8086 program to Multiply two 8 bit numbers using add and shift method
03:05ØØ Explanation :
· Consider that one byte is present in the AL register and another byte is present in the BL register.
· We have to multiply the byte in AL with the byte in BL.
· We will multiply the numbers using add and shift method. In this method, you add number with itself and rotate the other number each time and shift it by one bit to left alongwith carry. If carry is present add the two numbers.
· Initialise the count to 4 as we are scanning for 4 digits. Decrement counter each time the bits are added. The result is stored in AX. Display the result.
For example : AL = 11 H, BL = 10 H, Count = 4
Step I :
AX = | 11 | |
| + | 11 |
| | 22 H |
Rotate BL by one bit to left along with carry.
BL = 10 H | | 0 | ¬ | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | ||
| | CY | | | | | | | | | | ||
BL = | 0 | | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | |||
| CY | | | 2 | | | 0 | | |||||
Step II : Now decrement counter count = 3.
Check for carry, carry is not there so add number with itself.
AX = | | 22 |
| + | 22 |
| | 44 H |
Rotate BL to left,
BL = | 0 | ¬ | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 |
| CY | | | 4 | | | 0 | |
Carry is not there.
Decrement count, count=2
Step III : Add number with itself
AX = | | 44 |
| + | 44 |
| | 88 H |
Rotate BL to left,
BL = | 0 | | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| CY | | | 8 | | | 0 | |
Carry is not there.
Step IV : Decrement counter count = 1.
Add number with itself as carry is not there.
AX = | | 88 |
| + | 88 |
| | 110 H |
Rotate BL to left,
BL = | 1 | | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| CY | | | 0 | | | 0 | |
Carry is there.
Step V : Decrement counter = 0.
Carry is present.
\ add AX, BX
\ | | 0110 | | i.e. | | 11 H |
| + | 0000 | | | ´ | 10 H |
| | 0110 H | | | | 0110 H |
ØØ Algorithm :
Step I : Initialise the data segment.
Step II : Get the first number.
Step III : Get the second number.
Step IV : Initialize count = 04.
Step V : number 1 = number 1 ´ 2.
Step VI : Shift multiplier to left alongwith carry.
Step VII : Check for carry, if present goto step VIII else goto step IX.
Step VIII : number 1 = number1 + shifted number 2.
Step IX : Decrement counter.
Step X : If not zero, goto step V.
Step XI : Display the result.
Step XII : Stop.
ØØ Flowchart : Refer flowchart 12.
.model small
.data
a db 11H
b db 10H
.code
mov ax, @data ; Initialize data section
mov ds, ax
mov al, a ; Load number1 in al
mov bl, b ; Load number2 in bl
mov ah, 0
mov dl, 04h ; initialize counter
ad: add ax, ax ; add numbers. Result in dx
rcl bl, 01
jnc skip
add ax, bx
skip: dec dl ; dec number
jnz ad
mov ch, 04h ; Count of digits to be
; displayed
; displayed
mov cl, 04h ; Count to roll by 4 bits
mov bx, ax ; Result in reg bx
l2: rol bx, cl ; roll bl so that msb
; comes to lsb
; comes to lsb
mov dl, bl ; load dl with data to be
; displayed
; displayed
and dl, 0fH ; get only lsb
Flowchart 12
cmp dl, 09 ; check if digit is 0-9 or
; letter A-F
; letter A-F
jbe l4
add dl, 07 ; if letter add 37H else only
; add 30H
; add 30H
l4: add dl, 30H
mov ah, 02 ; Function 2 under INT 21H
; (Display character)
; (Display character)
int 21H
dec ch ; Decrement Count
jnz l2
mov ah, 4cH ; Terminate Program
int 21H
end
ØØ Result :
C:\programs>tasm shaddmul.asm
Turbo Assembler Version 3.0 Copyright (c) 1988, 1991 Borland International
Assembling file: shaddmul.asm
Error messages: None
Warning messages: None
Passes: 1
Remaining memory: 438k
C:\programs>tlink shaddmul.obj
Turbo Link Version 3.0 Copyright (c) 1987, 1990 Borland International
Warning: No stack
C:\programs>shaddmul
0110
C:\programs>
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