, 4C00h ; endINT 21h

END

• Displaying different characters (‘A to ‘Z’) using LOOP on the same line.

TITLE MYPROG.ASMDOSSEG. MODEL SMALL. STACK 0100h. DATA. CODE

MOV AX,@DATAMOV DS,AXMOV CX,001Ah ; setting the number of loopsMOV AH,02h ; request display characterMOV DL,‘A’ ; character ‘A’

B: INT 21h ; call interrupt serviceINC DL ; add to 1 to the value of DLLOOP B ; go to label BMOV AX,4C00h ; endINT 21h

END

MIDTERM

Week 7- Instructions And Directives

Week 8 – Introduction to Assembly Language

Week 9 – Using SK Editor

Week 10 – Keyboard handling

Week 11 – Screen Handling

Instructions And Directives

Topics:

• Instructions

• Categories of Instructions

• Directives

• Types of Memory Model

Objectives

To familiarize with the different instructions.

To identify the categories of instructions

To identify the different directives and their corresponding functions.

To identify the different memory modes.

Instructions

- These are statements that will be translated into machine language and executed by the computer.

Examples of instructions

MOV ADD SUB

MUL DIV INC

DEC LOOP CMP

Categories of Instructions1. DATA TRANSFER

To move data in the memory and in the CPU registers.

Format: Memory to register Register to registerRegister to stackStack to register

Syntax: Mov Destiny, SourceExample: mov ax,bx

Note: the list of symbolic instructions are for Intel processor family

There are two very important details to note about the mov

instruction

1. There is no memory to memory move operation.

2. You cannot move immediate data into a segment register.

LDS: Load Data Segment Register

LEA: Load Effective Address

LES Load Extra Segment Register

LODS: Load String

LSS: Load stack Segment Register

MOV: Move data

MOVS: Move String

MOVSX: Move with sign-extend

MOVZX: Move with Zero-extend

STOS: Store string

XCHG: Exchange

XLAT: Translate

2. ARITHMETIC

Syntax: Adc destiny, source

Example: Adc bh,bl

ADD: add binary numbers ADC: Add with carry

DEC: Decrement by 1 INC: Increment by 1

DIV: Unsigned Divide IDIV: Signed (integer) divide

MUL: Unsigned multiply IMUL: Signed multiply

SUB: Subtract binary values SBB: Subtract with borrow

XADD: exchange and add

3. COMPARISON

Syntax: CMP destiny, source

Example: cmp ax,bx

CMP: compare

CMPSn: Compare string

CMPXCHG: Compare and exchange

4. LOGICAL OPERATIONSSyntax: AND destiny,sourceExample: And ax,bx

AND: Logical ANDNOT: Logical NOTOR: Logical ORXOR: Exclusive XOR

5. INPUT/OUTPUTSyntax: OUT destiny,sourceExample: in ax,dx

IN: Input byte or wordINSn: Input stringOUT: Output byte or wordOUTSn: Output string

6. JUMP(conditional)Syntax: ja labelExample: ja start

JA: Jump if AboveJAE: Jump if above/equalJB: Jump if belowJBE: Jump if below/equalJE: Jump if EqualJG: Jump if GreaterJGE: Jump if Greater/EqualJL: Jump if LessJNE: Jump if Not Equal

JNG: Jump if Not Greater

JNGE: Jump if Not Greater/Equal

JNL: Jump if Not Less

JNLE: Jump if Not Less/Equal

JNZ: Jump if Not Zero

JLE: Jump if Less/Equal

JNA: Jump if Not Above

JNB: Jump if Not Below

JNBE: Jump if Not Below/Equal

JZ: Jump if Zero

JC: Jump if Carry

JCXZ: Jump if CX is Zero

JNO: Jump if No Overflow

JO: Jump if Overflow

JNP: Jump if No Parity

JP/JPO: Jump if Parity Odd

JPE: Jump if Parity Even

JS: Jump if Sign

JNS: Jump if No Sign

7. JUMP (Unconditional)

CALL: Call a procedure

INT: Interrupt

IRET: Interrupt Return

JMP: Unconditional Jump

RET: Return

RETN/RETF: Return Near/Return Far

8. LOOPING

Syntax: loop label

Example: loop a

LOOP: Loop until complete

LOOPE: Loop While Equal

LOOPZ: Loop While Zero

LOOPNE: Loop While Not Equal

LOOPNZ: Loop While Not Zero

9. STACK INSTRUCTIONS

Syntax: pop destiny

Example: pop ax

POP: Retrieve the content of the Stack

PUSH: Store the content of the Stack

POPF: Pop Flags off the Stack

PUSHF: Push Flags onto Stack

Directives

- These are statements that give information to the assembler.

- Sometimes called pseudo-ops (short term for pseudo operations and pronounced as”su-da-op”).

- Unlike instructions, directives are not translated is not machine language but they are necessary for your program to assemble properly.

Examples of Directives

TITLE DOSSEG .MODEL

.STACK .DATA .CODE

Most Common Directives

1. TITLE – It creates a title (up to 60 characters) of a source listing.

Format:

Example: TITLE PROGRAM1.ASM

TITLE <TEXT>

2. DOSSEG – It tells the assembler to ignore all other requests and to adopt the DOS segment sequence – stack, data and code.

Format:

Example: DOSSEG

DOSSEG

3. .MODEL – It specifies and initializes the memory model before defining any segment.

Format:

Example: .MODEL TINY.MODEL SMALL.MODEL MEDIUM

.MODEL SMALL <MEMORY-MODEL>

Types of Memory ModelMEMORY MODEL NO. OF DATA SEGMENT NO. OF CODE SEGMENT

Tiny 0 0

Small 1 1

Medium 1 More than 1

Compact More than 1 1

Large More than 1 More than 1

4. .STACK – It defines the size of the stack. The default size is 1,024 bytes which you can overrule.

Format:

Example: .STACK 0100h

.STACK <SIZE>

5. .DATA – It defines and marks the beginning of data segment.

Format:

Example: .DATA

.DATA

6. .CODE – It defines and marks the code segment which consists of a set of instructions.

Format:

Example: .CODE

.CODE

7. END – It is placed at the last line of the source code.

Format:

Example: END

END

8. For Defining Data

DIRECTIVE LENGTH DESCRIPTION

DB 1 Define Byte

DW 2 Define Word

DD 4 Define Doubleword

DF 6 Define Farword

DQ 8 Define Quadword

DT 10 Define Tenbytes

Introduction to Assembly Language

Topics:

• Requirements for Coding Assembly language

• Reserved Words

• Identifiers

Objectives

• To know the requirements in coding assembly language

• To identify the different reserved words.

• To know the categories of reserved words

• To identify the two types of identifiers.

• To know the general guidelines in coding assembly program.

Introduction to Assembly Language Using Turbo

Assembler

Requirements for Coding Assembly language

Comment

-Is a part of a program that is ignored by the assembler. Though Optional, the use of comments throughout a program can improve its readability and clarity especially in assembly language where the purpose of a set of instructions is often ambiguous.

-It begins with a semicolon(;)

Ways to include comment• Any Statement whose first non-blank character is a semicolon.

Example: ; This program displays Good Day!

•At the end of an instruction.

Example: MOV AX, 8053h ; initializes the value of ax to 8053h

Reserved Words

These are words in which the assembler assigns a special meaning and it cannot be used as identifiers. They are reserved for their own purpose to be used only under special condition.

Categories of Reserved Words

•Instructions

•Directives

•Operators

•Pre-defined Symbols

Identifiers

An identifier is a user-defined name (Variable) that you apply to an item in your program that you expect to reference.

Types of Identifiers

•Name – It refers to the address of a data item.

Example: x db 0

msg db “Aloha! &”

•Label – It refers to the address of an instruction or procedure.

Example: mov dl, 41h

A: int 21h

add dl, 20h

int 21h

sub dl, 20h

loop A

Rules in Constructing Valid Identifier

•It must be letters (A..Z, a..z), numbers (0-9) &/or special characters like underscore(_), question mark(?) and at sign (@).

•* It must always start with a letter.

•It must not use reserved words.

•It must not exceed to 31 characters.

Examples of valid identifiers: neym u2u2

num_1 msg8

Examples of Invalid identifiers: title num-1

4ever F-4

Statement

An assembly language statement program consists of a set of statements. The two types of statements are instructions and directives.

General Guidelines for coding statement

•A statement may begin anywhere on the line

•Each line can only contain one statement

•Unlike C language, assembly is not case sensitive. You can use either upper or lowercase letters.

•Take note that statements can have two (2) operands, a single and even none at all.

Examples: ADD AX,BX ;uses 2 operands

DEC CX ; uses single operand

RET ; no operand

String-is used for directive data such as person’s name or simply a message. It must end with dollar ($) symbol and defined in double quotation marks (“ ”). DB is the conventional format for defining string of any length.

Example:

Neym db “Mary’s Internet Café $”

Numeric Constant-They are used to define arithmetic values and memory addresses. It is identified with a radix specifier such as d for decimal, b for binary and h for hexadecimal.

Example:

Msg db “Good day!”, 0ah, 0dh, “$”

Is the same with

Msg db “Good day!”, 10d, 13d “$”Msg db “Good day!”, 00001010b, 00001101b, “$”

Using SideKick

Assembling and Linking Procedures

Topics:

• Using SideKick

• Creating Assembly Language

• Structure of Assembly language

• Assembling and Linking Assembly Language

Objectives• To familiarize with the SideKick

Environment as text editor.• To know the procedures in invoking

Sidekick and quitting from the SideKick window.

• To know the procedures in creating and saving assembly program.

• To know the procedures in assembling, linking and running assembly program.

Using Sidekick as Editor

Sidekick

- is an editor used for DOS programming like assembly language.

Creating Assembly Language Program

Step 1: Open the text editor.

C:\tasm>SK

Step2: Type the name of the program with .asm as file extension name.

Example: c:program1.asm

Step3: You may now start typing your source code.

Exiting from the Sidekick Window

* Press escape (esc )key to go back to DOS environment.

Structure of an Assembly Language ProgramTITLE PROGRAM1.ASM

DOSSEG

.MODEL SMALL

.STACK

.DATA

X DB “HELLO THERE!$”

.CODE

MOV AX, @DATA

MOV DS,AX

MOV AH,O9H

MOV DX,OFFSET X

INT 21H

MOV AX,4C00H

INT 21H

END

Initializes the program

Concludes the program

Body of the program

Assembling and Linking Procedures

Assembling:

From the c:\tasm> prompt type the following the filename and its extension name then press enter key.

Example: c:\tasm>program1.asm

Linking:

After a successful assembly (I.e., no errors encountered), type tlink, press space bar, filename and its extension name then press enter key

Example: c:\tasm>tlink program1.obj

Executing:

To run the program, just type the filename only at c:\tasm>.

Example: c:\tasm>program1

To recapitulate the overall process:1. Open the text editor.

2. Save the program.

3. Start keying your program.

4. Exit the editor and go back to DOS environment by pressing the Esc key.

5. Assemble the program by typing tasm filename.asm.

6. If there are no errors, do the next step otherwise, repeat steps 1-5 and rectify the program.

7. Link the object file by typing tlink filename.obj & press enter key.

8. Type filename and press enter key to run the program.

KEYBOARD HANDLING

Topics:

Interrupt and Service

Character Output

Character Input

String Display

String Input

Objectives

• To identify the function/service number used to input/output character.

• To identify the function/service number used to input/output string.

• To identify the corresponding interrupt used in character and string input/output.

Interrupt – execute commands

Service – set the command

Interrupt and Service

Character Output

Int 21h / 02

This function display one character on the screen.

AH = 02h

DL = holds the character to display on the screen.

Example:

. MODEL SMALL

. STACK 0100h

. DATA

. CODEMOV AX, @DATAMOV DS, AXMOV AH, 02h ; request display

characterMOV DL, ‘j’ ; character to displayINT 21h ; call interrupt serviceINT 21h ; call interrupt serviceMOV AX, 4C00h ; endINT 21h

END

Character InputInt 21h / 1 – Keyboard input

This function waits for you to type a character on the keyboard. It echoes the character to the screen and returns the ASCII character in the AL register.

AH = 01

AL = character read from the keyboard

Example:

.model small

.stack

.data

.code

mov ah, 01h ; request for character input

int 21h

Character Input

Int 21h / 7 – keyboard input without echo

This function reads a character from a keyboard but doesn’t display the character on the screen and wait for control break.


Character Input

Int 21h / 8 – keyboard input without echo

This function works the same way with service 7 but do not respond to control break.


Example:.MODEL SMALL.STACK 0100H.DATA

MSG DB “INPUT ANY CHARACTER: $”MSG2 DB 0AH, 0DH, “The character you type is: $”x db 0

.CODEMOV AX,@DATAMOV DS, AXMOV AH,09HMOV DX,OFFSET MSG1INT 21H

MOV AH,01H ;REQUEST FOR CHARACTER INPUTINT 21HMOV X,AL

MOV AH,09HMOV DX,OFFSET MSG2INT 21H

MOV AH,02HMOV DL,XINT 21H

MOV AX,4C00HINT 21H

String Display

Int 21h / 09

This function displays the string pointed by the DS:DX pair of registers. The end of the string should be marked with the $ sign.

DS:DX points to address of the string

String Display

Int 21h / 09h

Display string in the data area followed by a dollar sign ($ or 24h) delimiter, which the operation uses to end the display.

Example:

Nick_name db “Your nickname: $“ ;display string

Example fragment of a program

MOV AH,09H ;request display

LEA DX, NICK_NAME ;load address of prompt

INT 21H ;call interrupt service

String Input

Int 21h / 0A

This function reads the string from the keyboard.

DS:DX points to address of the string

Example:. MODEL SMALL. STACK 0100h. DATA

NEYM DB 08h, ?, 08h DUP (“$”)Q DB “What is your name? $”A1 DB 0Ah, 0Dh, “What a nice name ! $”A2 db 0Ah, 0Dh, “I’m glad to meet you ! $”

. CODEMOV AX, @DATAMOV DS, AXMOV AH, 09hMOV DX, OFFSET QMOV AH, 0Ah ; request for string inputMOV DX, OFFSET NEYM ; load addressINT 21hMOV AH, 09hMOV DX, OFFSET A1INT 21hMOV AH, 09hMOV DX, OFFSET NEYM+2INT 21hMOV DX, OFFSET A2INT 21hMOV AX, 4C00hINT 21h

END

DISPLAYING THE INPUT STRING

The following instructions display the input string starting at the third memory address.

MOV AH, 09h ; request for string output

MOV DX, OFFSET X+2 ; LOAD ADDRESS

INT 21h

Sample program that demonstrates string and displays it again on screen.

. MODEL SMALL

. STACK 0100h

. DATANEYM DB 08h, ?, 08h DUP (“$”)Q DB “What is your name? $”A1 DB 0Ah, 0Dh, “What a nice name ! $”A2 db 0Ah, 0Dh, “I’m glad to meet you ! $”

. CODEMOV AX, @DATAMOV DS, AXMOV AH, 09hMOV DX, OFFSET QMOV AH, 0Ah ; request for string inputMOV DX, OFFSET NEYM ; load addressINT 21hMOV AH, 09hMOV DX, OFFSET A1INT 21hMOV AH, 09hMOV DX, OFFSET NEYM+2INT 21hMOV DX, OFFSET A2INT 21hMOV AX, 4C00hINT 21h

END

Screen Handling

Topics:

• The Monitor

• GRAPHICS

Objectives

• To identify the instructions and directives used for screen handling applications

• To create an assembly program with the application of graphics.

THE MONITOR

• A typical video screen has eighty (80) columns numbered from 0 to 79 and twenty five (25) rows numbered from o to 24. The columns and rows provide a grid of addressable locations at any one of which the cursor can be set. The table below shows the examples of a cursor locations

Screen

Location

COLUMN ROW COLUMN ROW

In hexadecimal value In hexadecimal value

Upper Left Corner

0 0 00H 00H

Upper Right Corner

79 0 4FH 00H

Center of Screen

40 12 28H 0CH

Lower Left Corner

0 24 00H 18H

Lower Right Corner

79 24 4FH 18H

Screen Features

Clearing the Screen in Assembly Approach

Int 10h / 06h

Handles the process of clearing the screen and scrolling.

AH = Function in 06H

AL = number of lines to scroll, or 00H for the full screen

BH = attribute value (color blinking etc.)

CX = starting row column

DX = ending row column

sample code shows how to create the

clear screen:

MOV AX,0600h ;AH=06h (Scroll), AL=00h (Full Screen)

MOV BH,07h ;0 (Black Background), 7 (White Text Color)

MOV CX,0000h ;CH=00h (Row), CL=00h (Column)

MOV DX,184Fh ;DH=18h (Row), DL=4Fh (Column)INT 10h ;Call interrupt service

SETTING THE CURSOR POSITION

Int 10h / 02Used to set the cursor position.

AH = 02

MOV AH, 02h ; request to set the cursor positionMOV BH, 002h ; page number 0MOV DH, 0Ah ; Row = 10 in decimalMOV DL, 08h ; Column = 8 in decimalINT 10h ; call interrupt service

Alphanumeric

DISPLAYING THE STRING ‘FLOWERS’ AT THE CENTER OF THE SCREEN

.MODEL SMALL .STACK 0100h .DATA

A DB “FLOWERS$” .CODE

MOV AX,@DATAMOV DS,AXMOV AX,0600h ; AH=00h (Scroll0, AL=00h (Full Screen)MOV BH,07h ; 0 (Black Background), 7 (White Text Color)

MOV CX,0000h ; CH=00h (Row), CL 00h (Column)MOV DX,184Fh ; DH=18h (Row), DL=4Fh (Column)INT 10h ; Call interrupt ServiceMOV AH,02h ; request to set the cursor positionMOV BH,00h ; page number 0MOV DH,0Ch ; Row = 12 in decimalMOV DL,24h ; Column = 36 in decimalINT 10h ; Call interrupt service

MOV AH,09H ; request display string LEA DX,A ; load address of the string

MOV AX, 4C00hINT 21h

END

GraphicsTEXT AND BACKGROUND COLORS

BACKGROUND COLOR TEXT COLOR HEX CODE MEANINGBL R G B I R G B 0 0 1 0 1 0 0 0 28 Green Background Color

Gray Text Color

1 0 1 0 1 0 0 0 0A8 Green Background Color

Gray Black ColorBlinking

0 1 1 0 0 0 0 1 61 Brown Background Color Blue Text Color

1 1 1 0 0 0 0 1 0E1 Brown Background Color Blue Text Color Blinking

DESIGNING PROGRAMS USING PROCEDURE

Procedures make it possible to break a large program down into smaller pieces so that each piece can be shown to work correctly without the reference to any of the others.

DEFINING A PROCEDURE

Every procedure is defined using PROC and ENDP directives. Since these are directives, they don’t generate machine language instructions; rather, PROC and ENDP are directions to the assembler, making the beginning and end of a procedure. In assembly, the first declared procedure in the program is considered the major calling procedure.

Format & Example• Format:•• <procedure-name> PROC• <instruction-1>• <instruction-2>• …• <instruction-n>• <procedure-name> ENDP• Example:•• MAIN PROC• CALL P1• CALL P2• CALL QUITPROGRAM• MAIN ENDP

CALLING PROCEDURE

To call a procedure

Format:

CALL <procedure-name>

FINALS

Week 13 – File and Disk handling

Week 14 – Sample Programs in Assembly Language

Week 15 – Math Operation

Week 16 – In-line Assembly

Week 17 – Case Study

FILE AND DISK HANDLING

Topics:

•Common DOS File Handle Services

•File Attributes

•Modes of Accessing Files

Objectives:

• To identify and familiarize with the different services for file and disk handling.

• To create a program using the different services in file and disk handling.

• To identify the different file attributes

• To apply the different modes of accessing files.

Common DOS File Handle Services

INT 21H / 39H

Create SubdirectoryThis service creates a subdirectory as does

the system command MKDIR. To use it, load DX with the address of an ASCIIZ string containing the drive and directory pathname.

AH = 39HDS:DX = points to ASCIIZIf CX = 1, AX has error

Example fragment of a program:

ASCstrg DB ‘n:\pathname’, 00H ;ASCIIZ string

…MOV AH,39H ;request create subdirectory

LEA DX,ASCstrg ;Address of ASCIIZ string (DS:DX)

INT 21H

Remove SubdirectoryThis service deletes a subdirectory, just as

does the system command RMDIR. Note that you cannot delete the current (active) directory or a subdirectory containing files. Load DX with the address of an ASCIIZ string containing the drive and directory pathname.

AH = 3AHDS:DX = points to ASCIIZ stringIF CX = 1, AX has error

INT 21H / 3AH


ASCstrg DB ‘n:\pathname’,00H ;ASCIIZ string

…

MOV AH,3AH ;request delete subdirectory

LEA DX,ASCStrg ;address of ASCIIZ (DS:DX)


INT 21H / 56H

Rename File or DirectoryThis service name a file or directory from within a program. Load DX with the address of an ASCIIZ string containing the old drive, path and name of the file or directory to be renamed. Load DI (combined as ES:DI) with the address of an ASCIIZ string containing the new drive, path and name, with no wild cards.

Drive numbers, if used, must be the same in both strings. Because the paths need not be the same, the operation can both rename a file and move it to another directory on the same drive.


Oldstring db ‘n:\oldpath\oldname’,00h

Newstring db ‘n:\newpath]newname’,00h

…MOV AH,56H ;request rename file/directory

LEA DX,Oldstring ;DS:DX

LEA DI,Newstring ;ES:DI

INT 21H / 5BH

Create a new File

This service creates a file only if the named file does not already exist; otherwise it is identical to function 3CH (create file). You could use function 5BH whenever you don’t want to overwrite an existing file.

INT 21H / 3CH

Create File

AH = 3CH

DS:DX = points to ASCIIZ string

CX = attribute

File AttributesBIT ATTRIBUTE

0 Read only file

1 Hidden file

2 System file

3 Volume label

4 Subdirectory

5 Archive file

INT 21H / 43H

Get/Set File Attribute

You can use this operation either to get or set a file attribute in the directory. This operation requires the address of an ASCIIZ string containing the drive, path and filename for the requested file. (Or use the default directory if no path is given.)


ASCstrg db ‘n:\pathname’,00H ;ASCIIZ string

…MOV AH,43H ;request

MOV AL,00H ;get attribute

LEA DX,ASCstrg ;ASCIIZ string (DS:DX)


INT 21H / 3BH

Change Directory

This service changes directory.

AH = 3BH


If CX = 1, AX has error

Modes for Accessing Files

Access mode Means

0 Read Only

1 Write Only

2 Both Read and Write

INT 21H / 3DH

Open File

This function opens a file.

AH = 3DH


AL = MODE

INT 21H / 3EH

Close File

This service closes a file.

AH = 3EH

BX = file handle

If CX = 1, AX has error

INT 21H / 41H

Delete File

This service deletes a file (but not read-only) from within a program. Load the address in DX of an ASCIIZ string containing the device path and filename, with no wild cards references.


ASCstrg db ‘n:\pathname’,00H ;ASCIIZ string

…MOV AH,41H ;request delete file

LEA DX,ASCSTRG ;address of ASCIIZ string

(DS:DX)


Sample Programs In Assembly Language

Topic:

• Designing programs using procedure

• Basic Stack Instruction

• Loop Instruction

Objectives:

• Create an assembly program with procedures.

• Create an assembly program using stack instruction.

• Create an assembly program using loop instruction.

DESIGNING PROGRAMS USING PROCEDURE

• DEFINING A PROCEDURE

Every procedure is defined using PROC and ENDP directives. Since these are directives, they don’t generate machine language instructions; rather, PROC and ENDP are directions to the assembler, making the beginning and end of a procedure. In assembly, the first declared procedure in the program is considered the major calling procedure.

Format:<procedure-name> PROC

<instruction-1><instruction-2>…<instruction-n>

<procedure-name> ENDP

Example:

MAIN PROCCALL P1CALL P2CALL QUITPROGRAM

MAIN ENDP

CALLING PROCEDURE

To call a procedure

Format:

CALL <procedure-name>

Example:.model small.codeorg 100h

start: jmp mainx db “assembly$”y db “language$”z db “program$”

main proc nearmov dx,offset xcall proc1call proc2mov dx,offset ycall proc 1call proc2mov dx,offset zcall proc1call proc2

main endp

proc2 proc nearmov ah,2mov dl,13int 21hmov dl,10int 21hret

proc2 endp

proc1 proc nearmov ah,9int 21hret

proc1 endp

end start

Basic Stack Instruction

Push Instruction

puts/places data onto the stack segment

Pop Instruction

takes out data from the stack

Format:

• push reg16

• pop reg16

• push reg32 (3)

• pop reg32 (3)

• push segreg

• pop segreg (except CS) push memory

• pop memory

• The first two instructions push and pop a 16 bit general purpose register. This is a compact (one byte) version designed specifically for registers

• The second pair of instructions push or pop an 80386 32 bit general purpose register.

• The third pair of push/pop instructions let you push or pop an 80x86 segment register.

• The fourth pair of push/pop instructions allow you to push or pop the contents of a memory location.

Example:.model small.codeorg 100h

start: mov cx,5 x: mov ah,1

int 21hadd al,20hpush axloop x

mov ah,2mov dl,13int 21hmov dl,10int 21hmov cx, 5

y: pop dxint 21hloop yint 20h

end start

Loop Instruction

Loop allows a part of the program to be repeated for several times, the number of loops depend on the value of the CX register.

- each time the loop is executed, CX is decremented and the looping stops when CX becomes zero.

Format:

LOOP label

Example:

LOOP start

This will cause the program flow to go to the label start for CX times.

Example:

.model small

.codeorg 100hstart:mov cx,20

mov ah,2mov dl, ‘j’

z: int 21hloop z

end start

Math Operation

• BCD

• ASCII

Objectives:

• To create a program using addition, subtraction, multiplication and division of Binary Coded Decimal.

• To create a program that converts the addition, subtraction of ASCII codes to BCD.

• To identify the instructions and directives for addition, subtraction of BCD.

BCD• binary coded decimals

Binary coded decimal (BCD) is a method for implementing lossless decimal arithmetic (including decimal fractions) on a binary computer. The most obvious uses involve money amounts where round-off error from using binary approximations is unacceptable. Some early computers used BCD exclusively.

•

• Decimal digits (0-9) can be encoded in a nibble (half a byte), with some left over bit patterns (hexadecimal A-F). In BCD operations, the processor performs ordinary binary computations, then adjusts the result to conform to BCD.

• For example, if you add the binary number 5 (bit pattern 0101) to binary number 6 (bit pattern 0110), you get the binary result of 11 (bit pattern 1011, or hexadecimal B). With BCD arithmetic, the processor would adjust the result to make it into a valid BCD result (which in this case would be bit pattern 0001 0001).

• BCD arithmetic includes BCD addition, BCD subtraction, BCD multiplication, BCD division, and BCD negate.

The Intel 80x86 series uses a two step approach for BCD arithmetic. Instead of having separate BCD instructions, the normal binary addition and subtraction instructions are used, then hardware instructions are used to adjust the results to correct BCD results.

• There are instructions for both packed and unpacked adjustments.

• The advantage of this approach is greater flexibility (more addressing modes and choices of arithmetic operations because of the use of regular binary integer instructions in the first step).

• The disadvantage of this approach is that it is slower and takes more memory.

ASCII

• American Standard Character Information Interchange.

• Refer to ASCII Table

Adding ASCII Numbers

Consider the effect of the following three examples of adding ASCII numbers:

Ex. 1 35H Ex.2 38H Ex.3 39H +32H 34H +39HTotal 67H 6CH 72H

Example 1.

The sum of 35h and 32h is 67h. AAA checks the rightmost digit (7). Because it is not between A and F, nor is the AF flag set, AAA clears the leftmost digit (6) to 0.

Example 2. The sum of 38H and 34 H is 6CH. Because the right most digit (C) is between A and F, AAA performs the following:

AH AL

00 0C

Adds 6 to AL 00 72

Adds 1 to H 01 72Clears leftmost digit of AL 01 02 (sum = 12)

Example 3. The sum of 39H and 39H is 72H. Although the rightmost digit (2) is not between A and F, the AF flag is set because of the carry into the tens position. AAA performs the following:

AH AL

00 72

Adds 6 to AL 00 78

Adds 1 to AH 01 78

Clears leftmost digit of AL 01 08 (sum = 18)

Subtracting ASCII NUMBERSExample 1. ASCIIvalue1 contains 39h and ASCII value2

contains 39h. Subtracts ASCIIvalue2 (35h) from ASCIIvalue1 (39h). AAS does not need to make an adjustment because the rightmost hex digit is less than the hex A.

AX AF CF

MOV AL,ASCVALUE1 ; 0039

SUB AL,ASCVALUE2 ; 0004 0 0

AAS ; 0004 0 0

OR AL,30h ; 0034

Example 2. Subtracts ASCIIvalue1 (39h) from ASCII value2 (35h). Because the rightmost digit of the result is hex C, AAS subtracts 6 from AL, subtracts 1 from AH, and sets the AF and CF flags:

AX AF CF

MOV AL,ASCVALUE2 ; 0035

SUB AL,ASCVALUE1 ; 00FC 1 1

AAS ; FFO6 1 1

The answer, which should be -4, is FF06h, its ten’s complement; that is, decimal -10 + 6 = -4.

Multiplying ASCII Numbers

INSTRUCTION COMMENT AX CL. . . ; Initial values 0035 39

AND CL, 0FH ; Convert CL to 09 0035 09AND AL, 0FH ; Convert CL to 05 0005 09MUL CL ; Multiply AL by CL 002D 09AAM ; Convert AX to unpacked BCD

OR AX 3030H ; Convert AX to ASCII 3435

Dividing ASCII Numbers

INSTRUCTION COMMENT AX CL. . . ; Initial values 3238 37

AND CL,0FH ; Convert to unpacked BCD 3238 07AND AX,0FH0FH ; Convert to unpacked BCD 0208AAD ; Convert to binary 00ICDIV CL ; Divide by 7 0004

CONVERTING BINARY DATA TO ASCII FORMAT

• The following example converts binary number 0DF1H to ASCII format:

ASCVALUE DB 4 DUP (‘ ‘) ; Data Items

BINVALUE DW 0DF1H

. . .

MOV CX, 0010 ; Division factor

LEA SI, ASCVALUE ;Address of SCVALUE

MOV AX, BINALUE ; Get binary amount

L20:CMP AX,CX ; Value < 10?JB L30 ; yes, exitXOR DX,DX ; Clear upper quotientDIV CX ; Divided by 10OR DL,30HMOV [SI],DL ; Store ASCII characterDEC SIJMP L20

L30:OR AL,30H ;Store last quotientMOV [SI], AL ; as ASCII character

C20: mov al,[SI] ;get a byteShr al,04 ;shift off right hex digitLea bx,xlatab ;set table addressXlat ; translate hexStosb ;store AL in DISPAREAMov al,[SI] ;get other half of byteAnd al,0fh ;clear left hex digitXlat ;translate hexStosb ;store AL in DISPAREAInc SI ; incr address of IOBUFFERLea bx,IOBUFFER+512 ;Cmp SI,BX ;end of IOBUFFER?Jb c20 ;if no, repeatLea bp,disparea ;if yes, displayMov cx,1024 ;hex charactersCall D10DISPLAYret

C10CONVRT endp

; display data:

; - - - - - - - - - ;uses AX, BX, DXD10DISPLAY proc near ;BP, CX set on entry

MOV AX,1301 ;request display

MOV BX,0016H ;page:attribute

MOV DX,0500H ;row:column

RET

D10DISPLAY end p

End A10MAIN

In-Line Assembly

Topic:

• Comparison of C Language to Assembly

Objective:

• To construct a program with the integration of Pascal and C language programming.

Comparison of C Language To Assembly Language

If..Else ConstructA conditional statement of C language in the form:

If <condition>{<statement-1>;<statement-2>;. . .<statement-n>;}

Else{<statement-1>;<statement-2>;

. . . <statement-n>;}

Can be implemented by assembly language of the form:

<cmp instruction>

<conditional jump instruction> <label-1>

<instruction-1>

<instruction-2>

…

<instruction-n>

<jump instruction> <label-2>

Comparison

C language Assembly language

If (ax == 0)

{

cx = cx – ax;

ax = ax + 1;

}

Else

{

Cx = cx – 8;

}

CMP AX,0000H

JNZ ACTION1

SUB CX,AX

INC AX

JMP NEXT

ACTION1: SUB CX,0008H

NEXT:

Do..While Construct

A do..while loop similar to C language:

do{<statement-1>;<<statement-2>;…<statement-n>;} while <condition>;

Can be implemented by assembly in the form:

<label>: <instruction-1>

<instruction-2>

…

<instruction-n>

<cmp instruction>

<conditional jump instruction> <label>

Comparison

C language Assembly language

ax = 5;

do

{

ax = ax – 1;

bx = bx + 1;

} while (ax >= 0);

MOV AX,0005H

HERE: DEC AX

INC BX

CMP AX

JNZ HERE

While Construct

Similarly, C language while loop:

While <condition>{<statement-1>;<statement-2>;…<statement-n>;}

Is roughly equivalent to the assembly language structure:

<label-1>: <cmp instruction> <conditional jump instruction> <label> <instruction-1> <instruction-2> . . . <instruction-n> jmp <label-1>

<label-2>: <instruction-1><instruction-2> . . .<instruction-n>

ComparisonC language Assembly language

while (ax <= 10)

{

ax = ax + bx;

bx = bx + 1;

}

START: CMP AX,000AH

JGE NEXT

ADD AX,BX

INC BX

JMP START

NEXT: ADD CX,AX

Final Case Study

Design a problem that will utilize any of the constructs if-else, do-while and while. Please don’t forget to check the complexity and eventual completeness of your output.

Objective:

• To develop programs that apply the whole concepts of assembly programming.

Final Examinationand

Case Presentation

ENDEND

kaizen cso002 l1

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