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INTERMEDIATE 8086 ASSEMBLY LANGUAGE PROGRAMMING Cutajar & Cutajar © 2012

Assembly language 8086 intermediate

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Page 1: Assembly language 8086 intermediate

INTERMEDIATE 8086 ASSEMBLY

LANGUAGE PROGRAMMING

Cutajar & Cutajar

© 2012

Page 2: Assembly language 8086 intermediate

Machine Code 2

There are occasions when the programmer must program at the machine’s own level.

Machine Code programs are tedious to write and highly error prone.

0000111100001111

0010010101010100

1010101010100101

In situations where a high-level language is inappropriate we avoid working in machine code most of the time by making the computer do more of the work. Thus we write in assembly language and then the computer converts this assembly language program into machine code.

Page 3: Assembly language 8086 intermediate

Assembly Language 3

In assembly language, a mneumonic (i.e. memory aid) is used as a short notation for the instruction to be used.

Assembly Machine Code

Language

SUB AX,BX 001010111000011

MOV CX,AX 100010111001000

MOV DX,0 10111010000000000000000

Assembly language is an intermediate step between high level languages and machine code. Most features present in HLL are not present in Assembly Language as type checking etc.

Page 4: Assembly language 8086 intermediate

Compilers / Assemblers 4

High-level Languages such as Pascal programs are sometimes converted firstly to assembly language by a computer program called compiler and then into machine code by another program called assembler

Pascal Program

Assembler language Program

Machine Code Program

This version is actually loaded and

executed

Compiler

Assembler

Page 5: Assembly language 8086 intermediate

General Purpose Registers

There are 4 general

purpose registers in

the 8086.

They are all 16-bit

registers

Each byte can be

addressed individually

by specifying the High

order or the Low order

byte of the register.

5

AX

AH AL

BX

BH BL

CX

CH CL

DX

DH DL

Page 6: Assembly language 8086 intermediate

Some Simple Commands 6

MOV AX,3 ; Put 3 into register AX ADD AX,2 ; Add 2 to the contents of AX MOV BX,AX ; Copy the contents of AX in BX INC CX ; Add 1 to the contents of CX DEC DX ; Subtract 1 from the contents of DX SUB AX,4 ; Subtract 4 from the contents of AX MUL BX ; Multiply the contents of AX with BX leaving

; the answer in DX-AX DIV BX ; Divide the contents of DX-AX by BX leaving

; the quotient in AX and remainder in DX.

Page 7: Assembly language 8086 intermediate

Number Formats 7

0 1 0 1 0 1 0 1

AX

0 0 1 0 0 1 1 1 0 1 0 1 0 1 0 1 MOV AH,01010101B MOV AL,00100111B

0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 MOV AX,3 0 0 0 0 0 0 1 1 0 0 0 0 0 0 1 1 MOV AH,AL 0 0 0 0 1 0 1 0 0 0 0 0 0 0 1 1 MOV AL,10D 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 1 MOV AL,10H

In case a number is moved (copied) into the register the base of a is specified by a letter B for Binary, D for Decimal and H for Hex.

AH AL

Page 8: Assembly language 8086 intermediate

AMBIGUITY 8

Consider the instruction MOV DL, AH Does it mean ‘copy the contents of register AH to DL or Does it mean ‘copy A in hexadecimal into register DL

To avoid this ambiguity all hexadecimal numbers must start with a number. This can always be done by preceding a number starting with A,B,C,D,E and F with a preceding zero to remove ambiguity. Thus MOV DL, AH means copy AH to DL whilst MOV DL, 0AH means sore hexadecimal A to DL

Page 9: Assembly language 8086 intermediate

The Flags Register 9

Some of the instructions (but not all) affect the flag register. The flag register signals the status of the CPU after the last operation performed. For example if SUB AX,2 results in zero the ZF get 1 (lights on) indicating that

the result of the last operation was zero.

Page 10: Assembly language 8086 intermediate

JUMPS 10

Jump instructions allow the 8086 to take decisions according to information provided by the flag register.

For example, if AX and BX contain the ASCII code for the same letter then do one thing, if not then do another.

`

… CMP AX,BX ; Compares the contents of BX with that of AX JE SAME ; Jump if they are equal to the point ; in the code labeled SAME … ; Obey these instructions if the contents of AX … ; is not equal to that of BX SAME: MOV CX,AX ; Program continues from here if AX = BX. …

Page 11: Assembly language 8086 intermediate

Labels 11

We saw that the jump instruction has a general format JE <label> where <label> is a facility offered by the assembler.

These labels are converted by the assembler to exact address where the program is to continue. Labels must start with a letter and can contain thereafter letters, numbers and

underscores (_). Spaces and punctuation marks are not permitted Avoid using keywords in labels Once_again, Next, Name34, this_37 are permitted as labels 3rdday, tues+wed and semi;colons are not permitted as labels.

Page 12: Assembly language 8086 intermediate

JUMP Conditions 12

JA/JNBE (CF and ZF) = 0 Above / Not Below or Equal JAE/JNB CF = 0 Above or Equal / Not Below JB/JNAE/JC CF = 1 Below / Not Above or Equal / Carry JBE/JNA (CF or ZF) = 1 Below or Equal / Not Above JE/JZ ZF = 1 Equal / Zero JMP none Unconditionally JNC CF = 0 No Carry JNE/JNZ ZF = 0 Not Equal / Not Zero JNO OF = 0 No Overflow JNP/JPO PF = 0 No Parity / Parity Odd JNS SF = 0 No Sign / Positive JO OF = 1 Overflow JP/JPE PF = 1 Parity / Parity Even JS SF = 1 Sign JG/JNLE ZF = 0 and SF = OF Greater / Not Less nor Equal JGE/JNL SF = OF Grater or Equal / Not Less JL / JNGE SF <> OF Less / Not Greater nor Equal JLE/JNG (ZF = 1) or (SF <> OF) Less or equal / not greater JCXZ Register CX = 0 CX is equal to zero

Page 13: Assembly language 8086 intermediate

Example using Jumps 13

MOV CX, AX ; Keep a copy of AX before modification SUB AX,BX ; AX := AX – BX JZ MAKE1 ; This is instruction will cause execution ; to continue from MAKE1 if AX was ; equal to BX (subtraction resulted in Zero) MOV DX, 0 ; Otherwise store 0 in DX JMP RESET ; Jump to RESTORE where AX is restored ; thus avoiding the next instruction MAKE1: MOV DX, 1 ; If AX = BX then we set DX to 1 RESET: MOV AX, CX ; Restore the old value of AX

Note that in the Code a colon ends a label position

Page 14: Assembly language 8086 intermediate

The Logical Family 14

OR

Contents of AX = 0000101011100011

Contents of BX = 1001100000100001

Contents of AX = 1001101011100011

after OR AX,BX is executed

AND

Contents of AX = 0000101011100011

Contents of BX = 1001100000100001

Contents of AX = 0000100000100001

after AND AX,BX is executed

XOR

Contents of AX = 0000101011100011

Contents of BX = 1001100000100001

Contents of AX = 1001001011000010

after XOR AX,BX is executed

NOT (Invert: One’s Complement)

Contents of AX = 0000101011100011

Contents of AX = 1111010100011100

after NOT AX is executed

TEST

Contents of AX = 0000101011100011

Contents of BX = 1001100000100001

Contents of AX = 0000101011100011

after TEST AX,BX is executed

Similar to AND but the result is not stored in AX but only the Z-flag is changed

NEG (Two’s Complement)

Contents of AX = 0000101011100011

Contents of AX = 1111010100011101

after NEG AX is executed

Page 15: Assembly language 8086 intermediate

Use of Logical Family 15

Symbol ASCII (Dec) ASCII (Hex) 0 48 30 1 49 31 2 50 32 3 51 33 4 52 34 5 53 35 6 54 36 7 55 37 8 56 38 9 57 39

By Making an AND between an ASCII value and 0FH we can obtain the required number. Say we AND 33H = 00110011B

with 0FH = 00001111B

We obtain = 00000011B (3)

By Making an OR between a number value and 30H we can obtain its ASCII code. Say we OR 05H = 00000101B

with 30H = 00110101B

We obtain = 00110101B

(ASCII value for ‘5’)

Page 16: Assembly language 8086 intermediate

Masking 16

By the use of masking we can set or test individual bits of a register

Suppose we want to set the 3rd.

bit of AX to 1 leaving the

others unchanged.

AX = 0101010100011001

04H = 0000000000000100

OR AX,04H = 0101010100011101

Suppose we want to set the 5th.

bit of AX to 0 leaving the

others unchanged.

AX = 0101010100011001

0FFEFH = 1111111111101111

AND AX,0FFEFH = 0101010100001101

Suppose we want to test the if the

6th. bit of AX is 1 or 0:

AX = 0101010100011001

20H = 0000000000100000

AND AX,20H = 0000000000000000

So if the result is 0 then that

particular bit was 0, 1 otherwise

Page 17: Assembly language 8086 intermediate

Instructions which affect Memory 17

Computer memory is best thought of numbered pigeon holes (called locations), each capable of storing 8 binary digits (a byte)

[0000] [0001] [0002] [0003] [0004] [0005] [0006] [0007] [0008] [0009] [000A] [000B] [000C]

Data can be retrieved from memory, one or two bytes at a time: MOV AL, [20H] will transfer the Contents of location 20H to AL. MOV BX, [20H] will transfer the contents of locations 20H and 21H to BX. MOV [20H], AL will transfer the contents of AL to memory location 20H

Location ADDRESS Location CONTENTS

Page 18: Assembly language 8086 intermediate

Changing addresses 18

Varying an address whilst a program is running involves specifying the locations concerned in a register.

From all the general purpose registers BX is the only capable of storing such addresses. Thus MOV AX, [CX] is illegal Whilst MOV CL, [BX] copies the contents of memory location whose address is

specified by BX into the register CL. And MOV [BX], AL copies the contents of AL in the memory location whose

address is specified in BX

Page 19: Assembly language 8086 intermediate

Examples Affecting Memory 19

Consider the checkerboard memory test where a section of memory is filled with alternate 01010101 and 10101010.

The following program does the checkerboard test on locations 200H-300H inclusive.

MOV BX,200H

MOV AX,1010101001010101B

NEXT: MOV [BX],AX

INC BX

CMP BX,300H

JLE NEXT

Page 20: Assembly language 8086 intermediate

The Instruction Pointer (IP) 20

The computer keeps track of the next line to be executed by keeping its address in a special register called the Instruction Pointer (IP) or Program Counter.

This register is relative to CS as segment register and points to the next instruction to be executed.

The contents of this register is updated with every instruction executed.

Thus a program is executed sequentially line by line

MOV AX,BX

MOV CX,05H

MOV DX,AX

START

.

.

.

.

.

.

This is the

line which is

executing

IP

Page 21: Assembly language 8086 intermediate

The Stack 21

The Stack is a portion of memory which, like a stack of plates in a canteen, is organized on a Last-In-First-Out basis.

Thus the item which was put last on the stack is the first to be withdrawn

Page 22: Assembly language 8086 intermediate

The Stack Pointer 22

The Stack pointer keeps track of the position of the last item placed on the stack (i.e. the Top Of Stack)

The Stack is organized in words, (i.e. two

bytes at a time). Thus the stack pointer is incremented or decremented by 2.

The Stack Pointer points to the last occupied locations on the stack

[0000] [0002] [0004] [0006] [0008] [000A] [000C] [000E] [0010] [0012] [0014] [0016] [0018]

SP

Note that on placing items on the stack the address decreases

Page 23: Assembly language 8086 intermediate

PUSH & POP 23

The two set of instructions which explicitly modify the stack are the PUSH (which places items on the stack) and the POP (which retrieves items from the stack). In both cases, the stack pointer is adjusted accordingly to point always to the top of stack.

Thus PUSH AX means SP=SP-2 and AX -> [SP]

POP AX means [SP] -> AX and SP=SP+2.

[0000] [0002] [0004] [0006] [0008] [000A] [000C] [000E] [0010] [0012] [0014] [0016] [0018]

OLD SP

PUSH AX

OLD SP NEW SP

AX

[0000] [0002] [0004] [0006] [0008] [000A] [000C] [000E] [0010] [0012] [0014] [0016] [0018]

OLD SP

POP AX

OLD SP NEW SP

AX

Page 24: Assembly language 8086 intermediate

Subroutines 24

In high-level languages, procedures make it possible to break a large program down into smaller pieces so that each piece can be shown to work independently. In this way the final program is built up of a number of trusty bricks and is easier to debug because the error is either localized to one subprogram or its interlinking. This has also the advantage of re-usability of bricks.

CALL SUB1

START SUB1 PROC

RET

.

.

.

.

.

.

.

.

.

Page 25: Assembly language 8086 intermediate

The CALL Mechanism 25

Although at first sight the CALL and RET mechanism can be implemented by using two JMP’s. In fact this cannot be done since the CALL mechanism remembers the place where it was called from and returns to the line following it. Thus this is not a fixed address.

CALL SUB1

START SUB1 PROC

RET

.

.

.

.

.

.

.

.

.

CALL SUB1 .

.

.

1

2

Page 26: Assembly language 8086 intermediate

The Return Mechanism 26

When a CALL is encountered the current value of the instruction pointer is pushed on the stack and the it is filled with the address stated by the call.

Since the fetch cycle goes to search for the instruction pointed at by the instruction pointer, the program continues it’s execution from the first statement in the subroutine.

On encountering the RET instruction the contents of the IP is popped from the stack thus continuing the execution where it was suspended.

Thus care must be taken to leave the return address intact before leaving a subroutine. (i.e. a symmetrical number of pushes and pops within the subroutine)

Page 27: Assembly language 8086 intermediate

Software Interrupts 27

Software interrupts are like hardware interrupts which are generated by the program itself. From the interrupt number, the CPU derives the address of the Interrupt service routine which must be executed.

Software interrupts in assembly language can be treated as calls to subroutines of other programs which are currently running on the computer.

One of the most famous software interrupt is Interrupt No. 21H, which branches in the operating system, and permits the use of PC-DOS functions defined there. The function required to be performed by DOS is specified in AH prior to the the

interrupt. The functions return and accept values in various registers. AN interrupt is called using the instruction INT followed by the interrupt number

. For example: INT 21H

Page 28: Assembly language 8086 intermediate

Some INT 21H functions 28

Function Number

Description Explanation

1 Keyboard Input (echoed)

Waits until a character is typed at the keyboard and then puts the ASCII code for that character in register AL and echoed to screen

2 Display Output

Prints the character whose ASCII code is in DL

8 Keyboard Input (No echo)

Waits until a character is typed at the keyboard and then puts the ASCII code for that character in register AL and NOT echoed to screen

9 Display String

Prints a series of characters stored in memory starting with the one in the address given in DX (relative to DS).Stop when the ASCII code for $ is encountered

Page 29: Assembly language 8086 intermediate

INT 21H Example 29

Prompt DB ‘Please enter 1 or 2: ‘,13D,10D,’$’

Song1 DB ‘So you think you can tell heaven from hell’

Song2 DB ‘Blue Sky is in pain’,13D,10D,’$’

ASK: MOV DX, OFFSET Prompt

MOV AH,09H

INT 21H

GET: MOV AH,01H

INT 21H

CMP AL,01H

JE NEXT

MOV DX, OFFSET Song1

MOV AH,09H

INT 21H

This is only a

program fragment to

illustrate the use of

interrupt 21H – For

full details consult the

MASM notes

Page 30: Assembly language 8086 intermediate

Addition and Subtraction with carry or

borrow 30

In assembly language there are two versions of addition and two versions of subtraction. ADD - Simple addition of two numbers ADC - Adds two numbers together with

the carry flag SUB – Simple subtraction of two

numbers SBB – Subtracts the second number and

the carry flag (borrow) This provides a means of adding numbers

greater than 32-bits. CLC clears the carry for the first digit

addition

CF CF

0

0

0 1 1

00 01 98 41 +

00 02 71 64

00 04 70 05

Last

addition in

case of an

outgoing

carry

Page 31: Assembly language 8086 intermediate

The Compare Instruction 31

The compare instruction does not change the contents of the registers involved but only sets the flag register accordingly.

The actual operation performed by the compare is a subtraction, leaving the source and destination registers intact

Consider CMP AX,BX : Flags are set according to the result of subtracting BX from AX: If AX = BX then the ZF is set to 1 If AX > BX then the ZF is set to 0 and CF is set to 0 too If AX < BX then we need an external borrow, which is reflected in CF = 1 These flags are tested in the ABOVE or BELOW jumps which test unsigned numbers The GREATER and LESS jumps are for signed numbers and work on the SF, OF

and the ZF instead

Page 32: Assembly language 8086 intermediate

Addressing Modes

Computer Logic II

32

The addressing modes deal with the source and destination of the data required by the instruction. This can be either a register or a location in memory, or even a port.

Various addressing modes exist: Register Addressing Immediate and Direct Addressing Indirect Addressing Indexed Addressing Based Addressing Based-Indexed Addressing

Page 33: Assembly language 8086 intermediate

Register Addressing 33

This addressing mode involves the contents of the register directly as for example: MOV AX, BX MOV CL, DL

Note that the IP and Flags register cannot be accessed directly by the programmer

AX AH AL

BX BH BL

CX CH CL

DX DH DL

SI

DI

SP

BP

IP

FLAGS

General Purpose

SS

CS

DS

ES

Segment Registers

Ex. MOV AX,BX AX BX

Page 34: Assembly language 8086 intermediate

Immediate and Direct Addressing 34

In Immediate addressing – for example MOV CL,61H – the immediate operand 61H is stored as part of the instruction. Thus the number 61H is loaded directly in CL.

Direct addressing is similar except that in this case the effective address of one of the operands is taken directly from the instruction. Thus in MOV AL, [210H] the contents of location 210H relative to DS is put in AL

CL

Ex. MOV CL,61H

61H

Ex. MOV AL,[210H]

AL

(DS:210H) 75H

Page 35: Assembly language 8086 intermediate

Default Segment Register 35

AX AH AL

BX BH BL

CX CH CL

DX DH DL

SI

DI

SP

BP

IP

General Purpose

Relative to DS by default

Relative to DS by default

Relative to SS by default

Relative to SS by default

Relative to CS by default

Relative to DS by default

NORMALLY FOR STRINGS

DS

ES

Note that the default segment register can be changed using the segment override, i.e. stating the whole address in the form DS: Offset

Page 36: Assembly language 8086 intermediate

Some other useful Instructions 36

CLC: Clear Carry Flag (CF = 0) STC: Set Carry Flag (CF = 1) CMC : Complement Carry Flag (CF = CF) CBW: Convert Byte to Word CWD: Convert Word to Double-Word NEG: Negate (2’s Complement) NOT: Compliment (1’s Complement)

Page 37: Assembly language 8086 intermediate

Reference Books 37

Programming the 8086/86 for the IBM PC and Compatibles . Michael Thorne Microprocessors and Interfacing – Programming and Hardware – Douglas V.Hall Microsoft Macro Assembler – for the MS-DOS Operating Systems – Reference

Manual