8086 Program to find 2’s complement of a number

11:28


ØØ       Explanation :
·          We have a number. Let this number be stored in register AX. Our task is to find 2’s complement of this number. We use NEG instruction. It replaces the number in AX with 2’s complement of the number in AX, directly.




                        eg. :  AX = 1234 H.
NEG AX  =
1110
1101
1100
1100
= EDCC H
·          2’s complement of 1234 H = EDCC H
ØØ       Algorithm :
Step I       :    Initialize the data memory.
Step II     :    Load the number in AX.
Step III   :    Find 2’scomplement of number.
Step IV    :    Display the result.
Step V     :    Stop.
ØØ       Flowchart : Refer flowchart 2.
ØØ       Program :                                                                                           

.model small                                                            
.data
a dw 1234H
.code
       mov     ax, @data               ; Initialize data section 
       mov     ds, ax
       mov     ax, a                      ; Load number1 in ax                                                                   Flowchart 2
       neg      ax                         ; find 2's compement. Result in ax
       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
       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 2'scomp.asm
Turbo Assembler  Version 3.0  Copyright (c) 1988, 1991 Borland International
Assembling file:   2'scomp.asm
Passes:            1
Remaining memory:  438k
C:\programs>tlink 2'scomp.obj
Turbo Link  Version 3.0 Copyright (c) 1987, 1990 Borland International
Warning: No stack
C:\programs>2'scomp
EDCC
C:\programs>