Showing posts with label 8086 programs. Show all posts

8086 program to mask the lower nibble

, by Engineer's Vision




· Let the 8 bit number be in the AL register. We have to mask the lower nibble, ie we have to separate the lower nibble. In the result only MSB number should remain.
For example :
 eg. :        AL = 5B H.
0101
1011

Logically AND with F0 H
1111
0000


0101
0000
= 50 H.
                                  Result = 50 H. B is masked
                                  Display the result.
Algorithm :
Step I            :   Load the number in AL.
Step II          :   Mask tower  nibble i.e. AND AL, 0F0 H.
Step III        :   Display result.
Step IV         :   Stop.
 
Program :
.model small                                                                        
.data
a dw 0012H
.code
       mov     ax, @data               ; Initialize data section
       mov     ds, ax
       mov     ax, a                      ; Load number1 in ax
       and      al, 0f0h                  ; mask lower nibble.Result in al
       mov     ch, 02h                  ; Count of digits to be displayed
       mov     cl, 04h          ; Count to roll by 4 bits
       mov     bh, al                     ; Result in reg bh
l2:    rol       bh, cl                     ; roll bl so that msb comes to lsb
       mov     dl, bh                    ; 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
       int       21h
       end

  Result :
C:\programs>tasm lownib.asm
Turbo Assembler  Version 3.0  Copyright (c) 1988,  1991 Borland International
Assembling file:   lownib.asm
Error messages:    None
Warning messages:  None
Passes:            1
Remaining memory:  438k
C:\programs>tlink lownib.obj
Turbo Link  Version 3.0 Copyright (c) 1987, 1990 Borland International
Warning: No stack
C:\programs>lownib
10
C:\programs>



Flowchart : 
  
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8086 Program for 2's Compliment of a number

, by Engineer's Vision


    
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
     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.

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
       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 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>
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8086 Program to find 1's compliment

, by Engineer's Vision



·            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>
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