8086 program to Divide 16 bit numbers by an 8 bit numbers

03:33


ØØ       Explanation :
·            Consider that a word of data is present in the AX register and byte of data is present in the BL register, we have to divide word in AX with the byte in BL.
·            Using DIV instruction, divide the contents of two register. Result of division is stored in the AX register. AL contains the quotient and AH contains the remainder.
·            Display the result using display routine.
AX = 000F H (i.e. 15)
08
) 000 F (
01
BL = 08 H

–    08
¯ Quotient

Remainder
07

ØØ       Algorithm :
Step I       :    Initialise the data segment.
Step II     :    Get the first number in AX register i.e. dividend.
Step III   :    Get the second number in BL i.e. divisor.                                                
Step IV    :    Divide the two numbers.
Step V     :    Display the result.
Step VI    :    Stop
ØØ       Flowchart : Refer flowchart

ØØ       Program :
.model small
.data                                                                                                       
a dw 000FH
b db 08H
.code
       mov     ax, @data      ; Initialize data section
       mov     ds, ax
       mov     ax, a             ; Load number1 in ax
       mov     bl, b             ; Load number2 in bl                                                      
       div       bl                 ; divide numbers. Quotient in al and Rem in ah
       mov     ch, 04h         ; Count of digits to be displayed
       mov     cl, 04h ; Count to roll by 4 bits
       mov     bx, ax           ; Result in reg bh
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
       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:\mbs>tasm 16by8div.asm
Turbo Assembler  Version 3.0  Copyright (c) 1988, 1991 Borland International
Assembling file:   16by8div.asm
Error messages:    None
Warning messages:  None
Passes:            1
Remaining memory:  438k
C:\mbs>tlink 16by8div
Turbo Link  Version 3.0 Copyright (c) 1987, 1990 Borland International
Warning: No stack
C:\mbs>16by8div
0701