8086 Program to find 1's compliment

·            We have a number. Let the number be loaded in the register AX. Now, we have to find 1’s complement of this number. One’s complement of a number means to invert each bit of a number. NEG instruction in 8086 allows us, to find 2’s complement of a number, subtracting 1 from 2’s complement gives the 1’s complement of the number.

                                  eg. :       AX = 1234 H.

		0001	0010	0011	0100	= 1234 H
NEG AX		1110	1101	1100	1100	= EDCC
SUB AX, 1	–				1
		1110	1101	1100	1011	= EDCB

·            i.e. 1’s complement of 1234 H = EDCB.

  Algorithm :

Step I       :    Initialize the data memory.

Step II     :    Load the number in AX.

Step III   :    Find 2’s complement of number.

Step IV    :    1’s comp = 2’s comp – 1.

Step V     :    Display the result.

Step VI    :    Stop.

Flowchart :

 

  Program :

.model small                                                                                         

.data

a dw 1234H

.code

       mov     ax, @data      ; Initialize data section

       mov     ds, ax

       mov     ax, a             ; Load number1 in ax

       neg      ax                ; find 2's compement. Result in ax

       sub      ax, 1             ; 1's complement=2's comp-1

       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

   Result :

C:\programs>tasm 1'scomp.asm

Turbo Assembler  Version 3.0  Copyright (c) 1988, 1991 Borland International

Assembling file:   1'scomp.asm

Error messages:    None

Warning messages:  None

Passes:            1

Remaining memory:  438k

C:\programs>tlink 1'scomp.obj

Turbo Link  Version 3.0 Copyright (c) 1987, 1990 Borland International

Warning: No stack

C:\programs>1'scomp

EDCB

C:\programs>