8086 ASSEMBLY LANGUAGE

PROGRAMMING

Cutajar & Cutajar

© 2012

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.

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.

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

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

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.

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

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

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.

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. …

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.

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

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

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

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’)

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

The Shift Family 17

There are two different sets of shift instructions One set for doubling and halving unsigned binary numbers

SHL (Shift Left) – doubles SHR (Shift Right) - halves

The other for doubling and halving signed binary numbers SAL (Arithmetic Shift Left) – doubles SAR (Arithmetic Shift Right) – halves

CF 0

MSB

CF 0

MSB

SHL/SAL

SHR

CF MSB

SAR

Shift Examples 18

Instruction CL Initial Contents Final Contents

Decimal Binary Decimal Binary CF

SHR AL,1 250 11111010 125 01111101 0

SHR AL,CL 3 250 11111010 31 00011111 0

SHL AL,1 23 00010111 46 00101110 0

SHL BL,CL 2 23 00010111 92 01011100 0

SAL BL,1 +23 00010111 +46 00101110 0

SAL DL,CL 4 +3 00000011 +48 00110000 0

SAR AL,1 -126 10000010 -63 11000001 0

SAR AL,CL 2 -126 10000010 -32 11100000 1

The Rotate Family 19

The rotate is similar to the shift with the exception that the outgoing bit is not lost but rotated back into the shifted register.

An Alternative is to rotate through carry, which includes the carry in the rotation process.

CF LSB

ROL

CF MSB

ROR

CF MSB

RCR

CF LSB

RCL

Rotates Examples 20

Instruction CL Initial Contents Final Contents

CF Binary Binary CF

ROR AL,1 0 11111010 01111101 0

ROR AL,CL 3 1 11111010 01011111 0

ROL AL,1 0 00010111 00101110 0

ROL BL,CL 2 1 00010111 01011100 0

RCL BL,1 0 00010111 00101110 0

RCL DL,CL 4 1 00000011 00111000 0

RCR AL,1 1 10000010 11000001 0

RCR AL,CL 2 0 10000010 00100000 1

Instructions which affect Memory 21

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

Changing addresses 22

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

Examples Affecting Memory 23

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

The DS Register 24

The 8086 can address a total of 1 Megabyte. Rather than representing each address as a 20-bit unsigned number, memory is thought of as being divided uo into segments each of which contains 216 locations.

In this way an address can be thought of as consisting of two parts: a 16-bit segment address and a 16-bit offset from the start of the segment.

Thus , 020A:1BCD denotes offset 1BCDH from the start of segment 020AH.

Effective Address 25

The actual address is calculated from the segment and offset values as follows:

1. Add a zero to the right-hand side of the segment register.

2. Add the offset to this.

Logical Address

Segment Register 0000

ADDER

0 15

15 0

Physical Memory Address

0 19

Offset

Example = 020A:1BCD

Segment = 020AH -> 020A0H +

Offset = 1BCDH -> 1BCDH

Address = 03C6DH

Thus if DS = 500H the instruction MOV AX,[200H] would actually move The contents of location 5200H

The Instruction Pointer (IP) 26

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

The Stack 27

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

The Stack Pointer 28

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

PUSH & POP 29

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

Subroutines 30

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

.

.

.

.

.

.

.

.

.

The CALL Mechanism 31

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

The Return Mechanism 32

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)

NEAR and FAR 33

When a procedure lies within the same segment as the calling program (intra-segment) it can be declared as NEAR. Thus the return address can be specified as just an offset thus needing 1 word

When the does not lie in the same segment as the

calling program (inter-segment) it is declared as FAR. Thus the return address must specify the segment and the offset, thus occupying 2 words.

Offset

Offset

Segment

Register Parameters 34

The easiest way to pass a parameter to and fro a subprogram is by the use of the general purpose registers.

ADDITION PROC NEAR ;Procedure to add two numbers

MOV AX,BX ;First parameter in BX

ADD AX,CX ;Add second parameter in CX

RET ;Return ( result in AX )

ADDITION ENDP ;End of procedure definition

…

MOV BX,4 ;Assign first parameter

MOV CX,7 ;Assign second parameter

CALL ADDITION ;Call their addition

MOV [RES],AX ;Store the result returned

Parameters on STACK 35

Passing parameters in registers is very straightforward but limiting the number of parameters passed.

When several parameters are to be passed to the subroutine, they are pushed on the stack prior to the subroutine call.

SP Return Address

For the CALL

Parameters

Items on the

stack prior to

CALL

Retrieving parameters from the Stack 36

Accessing parameters placed on the stack cannot easily be done using PUSH and POP since the Stack Pointer must point to the return address.

This is done using the Base Pointer (BP) which uses Stack Segment by default.

First the SP is copied in the BP then the parameter is accessed using an offset from BP

Ex. MOV AX,[BP+2]

SP/BP Return Address

For the CALL

Parameters

Items on the

stack prior to

CALL

[BP+2]

[BP+4]

[BP+6]

Local Variables 37

Local variables are pushed on the stack and can be accessed using a negative displacement

Ex MOV AX, [BP-2] retrieves the first local parameter from the stack.

SP/BP Return Address

For the CALL

Parameters

Items on the

stack prior

to CALL

[BP+2]

[BP+4]

[BP+6]

Local

Variables

[BP-2]

[BP-4]

[BP-6]

Discarding Parameters after CALL 38

The RET command has an option to discard the parameters previously pushed on the stack

Example: RET 6 This discards the return

address and an additional 6 bytes.

In this manner the SP returns to the position it occupied before the parameters were pushed.

Return Address

For the CALL

Parameters

Items on the

stack prior to

CALL

NEW SP

OLD SP

Example – Factorial (recursive call) 39

FACT PROC NEAR

CMP BX,1 ;Input parameter (n) is in BX

JNE RECUR ;If n <> 1 then recurse

MOV AX,1 ;else return(1)

JMP DONE ;

RECUR: PUSH BX ;Store temporarily the value of n

DEC BX ;

CALL FACT ;Call FACT(n-1) returning value AX

POP BX ;Recall n

IMUL BX ;FACT:= n*FACT(n-1)

DONE: RET

FACT ENDP

Software Interrupts 40

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

Some INT 21H functions 41

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

INT 21H Example 42

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

Number Representation 43

One of the disadvantage of reading numbers as a series of ASCII codes for digits is that, before any arithmetic can be performed on such numbers, their binary equivalents have to be calculated.

Thus, if the decimal number 37 was typed at the keyboard, the ASCII codes for 3 and 7 (33H and 37H) would have to be converted to their binary equivalents, and then the binary equivalents, and then the binary equivalent of the first digit multiplied by ten and added to the second.

BCD is a way of representing numbers which avoids the need for conversions of this sort.

3 7

33H 37H

3 7

30

+

37

x10

ASCII

KEYS

BINARY

Binary Coded Decimal (BCD) 44

BCD is a way of representing numbers which avoids the need of a lot of conversions.

The principle used is to encode each decimal digit separately in their unsigned 4-bit equivalents.

Since the computer memory is organized in bytes of 8 bits, we can represent BCD digits as: Packed BCD : where two digits are packed in a byte Ex.

37 = 0011 0111 Unpacked BCD: where each digit is expanded on 8 bits.

Ex 37 = 00000011 000001111

BCD Binary 0 0000 1 0001 2 0010 3 0011 4 0100 5 0101 6 0110 7 0111 8 1000 9 1001

Arithmetic Operations 45

At first sight we can see that there are 6 unused binary patterns in BCD corresponding to the hex 6 letter digits A, B, C, D, E and F.

Thus this system is less compact. In the arithmetic operations it also involves some

complications if the answer contains any of the bit patterns not represented in BCD

Not all is lost however, additional instructions exist to overcome this problem

Binary BCD 0000 0 0001 1 0010 2 0011 3 0100 4 0101 5 0110 6 0111 7 1000 8 1001 9 1010 Unused 1011 Unused 1100 Unused 1101 Unused 1110 Unused 1111 Unused

BCD Addition 46

If the addition doesn’t produce a result which passes through the forbidden range, no problem arises.

If however the result produced contains a forbidden digit, it has to be adjusted.

This is done by adding another 6 to the result to overcome the forbidden range.

This adjustment is done automatically using the DAA instruction for packed and the AAA instruction for unpacked BCD via the AL register.

The Auxiliary carry exist in the flag register to indicate a carry from the least significant BCD digit to the most significant.

24 0010 0100 +

13 0001 0011

37 0011 0111 OK

19 0001 1001 +

24 0010 0100

3? 0011 1101 + NOT OK

06 0000 0110 ADJUST

43 0100 0011

AUXILIARY CARRY

1

19 0001 1001 +

18 0001 1000

31 0011 0001

BCD Subtraction 47

If the subtraction doesn’t produce a result which passes through the forbidden range or no borrow is required no problem arises.

If however the result produced contains a forbidden digit, it has to be adjusted.

This is done by subtracting another 6 to the result to overcome the forbidden range.

This adjustment is done automatically using the DAS instruction for packed and the AAS instruction for unpacked BCD

The Auxiliary carry here become a borrow and the result must be adjusted whenever this carry is set since the borrow is worth 16 not 10.

24 0010 0100 -

13 0001 0011

11 0001 0001 OK

35 0011 0101 -

16 0001 0110

1? 0001 1111 - NOT OK

06 0000 0110 ADJUST

19 0001 1001

1 AUXILIARY

21 0010 0001 -

19 0001 1001

08 0000 1000 - BORROW

06 0000 0110 ADJUST

02 0000 0010

Example 1 : Addition 48

Suppose we have a long packed BCD number contained in a number of memory locations, the number of these locations is contained in CX and the first number starts at 400H and the second at 500H. And the result is to be stored in the locations at 600H.

Here we use SI (Source Index) and DI (Destination Index) registers which are two 16-bit registers which like BX can address the memory.

CX:0003H Length of numbers

SI:0400H Number 1

DI:0500H Number 2

BX:0600H Answer

Location Contents

400H 41 L.S.D.

401H 98

402H 01

403H ??

.

500H 64 L.S.D.

501H 71

502H 02

.

600H 05

601H 70

602H 04

603H 00

After

execution

Example 1: Code 49

CLC ;Clear carry for first digit

NEXT: MOV AL,[SI] ;Get digit

ADC AL,[DI] ;Add corresponding digit

DAA ;Adjust for BCD

MOV [BX],AL ;Store answer digit

INC SI ;Increment pointers

INC DI

INC BX

DEC CX ;decrement counter

JNZ NEXT ;do next digit

MOV AL,0 ;adjust last digit

ADC AL,0

MOV [BX],AL

Addition and Subtraction with carry or

borrow 50

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 SBC – 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

Example 1 : Multiplication 51

Suppose we have a long unpacked BCD number contained in a number of memory locations, the number of these locations is contained in CX and the first number starts at 400H and the second is in DL. And the result of their multiplication is to be stored in the locations at 600H.

CX:0004H Length of numbers

SI:0400H Multiplicand

DL:06H Multiplier

DI:0600H Answer

Location Contents

400H 01 L.S.D.

401H 09

402H 02

403H 08

.

600H 06

601H 04

602H 07

603H 09

604H 04

After

execution

Example 2: Procedure 52

Remembering the multiplication algorithm, applied to 8291*6 is: 1. 6 times 1 is 6, we write down 6 and carry 0 2. 6 times 9 is 54, we add the previous carry (0) get 54, we write down 4 and carry 5. 3. 6 times 2 is 12, we add the previous carry (5) get 17, we write down 7 and carry 1. 4. 6 times 8 is 48, we add the previous carry (1) get 49, we write down 9 and carry 4 5. Since it was the last digit to be multiplied we just write down 4 4 1 5 0

8 2 9 1 x 6

6

4 9 7 4 6

Example 2: Code 53

MOV AL,00H

MOV [DI],AL ;Set “previous carry” to zero

NEXT: MOV AL,[SI] ;Get digit of Multiplicand

MUL DL ;Multiply by Multiplier

AAM ;ASCII adjust digit

ADD AL,[DI] ;Add “previous carry”

AAA ;ASCII adjust

MOV [DI],AL ;Write down Digit

INC DI ;point to next digits

INC SI

MOV [DI],AH ;Store “previous carry”

DEC CX ;Check if ready

JNZ NEXT

The Compare Instruction 54

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

Addressing Modes

Computer Logic II

55

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

Register Addressing 56

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

Immediate and Direct Addressing 57

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

Indirect Addressing 58

With indirect addressing, the effective address is found in either the BX, SI or DI registers. I.e. the effective address is not found directly in the instruction itself but indirectly by accessing a register, as in: MOV DL, [BX]

Note that this method is useful to pass parameters to subroutines by reference instead of by value.

Example MOV DL,[BX]

Say BX contains the address 0200H

relative to DS

DL

(DS:200H) 69H BX

Indexed Addressing 59

With indexed addressing, the effective address is calculated by the addition of an index register + displacement.

For this purpose two index registers exist SI (source index) and DI (destination index) By default SI and DI relative to DS if not

for string handling, in which case SI is relative to DS and DI is relative to ES.

Example MOV AL, ACCOUNT[SI] This adds the address of account to SI to obtain the effective address where data is to be retrieved from.

An alternative notation is MOV AL,[SI+BALANCE]

Ex. MOV AL,ACCOUNT[SI]

Assume that ACCOUNT has

an offset of 0200H relative to

DS and SI contains 05H

AL

205H 75H

204H

203H

202H

201H

ACCOUNT (DS:200H)

BASED ADDRESSING 60

In based addressing BX or BP are used as a variable base of the address from where the data is to be retrieved or stored. An offset can be added to this base address

Addresses given in BX are taken relative to DS whereas those in BP are taken relative to SS.

Example MOV AL, [BX+05H]

Ex. MOV AL,[BX+05H]

Assume that BX contains

0200H

AL

205H 75H

204H

203H

202H

201H

(DS:200H) BX

Based-Indexed Addressing 61

Based-Indexed addressing is a combination of the previous two addressing modes.

The effective address is calculated by summing up the contents of the base register together with the contents of the index register and the given displacement.

Example: MOV AL, [BX+SI+2]

Ex. MOV AL,[BX+SI+02H]

Assume that BX contains

0200H and SI contains 3

AL

205H 75H

204H

203H

202H

201H

(DS:200H) BX

SI

02H

Note that although in these examples the MOV instruction is always considered, this can apply to several instructions and the address calculation can apply both to the source and destination of data.

Default Segment Register 62

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

Some other useful Instructions 63

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)

Reference Books 64

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