Assembly language 8086

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
In situations where a high-level
0010010101010100
language is inappropriate we
1010101010100101
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 Program
Pascal programs are sometimes
Compiler
converted firstly to assembly
language by a computer program Assembler language Program
called compiler and then into
machine code by another Assembler
program called assembler Machine Code Program

This version is
actually loaded and
executed

General Purpose Registers
5

AH AL
 There are 4 general
purpose registers in AX
the 8086.
BH BL
 They are all 16-bit
registers
 Each byte can be BX
addressed individually CH CL
by specifying the High
order or the Low order
byte of the register. CX
DH DL

DX

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
AX
AH AL
MOV AH,01010101B 0 1 0 1 0 1 0 1
MOV AL,00100111B 0 1 0 1 0 1 0 1 0 0 1 0 0 1 1 1
MOV AX,3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1
MOV AH,AL 0 0 0 0 0 0 1 1 0 0 0 0 0 0 1 1
MOV AL,10D 0 0 0 0 0 0 1 1 0 0 0 0 1 0 1 0
MOV AL,10H 0 0 0 0 0 0 1 1 0 0 0 1 0 0 0 0

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.

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

AND NOT (Invert: One’s Complement)
Contents of AX = 0000101011100011 Contents of AX = 0000101011100011
Contents of BX = 1001100000100001 Contents of AX = 1111010100011100
Contents of AX = 0000100000100001 after NOT AX is executed
after AND AX,BX is executed

OR TEST
Contents of BX = 1001100000100001 Contents of BX = 1001100000100001
after OR AX,BX is executed after TEST AX,BX is executed

XOR Similar to AND but the result is not stored in AX
Contents of AX = 0000101011100011 but only the Z-flag is changed
Contents of BX = 1001100000100001 NEG (Two’s Complement)
after XOR AX,BX is executed Contents of AX = 1111010100011101
after NEG AX is executed

Use of Logical Family
15

Symbol ASCII (Dec) ASCII (Hex) By Making an AND between an ASCII value and 0FH
0 48 30 we can obtain the required number.
1 49 31 Say we AND 33H = 00110011B
2 50 32 with 0FH = 00001111B
3 51 33
We obtain = 00000011B (3)
4 52 34
5 53 35 By Making an OR between a number value and 30H we
6 54 36 can obtain its ASCII code.
7 55 37 Say we OR 05H = 00000101B
8 56 38 with 30H = 00110101B
We obtain = 00110101B
9 57 39
(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.
Suppose we want to test the if the
AX = 0101010100011001
6th. bit of AX is 1 or 0:
04H = 0000000000000100
AX = 0101010100011001
OR AX,04H = 0101010100011101 20H = 0000000000100000
AND AX,20H = 0000000000000000
Suppose we want to set the 5th. So if the result is 0 then that
bit of AX to 0 leaving the particular bit was 0, 1 otherwise
others unchanged.
AX = 0101010100011001
0FFEFH = 1111111111101111
AND AX,0FFEFH = 0101010100001101

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

0
SHL/SAL CF
MSB
MSB
0
CF CF
MSB
SHR 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.

ROR ROL
MSB LSB
CF CF

RCR RCL
MSB LSB
CF CF

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)
Data can be retrieved from memory, one or
[0000]
two bytes at a time: [0001]
[0002]
MOV AL, [20H] will transfer the [0003]
Contents of location 20H to AL. [0004]
[0005]
MOV BX, [20H] will transfer the contents of [0006]
[0007]
locations 20H and 21H to BX. [0008]
MOV [20H], AL will transfer the contents of [0009]
[000A]
AL to memory location 20H [000B]
Location ADDRESS [000C]
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 Offset
segment and offset values as follows: 15 0
Logical Address
1. Add a zero to the right-hand 15 0
side of the segment register. Segment Register 0000
2. Add the offset to this.
Example = 020A:1BCD
Segment = 020AH -> 020A0H + ADDER
Offset = 1BCDH -> 1BCDH
Address = 03C6DH
19 0
Thus if DS = 500H the instruction Physical Memory Address
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 START
. This is the
register called the Instruction Pointer (IP) or line which is
Program Counter. .
. executing
 This register is relative to CS as segment
register and points to the next instruction to MOV AX,BX
be executed. MOV CX,05H
 The contents of this register is updated with MOV DX,AX IP
every instruction executed.
.
 Thus a program is executed sequentially line
.
by line
.

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

[0000]
[0002]
 The Stack pointer keeps track of the [0004]
position of the last item placed on the [0006]
[0008]
stack (i.e. the Top Of Stack) [000A]
[000C]
SP [000E]
[0010]
[0012]
[0014]
 The Stack is organized in words, (i.e. two [0016]
bytes at a time). Thus the stack pointer is [0018]
incremented or decremented by 2. Note that on placing items on the
 The Stack Pointer points to the last stack the address decreases
occupied locations on the stack

PUSH & POP
29

PUSH AX AX
 The two set of instructions which [0000]
[0002]
[0004]
explicitly modify the stack are the [0006]
[0008] NEW SP
PUSH (which places items on the OLD SP [000A]
[000C]
stack) and the POP (which [000E]
[0010]
retrieves items from the stack). In [0012]
[0014]
both cases, the stack pointer is [0016]
[0018]
adjusted accordingly to point
POP AX
always to the top of stack. [0000]
AX
[0002]
 Thus PUSH AX means SP=SP-2 [0004]
[0006]
and AX -> [SP] [0008]
[000A]
OLD SP NEW SP
[000C]
 POP AX means [SP] -> AX and [000E]
[0010]
SP=SP+2. [0012]
[0014]
[0016]
[0018]

Subroutines
30

 In high-level languages, procedures START SUB1 PROC
.
make it possible to break a large . .
program down into smaller pieces so . .
. RET
that each piece can be shown to work
independently. In this way the final CALL SUB1
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.

The CALL Mechanism
31

 Although at first sight the CALL START SUB1 PROC
and RET mechanism can be .
. .
implemented by using two JMP’s. . .
In fact this cannot be done since . RET
the CALL mechanism remembers CALL SUB1
the place where it was called from 1
.
and returns to the line following it. .
Thus this is not a fixed address. .
CALL SUB1
.
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 Offset
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 Offset
segment and the offset, thus occupying 2 words. 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 SP
Return Address
For the CALL
the number of parameters
passed.
 When several parameters are to
Parameters
be passed to the subroutine, they
are pushed on the stack prior to
the subroutine call.
Items on the
stack prior to
CALL

Retrieving parameters from the Stack
36

 Accessing parameters placed
on the stack cannot easily be
Return Address
done using PUSH and POP SP/BP For the CALL
since the Stack Pointer must
point to the return address. [BP+2]
 This is done using the Base
Pointer (BP) which uses [BP+4] Parameters
Stack Segment by default.
 First the SP is copied in the [BP+6]
BP then the parameter is Items on the
accessed using an offset from stack prior to
BP CALL
 Ex. MOV AX,[BP+2]

Local Variables
37

 Local variables are pushed [BP-6]
on the stack and can be [BP-4] Local
accessed using a negative Variables
[BP-2]
displacement Return Address
SP/BP
 Ex MOV AX, [BP-2] For the CALL

retrieves the first local [BP+2]

parameter from the stack. [BP+4] Parameters

[BP+6]

Items on the
stack prior
to CALL

Discarding Parameters after CALL
38

 The RET command has an
option to discard the Return Address
OLD SP
parameters previously pushed For the CALL
on the stack
 Example: RET 6
Parameters
 This discards the return
address and an additional 6
bytes.
Items on the
 In this manner the SP NEW SP
stack prior to
returns to the position it CALL

occupied before the
parameters were pushed.

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 Description Explanation
Number
1 Keyboard Waits until a character is typed at the keyboard and then puts the ASCII
Input code for that character in register AL and echoed to screen
(echoed)
2 Display Prints the character whose ASCII code is in DL
Output

8 Keyboard Waits until a character is typed at the keyboard and then puts the ASCII
Input code for that character in register AL and NOT echoed to screen
(No echo)
9 Display Prints a series of characters stored in memory starting with the one in the
String 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
This is only a
INT 21H
program fragment to
illustrate the use of
GET: MOV AH,01H
interrupt 21H – For
INT 21H
full details consult the
MASM notes
CMP AL,01H
JE NEXT
MOV DX, OFFSET Song1
MOV AH,09H
INT 21H

Number Representation
43

3 7 KEYS
 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, 33H 37H ASCII
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 3 7 BINARY
37H) would have to be converted to their binary x10
equivalents, and then the binary equivalents, and
then the binary equivalent of the first digit 30
multiplied by ten and added to the second. +
 BCD is a way of representing numbers which
avoids the need for conversions of this sort.
37

Binary Coded Decimal (BCD)
44

BCD Binary
 BCD is a way of representing numbers which avoids the
need of a lot of conversions. 0 0000
 The principle used is to encode each decimal digit separately 1 0001
in their unsigned 4-bit equivalents. 2 0010
 Since the computer memory is organized in bytes of 8 bits, 3 0011
we can represent BCD digits as: 4 0100
 Packed BCD : where two digits are packed in a byte Ex. 5 0101
37 = 0011 0111 6 0110
 Unpacked BCD: where each digit is expanded on 8 bits. 7 0111
Ex 37 = 00000011 000001111 8 1000
9 1001

Arithmetic Operations Binary BCD

45
0000 0
0001 1
0010 2
 At first sight we can see that there are 6 unused 0011 3
binary patterns in BCD corresponding to the hex 0100 4
6 letter digits A, B, C, D, E and F. 0101 5
 Thus this system is less compact. 0110 6
0111 7
 In the arithmetic operations it also involves some
1000 8
complications if the answer contains any of the 1001 9
bit patterns not represented in BCD 1010 Unused
 Not all is lost however, additional instructions 1011 Unused
exist to overcome this problem 1100 Unused
1101 Unused
1110 Unused
1111 Unused

BCD Addition
46

 If the addition doesn’t produce a result which passes 24 0010 0100 +
through the forbidden range, no problem arises. 13 0001 0011
 If however the result produced contains a forbidden 37 0011 0111 OK
digit, it has to be adjusted.
19 0001 1001 +
 This is done by adding another 6 to the result to 24 0010 0100
overcome the forbidden range. 3? 0011 1101 + NOT OK
 This adjustment is done automatically using the DAA 06 0000 0110 ADJUST
43 0100 0011
instruction for packed and the AAA instruction for
unpacked BCD via the AL register. AUXILIARY CARRY
 The Auxiliary carry exist in the flag register to indicate 1
a carry from the least significant BCD digit to the 19 0001 1001 +
most significant. 18 0001 1000
31 0011 0001

BCD Subtraction
47

24 0010 0100 -
 If the subtraction doesn’t produce a result which passes 13 0001 0011
through the forbidden range or no borrow is required no 11 0001 0001 OK
problem arises.
 If however the result produced contains a forbidden digit, it 35 0011 0101 -
has to be adjusted. 16 0001 0110
 This is done by subtracting another 6 to the result to 1? 0001 1111 - NOT OK
overcome the forbidden range. 06 0000 0110 ADJUST
 This adjustment is done automatically using the DAS 19 0001 1001
instruction for packed and the AAS instruction for
unpacked BCD 1 AUXILIARY
21 0010 0001 -
 The Auxiliary carry here become a borrow and the result
19 0001 1001
must be adjusted whenever this carry is set since the borrow
08 0000 1000 - BORROW
is worth 16 not 10.
06 0000 0110 ADJUST
02 0000 0010

Example 1 : Addition
48

 Suppose we have a long packed BCD CX:0003H Length of numbers
SI:0400H Number 1
number contained in a number of DI:0500H Number 2
memory locations, the number of BX:0600H Answer
these locations is contained in CX
and the first number starts at 400H Location Contents
and the second at 500H. And the 400H 41 L.S.D.
401H 98
result is to be stored in the locations 402H 01
at 600H. 403H ??
 Here we use SI (Source Index) and .
500H 64 L.S.D.
DI (Destination Index) registers 501H 71
which are two 16-bit registers which 502H 02
like BX can address the memory. .
600H 05
601H 70 After
602H 04 execution
603H 00

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
50
borrow
 In assembly language there are two versions
of addition and two versions of subtraction. CF CF
 ADD - Simple addition of two numbers
0
 ADC - Adds two numbers together with
the carry flag 0

 SUB – Simple subtraction of two
numbers
 SBC – Subtracts the second number and Last 0 1 1
the carry flag (borrow) addition in 00 01 98 41 +
 This provides a means of adding numbers case of an 00 02 71 64
outgoing 00 04 70 05
greater than 32-bits.
carry
 CLC clears the carry for the first digit
addition

Example 1 : Multiplication
51

 Suppose we have a long unpacked
CX:0004H Length of numbers
BCD number contained in a number SI:0400H Multiplicand
of memory locations, the number of DL:06H Multiplier
these locations is contained in CX DI:0600H Answer
and the first number starts at 400H Location Contents
and the second is in DL. And the 400H 01 L.S.D.
result of their multiplication is to be 401H 09
402H 02
stored in the locations at 600H. 403H 08
.
600H 06
601H 04 After
602H 07 execution
603H 09
604H 04

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
8291x6
6
49746

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
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

Computer Logic II

Register Addressing
56

 This addressing mode General Purpose Segment Registers
involves the contents of AX AH AL CS
the register directly as
BX BH BL DS
for example:
CX CH CL SS
 MOV AX, BX
DX DH DL ES
 MOV CL, DL
 Note that the IP and SI FLAGS
Flags register cannot be DI
IP
accessed directly by the SP
programmer BP

AX BX
Ex. MOV AX,BX

Immediate and Direct Addressing
57

 In Immediate addressing – for example Ex. MOV CL,61H
MOV CL,61H – the immediate operand CL
61H
61H is stored as part of the instruction.
Thus the number 61H is loaded directly in
CL.
 Direct addressing is similar except that in Ex. MOV AL,[210H]
this case the effective address of one of the AL
operands is taken directly from the
instruction. Thus in MOV AL, [210H] the
contents of location 210H relative to DS is
put in AL (DS:210H) 75H

Indirect Addressing
58

 With indirect addressing, the Example MOV DL,[BX]
effective address is found in either Say BX contains the address 0200H
the BX, SI or DI registers. I.e. the relative to DS
effective address is not found DL
directly in the instruction itself but
indirectly by accessing a register, as
in: MOV DL, [BX]
(DS:200H) 69H BX
 Note that this method is useful to
pass parameters to subroutines by
reference instead of by value.

Indexed Addressing
59

 With indexed addressing, the effective address is
calculated by the addition of an index register + Ex. MOV AL,ACCOUNT[SI]
Assume that ACCOUNT has
displacement.
an offset of 0200H relative to
 For this purpose two index registers exist SI DS and SI contains 05H
(source index) and DI (destination index)
 By default SI and DI relative to DS if not
ACCOUNT (DS:200H)
for string handling, in which case SI is
AL 201H
relative to DS and DI is relative to ES.
 Example MOV AL, ACCOUNT[SI] This adds 202H
the address of account to SI to obtain the 203H
effective address where data is to be retrieved 204H
from.
205H 75H
 An alternative notation is
MOV AL,[SI+BALANCE]

BASED ADDRESSING
60

Ex. MOV AL,[BX+05H]
 In based addressing BX or BP are Assume that BX contains
used as a variable base of the 0200H
address from where the data is to
be retrieved or stored. An offset
(DS:200H) BX
can be added to this base address
AL 201H
 Addresses given in BX are taken
202H
relative to DS whereas those in
203H
BP are taken relative to SS.
204H
 Example MOV AL, [BX+05H] 205H 75H

Based-Indexed Addressing
61

 Based-Indexed addressing is a combination Ex. MOV AL,[BX+SI+02H]
of the previous two addressing modes. Assume that BX contains
 The effective address is calculated by 0200H and SI contains 3
summing up the contents of the base
register together with the contents of the
index register and the given displacement. (DS:200H) BX
 Example: MOV AL, [BX+SI+2] AL 201H
SI
202H
Note that although in these examples the MOV 203H
instruction is always considered, this can apply 204H
to several instructions and the address 02H
calculation can apply both to the source and 205H 75H
destination of data.

Default Segment Register
62

Note that the default segment register can be changed
using the segment override, i.e. stating the whole
General Purpose
address in the form DS: Offset
AX AH AL
BX BH BL Relative to DS by default
CX CH CL
DX DH DL NORMALLY FOR STRINGS

SI Relative to DS by default DS
DI Relative to DS by default ES
SP Relative to SS by default

BP Relative to SS by default

IP Relative to CS by default

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

Assembly language 8086

Recommended

More Related Content

What's hot (20)

Viewers also liked (13)

Similar to Assembly language 8086 (20)

More from John Cutajar (14)

Recently uploaded (20)

Assembly language 8086