8086 program to Add 8 bit BCD numbers

03:08


ØØ       Explanation :
·            Consider that a byte of data is present in the AL register and a second byte of data is present in the BL register. We have to add byte in AL with the byte in BL. Using add instruction, add the contents of 2 registers.
·            Result will be stored in the AL register.
·            Use DAA instruction that will check if BCD is valid, if it is not valid then 6 is added to give proper BCD result. Display the result.

ØØ       Algorithm :
Step I       :    Initialize the data memory.
Step II     :    Get the first BCD number in AL.
Step III   :    Get the second BCD number in BL.
Step IV    :    Add the two BCD numbers.
Step V     :    Using DAA, adjust result to valid BCD number.
Step VI    :    Display the result.
Step VII  :    Stop.
ØØ       Flowchart : Refer flowchart 13.

ØØ       Program :
.model small
.data
a db 09H
b db 02H
.code
       mov     ax, @data      ; Initialize data section
       mov     ds, ax
       mov     al, a              ; Load number1 in al
       mov     bl, b             ; Load number2 in bl
       add      al, bl             ; add numbers and result in al
       daa                         ; adjust result to valid BCD number    
       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         ; Terminate Program
       int       21H
       end
ØØ       Result :
C:\programs>tasm add-bcd8.asm
Turbo Assembler  Version 3.0  Copyright (c) 1988, 1991 Borland International
Assembling file:   add-bcd8.asm
Error messages:    None
Warning messages:  None
Passes:            1
Remaining memory:  438k
C:\programs>tlink add-bcd8
Turbo Link  Version 3.0 Copyright (c) 1987, 1990 Borland International
Warning: No stack
C:\programs>add-bcd8
11
C:\programs>


ØØ       Explanation :
·            Consider that a byte of data is present in the AL register and a second byte of data is present in the BL register. We have to add byte in AL with the byte in BL. Using add instruction, add the contents of 2 registers.
·            Result will be stored in the AL register.
·            Use DAA instruction that will check if BCD is valid, if it is not valid then 6 is added to give proper BCD result. Display the result.

ØØ       Algorithm :
Step I       :    Initialize the data memory.
Step II     :    Get the first BCD number in AL.
Step III   :    Get the second BCD number in BL.
Step IV    :    Add the two BCD numbers.
Step V     :    Using DAA, adjust result to valid BCD number.
Step VI    :    Display the result.
Step VII  :    Stop.
ØØ       Flowchart : Refer flowchart 13.

ØØ       Program :
.model small
.data
a db 09H
b db 02H
.code
       mov     ax, @data      ; Initialize data section
       mov     ds, ax
       mov     al, a              ; Load number1 in al
       mov     bl, b             ; Load number2 in bl
       add      al, bl             ; add numbers and result in al
       daa                         ; adjust result to valid BCD number    
       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         ; Terminate Program
       int       21H
       end
ØØ       Result :
C:\programs>tasm add-bcd8.asm
Turbo Assembler  Version 3.0  Copyright (c) 1988, 1991 Borland International
Assembling file:   add-bcd8.asm
Error messages:    None
Warning messages:  None
Passes:            1
Remaining memory:  438k
C:\programs>tlink add-bcd8
Turbo Link  Version 3.0 Copyright (c) 1987, 1990 Borland International
Warning: No stack
C:\programs>add-bcd8
11
C:\programs>


8086 program to Multiply two 8 bit numbers using add and shift method

03:05


ØØ       Explanation :
·            Consider that one byte is present in the AL register and another byte is present in the BL register.
·            We have to multiply the byte in AL with the byte in BL.
·            We will multiply the numbers using add and shift method. In this method, you add number with itself and rotate the other number each time and shift it by one bit to left alongwith carry. If carry is present add the two numbers.

·            Initialise the count to 4 as we are scanning for 4 digits. Decrement counter each time the bits are added. The result is stored in AX. Display the result.
                        For example :   AL = 11 H,  BL = 10 H, Count = 4
Step I       :
AX =
11

+
11


22 H
                               Rotate BL by one bit to left along with carry.
BL = 10 H

0
¬
0
0
0
1
0
0
0
0


 ­
CY









BL =
0

0
0
1
0
0
0
0
0

CY



2


0

Step II     :    Now decrement counter count = 3.
                               Check for carry, carry is not there so add number with itself.
AX =

22

+
22


44 H
                               Rotate BL to left,
BL =
0
¬
0
1
0
0
0
0
0
0

CY



4


0

                               Carry is not there.
                               Decrement count, count=2
Step III   :    Add number with itself
AX =

44

+
44


88 H




                      Rotate BL to left,
BL =
0

1
0
0
0
0
0
0
0

CY



8


0

                               Carry is not there.
Step IV :      Decrement counter count = 1.
                               Add number with itself as carry is not there.
AX =

88

+
88


110 H
                               Rotate BL to left,
BL =
1

0
0
0
0
0
0
0
0

CY



0


0

                               Carry is there.
Step V :        Decrement counter = 0.
                               Carry is present.
                               \ add AX, BX
\

0110

i.e.

11 H

+
0000


´
10 H


0110 H



0110 H
ØØ       Algorithm :
Step I          :    Initialise the data segment.
Step II        :    Get the first number.
Step III      :    Get the second number.
Step IV       :    Initialize count = 04.
Step V        :    number 1 = number 1 ´ 2.
Step VI       :    Shift multiplier to left alongwith carry.
Step VII     :    Check for carry, if present goto step VIII else goto step IX.
Step VIII   :    number 1 = number1 + shifted number 2.
Step IX      :    Decrement counter.
Step X        :    If not zero, goto step V.
Step XI      :    Display the result.
Step XII     :    Stop.
ØØ       Flowchart : Refer flowchart 12.

.model small
.data
a db 11H
b db 10H
.code
           mov      ax, @data     ; Initialize data section
           mov      ds, ax
           mov      al, a             ; Load number1 in al
           mov      bl, b            ; Load number2 in bl
           mov      ah, 0           
           mov      dl, 04h         ; initialize counter
ad:       add       ax, ax           ; add numbers. Result in dx                                                    
           rcl        bl, 01
           jnc        skip
           add       ax, bx
skip:     dec       dl                ; dec number
           jnz        ad                                 
           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
                                                                                                      Flowchart 12
           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 shaddmul.asm
Turbo Assembler  Version 3.0  Copyright (c) 1988, 1991 Borland International
Assembling file:   shaddmul.asm
Error messages:    None
Warning messages:  None
Passes:            1
Remaining memory:  438k
C:\programs>tlink shaddmul.obj
Turbo Link  Version 3.0 Copyright (c) 1987, 1990 Borland International
Warning: No stack
C:\programs>shaddmul
0110
C:\programs>


ØØ       Explanation :
·            Consider that one byte is present in the AL register and another byte is present in the BL register.
·            We have to multiply the byte in AL with the byte in BL.
·            We will multiply the numbers using add and shift method. In this method, you add number with itself and rotate the other number each time and shift it by one bit to left alongwith carry. If carry is present add the two numbers.

·            Initialise the count to 4 as we are scanning for 4 digits. Decrement counter each time the bits are added. The result is stored in AX. Display the result.
                        For example :   AL = 11 H,  BL = 10 H, Count = 4
Step I       :
AX =
11

+
11


22 H
                               Rotate BL by one bit to left along with carry.
BL = 10 H

0
¬
0
0
0
1
0
0
0
0


 ­
CY









BL =
0

0
0
1
0
0
0
0
0

CY



2


0

Step II     :    Now decrement counter count = 3.
                               Check for carry, carry is not there so add number with itself.
AX =

22

+
22


44 H
                               Rotate BL to left,
BL =
0
¬
0
1
0
0
0
0
0
0

CY



4


0

                               Carry is not there.
                               Decrement count, count=2
Step III   :    Add number with itself
AX =

44

+
44


88 H




                      Rotate BL to left,
BL =
0

1
0
0
0
0
0
0
0

CY



8


0

                               Carry is not there.
Step IV :      Decrement counter count = 1.
                               Add number with itself as carry is not there.
AX =

88

+
88


110 H
                               Rotate BL to left,
BL =
1

0
0
0
0
0
0
0
0

CY



0


0

                               Carry is there.
Step V :        Decrement counter = 0.
                               Carry is present.
                               \ add AX, BX
\

0110

i.e.

11 H

+
0000


´
10 H


0110 H



0110 H
ØØ       Algorithm :
Step I          :    Initialise the data segment.
Step II        :    Get the first number.
Step III      :    Get the second number.
Step IV       :    Initialize count = 04.
Step V        :    number 1 = number 1 ´ 2.
Step VI       :    Shift multiplier to left alongwith carry.
Step VII     :    Check for carry, if present goto step VIII else goto step IX.
Step VIII   :    number 1 = number1 + shifted number 2.
Step IX      :    Decrement counter.
Step X        :    If not zero, goto step V.
Step XI      :    Display the result.
Step XII     :    Stop.
ØØ       Flowchart : Refer flowchart 12.

.model small
.data
a db 11H
b db 10H
.code
           mov      ax, @data     ; Initialize data section
           mov      ds, ax
           mov      al, a             ; Load number1 in al
           mov      bl, b            ; Load number2 in bl
           mov      ah, 0           
           mov      dl, 04h         ; initialize counter
ad:       add       ax, ax           ; add numbers. Result in dx                                                    
           rcl        bl, 01
           jnc        skip
           add       ax, bx
skip:     dec       dl                ; dec number
           jnz        ad                                 
           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
                                                                                                      Flowchart 12
           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 shaddmul.asm
Turbo Assembler  Version 3.0  Copyright (c) 1988, 1991 Borland International
Assembling file:   shaddmul.asm
Error messages:    None
Warning messages:  None
Passes:            1
Remaining memory:  438k
C:\programs>tlink shaddmul.obj
Turbo Link  Version 3.0 Copyright (c) 1987, 1990 Borland International
Warning: No stack
C:\programs>shaddmul
0110
C:\programs>


8086 program to Multiply two 8 bit numbers using successive addition method

03:01



ØØ       Explanation :
·           Consider that a byte is present in the AL register and second byte is present in the BL register.
·           We have to multiply the byte in AL with the byte in BL.
·           We will multiply the numbers using successive addition method.
·           In successive addition method, one number is accepted and other number is taken as a counter. The first number is added with itself, till the counter decrements to zero.



ØØ       Explanation :
·           Consider that a byte is present in the AL register and second byte is present in the BL register.
·           We have to multiply the byte in AL with the byte in BL.
·           We will multiply the numbers using successive addition method.
·           In successive addition method, one number is accepted and other number is taken as a counter. The first number is added with itself, till the counter decrements to zero.

8086 program To multiply two 8 bit numbers

02:59



ØØ       Explanation :
·           Consider that a byte of data is present in the AL register and second byte of data is present in the BL register.
·           We have to multiply the byte in AL with the byte in BL.
·           Using MUL instruction, multiply the contents of two registers.



ØØ       Explanation :
·           Consider that a byte of data is present in the AL register and second byte of data is present in the BL register.
·           We have to multiply the byte in AL with the byte in BL.
·           Using MUL instruction, multiply the contents of two registers.

8086 program to Subtract two 8 bit numbers

13:07


ØØ       Explanation :
·            Consider that a byte of data is present in the AL register and second byte of data is present in the BL register.
·            We have to subtract the byte in BL from the byte in AL. Using sub instruction subtract the contents of two registers. Result will be stored in the AL register.



ØØ       Explanation :
·            Consider that a byte of data is present in the AL register and second byte of data is present in the BL register.
·            We have to subtract the byte in BL from the byte in AL. Using sub instruction subtract the contents of two registers. Result will be stored in the AL register.

8086 program to Add two 8 bit numbers

13:05


ØØ       Explanation :
·            Consider that a byte of data is present in the AL register and second byte of data is present in the BL register.
·            We have to add the byte in AL with the byte in BL.
·            Using ADD instruction add the contents of 2 registers.
·            Result will be stored in the AL register.


ØØ       Explanation :
·            Consider that a byte of data is present in the AL register and second byte of data is present in the BL register.
·            We have to add the byte in AL with the byte in BL.
·            Using ADD instruction add the contents of 2 registers.
·            Result will be stored in the AL register.

8086 program to Count the number of 1’s in a register

13:02



ØØ       Explanation :
·           We have a word that is stored in AX register.
·           Initialize the counter 1 = 16.
·           Initialize counter 2 = 0 to count the number of 1’s.
·           We will rotate the number in AX alongwith carry by 1 bit to the right. If there is a carry we will increment counter 2. Decrement counter 1. This process with continue till all the bits are checked. The counter 2 will indicate the number of 1’s present in the word. The result of counter 2 is stored in BL. Display the result.



ØØ       Explanation :
·           We have a word that is stored in AX register.
·           Initialize the counter 1 = 16.
·           Initialize counter 2 = 0 to count the number of 1’s.
·           We will rotate the number in AX alongwith carry by 1 bit to the right. If there is a carry we will increment counter 2. Decrement counter 1. This process with continue till all the bits are checked. The counter 2 will indicate the number of 1’s present in the word. The result of counter 2 is stored in BL. Display the result.
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