8086 program to Multiply two 8 bit numbers using add and shift method

03:05


ØØ       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
           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
           mov      dl, bl            ; load dl with data to be         
                                        ; displayed
           and       dl, 0fH         ; get only lsb
                                                                                                      Flowchart 12
           cmp      dl, 09          ; check if digit is 0-9 or
                                        ; letter A-F
           jbe        l4
           add       dl, 07           ; if letter add 37H else only
                                        ; add 30H
l4:       add       dl, 30H
           mov      ah, 02          ; Function 2 under INT 21H
                                        ; (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>