When working with assembly language, one of the fundamental operations is division. In this blog post, we will explore an 8086 assembly program that divides two 16-bit numbers using the DIV instruction.

The following program takes two 16-bit numbers stored in memory, performs division, and stores the result.

data segment
a dw 4444h
b dw 0002h
c dw ?
data ends

code segment
assume ds:data, cs:code
start:
mov ax,data
mov ds,ax
mov ax,a
mov bx,b
div bx
mov c,ax
int 3
code ends
end start

Step-by-Step Explanation:

1. Data Segment:
   • a is initialized with 4444h (17476 in decimal).
   • b is initialized with 0002h (2 in decimal).
   • c is left uninitialized to store the result.
2. Code Segment:
   • The MOV AX, DATA and MOV DS, AX instructions set up the data segment.
   • The dividend is moved to the AX register, and the divisor is placed in the BX register.
   • The DIV BX instruction divides AX by BX and stores the quotient in AX and the remainder in DX.
   • The quotient is then stored in memory location c.
   • INT 3 is used to pause execution for debugging.

Flowchart:

Output

C:\TASM>masm an16div.asm
Microsoft (R) Macro Assembler Version 5.00
Copyright (C) Microsoft Corp 1981-1985, 1987.  All rights reserved.
 
Object filename [an16div.OBJ]:
Source listing  [NUL.LST]:
Cross-reference [NUL.CRF]:
 
  50402 + 450254 Bytes symbol space free
 
      0 Warning Errors
      0 Severe  Errors
 
C:\TASM>link an16div.obj
 
Microsoft (R) Overlay Linker  Version 3.60
Copyright (C) Microsoft Corp 1983-1987.  All rights reserved.
 
Run File [AN16DIV.EXE]:
List File [NUL.MAP]:
Libraries [.LIB]:
LINK : warning L4021: no stack segment
 
C:\TASM>debug an16div.exe
-g
 
AX=2222  BX=0002  CX=0022  DX=0000  SP=0000  BP=0000  SI=0000  DI=0000
DS=0B97  ES=0B87  SS=0B97  CS=0B98  IP=0011   NV UP EI PL NZ NA PO NC
0B98:0011 CC            INT     3
-d 0B97:0000
0B97:0000  44 44 02 00 22 22 00 00-00 00 00 00 00 00 00 00   DD..""..........
0B97:0010  B8 97 0B 8E D8 A1 00 00-8B 1E 02 00 F7 F3 A3 04   ................
0B97:0020  00 CC 86 72 FF 77 15 8A-86 70 FF 2A E4 50 B8 FD   ...r.w...p.*.P..
0B97:0030  05 50 FF 36 24 21 E8 77-63 83 C4 06 FF 36 24 21   .P.6$!.wc....6$!
0B97:0040  B8 0A 00 50 E8 47 5E 83-C4 04 5E 8B E5 5D C3 90   ...P.G^...^..]..
0B97:0050  55 8B EC 81 EC 84 00 C4-5E 04 26 80 7F 0A 00 74   U.......^.&....t
0B97:0060  3E 8B 46 08 8B 56 0A 89-46 FC 89 56 FE C4 5E FC   >.F..V..F..V..^.
0B97:0070  26 8A 47 0C 2A E4 40 50-8B C3 05 0C 00 52 50 E8   &.G.*[email protected].
-q
 
C:\TASM>

Understanding the Memory Dump

The memory dump displayed in the DEBUG session shows the contents of memory after running the program. Let’s analyze the key parts.

Memory Dump Output:

0B97:0000  44 44 02 00 22 22 00 00-00 00 00 00 00 00 00 00   DD..""..........

Summary of the Memory Dump Analysis: