Showing posts with label 8086 program. Show all posts

8086 program to Multiply two 16 bit numbers

03:56



ØØ       Explanation :
·           Consider that a word of data is present in the AX register and 2nd word of data is present in the BX register.
·           We have to multiply the word in AX with the word in BX. Using MUL instruction, multiply the contents of the 2 registers.
·           The multiplication of the two 16 bit numbers may result into a 32 bit number. So result is stored in the DX and AX register.



ØØ       Explanation :
·           Consider that a word of data is present in the AX register and 2nd word of data is present in the BX register.
·           We have to multiply the word in AX with the word in BX. Using MUL instruction, multiply the contents of the 2 registers.
·           The multiplication of the two 16 bit numbers may result into a 32 bit number. So result is stored in the DX and AX register.

8086 program to Subtract two 16 bit numbers

03:52


ØØ       Explanation :
·            Consider that a word of data is present in the AX register and second word of data is present in the BX register.
·            We have to subtract the word in BX with the word AX.
·            Using subtract (SUB) instruction, subtract the contents of two registers.
·            Result will be stored in the AX register.


ØØ       Explanation :
·            Consider that a word of data is present in the AX register and second word of data is present in the BX register.
·            We have to subtract the word in BX with the word AX.
·            Using subtract (SUB) instruction, subtract the contents of two registers.
·            Result will be stored in the AX register.

8086 program to Add two 16 bit numbers

03:49


ØØ       Explanation :
·            Consider that a word of data is present in the AX register and a 2nd word of data is present in the BX register.
·            We have to add word in AX with the word in BX. Using ADD instruction, add the contents.
·            Result will be stored in the AX register. Display the result using display routine.


ØØ       Explanation :
·            Consider that a word of data is present in the AX register and a 2nd word of data is present in the BX register.
·            We have to add word in AX with the word in BX. Using ADD instruction, add the contents.
·            Result will be stored in the AX register. Display the result using display routine.

8086 program to Divide 16 bit numbers by an 8 bit numbers

03:33


ØØ       Explanation :
·            Consider that a word of data is present in the AX register and byte of data is present in the BL register, we have to divide word in AX with the byte in BL.
·            Using DIV instruction, divide the contents of two register. Result of division is stored in the AX register. AL contains the quotient and AH contains the remainder.
·            Display the result using display routine.
AX = 000F H (i.e. 15)
08
) 000 F (
01
BL = 08 H

–    08
¯ Quotient

Remainder
07

ØØ       Algorithm :
Step I       :    Initialise the data segment.
Step II     :    Get the first number in AX register i.e. dividend.
Step III   :    Get the second number in BL i.e. divisor.                                                
Step IV    :    Divide the two numbers.
Step V     :    Display the result.
Step VI    :    Stop
ØØ       Flowchart : Refer flowchart

ØØ       Program :
.model small
.data                                                                                                       
a dw 000FH
b db 08H
.code
       mov     ax, @data      ; Initialize data section
       mov     ds, ax
       mov     ax, a             ; Load number1 in ax
       mov     bl, b             ; Load number2 in bl                                                      
       div       bl                 ; divide numbers. Quotient in al and Rem in ah
       mov     ch, 04h         ; Count of digits to be displayed
       mov     cl, 04h ; Count to roll by 4 bits
       mov     bx, ax           ; Result in reg bh
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:\mbs>tasm 16by8div.asm
Turbo Assembler  Version 3.0  Copyright (c) 1988, 1991 Borland International
Assembling file:   16by8div.asm
Error messages:    None
Warning messages:  None
Passes:            1
Remaining memory:  438k
C:\mbs>tlink 16by8div
Turbo Link  Version 3.0 Copyright (c) 1987, 1990 Borland International
Warning: No stack
C:\mbs>16by8div
0701


ØØ       Explanation :
·            Consider that a word of data is present in the AX register and byte of data is present in the BL register, we have to divide word in AX with the byte in BL.
·            Using DIV instruction, divide the contents of two register. Result of division is stored in the AX register. AL contains the quotient and AH contains the remainder.
·            Display the result using display routine.
AX = 000F H (i.e. 15)
08
) 000 F (
01
BL = 08 H

–    08
¯ Quotient

Remainder
07

ØØ       Algorithm :
Step I       :    Initialise the data segment.
Step II     :    Get the first number in AX register i.e. dividend.
Step III   :    Get the second number in BL i.e. divisor.                                                
Step IV    :    Divide the two numbers.
Step V     :    Display the result.
Step VI    :    Stop
ØØ       Flowchart : Refer flowchart

ØØ       Program :
.model small
.data                                                                                                       
a dw 000FH
b db 08H
.code
       mov     ax, @data      ; Initialize data section
       mov     ds, ax
       mov     ax, a             ; Load number1 in ax
       mov     bl, b             ; Load number2 in bl                                                      
       div       bl                 ; divide numbers. Quotient in al and Rem in ah
       mov     ch, 04h         ; Count of digits to be displayed
       mov     cl, 04h ; Count to roll by 4 bits
       mov     bx, ax           ; Result in reg bh
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:\mbs>tasm 16by8div.asm
Turbo Assembler  Version 3.0  Copyright (c) 1988, 1991 Borland International
Assembling file:   16by8div.asm
Error messages:    None
Warning messages:  None
Passes:            1
Remaining memory:  438k
C:\mbs>tlink 16by8div
Turbo Link  Version 3.0 Copyright (c) 1987, 1990 Borland International
Warning: No stack
C:\mbs>16by8div
0701


8086 program to Add two 16 bit BCD numbers

03:29


ØØ       Explanation :
·            Consider that two words are available in registers AX and BX. We have to add these two words.
·            Using add instruction, add the contents, of the lower two bits i.e. add AL and BL.
·            The result of this addition is stored in the AL register.
·            DAA instruction is then used to convert the result to valid BCD. Now, add the contents of MSB alongwith carry if generated in LSB addition.


ØØ       Explanation :
·            Consider that two words are available in registers AX and BX. We have to add these two words.
·            Using add instruction, add the contents, of the lower two bits i.e. add AL and BL.
·            The result of this addition is stored in the AL register.
·            DAA instruction is then used to convert the result to valid BCD. Now, add the contents of MSB alongwith carry if generated in LSB addition.

8086 Program to multiply two 8 bit BCD numbers

03:26


ØØ       Explanation :
·            Consider that two unpacked numbers are present in the AL and BL registers. We have to multiply the two numbers. The result is stored in the AX register. The AAM (BCD adjust after Multiply) is used to adjust the product to two unpacked BCD digits in AX.


ØØ       Explanation :
·            Consider that two unpacked numbers are present in the AL and BL registers. We have to multiply the two numbers. The result is stored in the AX register. The AAM (BCD adjust after Multiply) is used to adjust the product to two unpacked BCD digits in AX.

8086 program to Subtract 8 bit BCD numbers

03:20


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 subtract byte in BL with byte in AL. Using SUB instruction, we will subtract the contents. The result is stored in the AL register. Using DAS instruction we will make sure that the result is valid BCD.


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 subtract byte in BL with byte in AL. Using SUB instruction, we will subtract the contents. The result is stored in the AL register. Using DAS instruction we will make sure that the result is valid BCD.

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.


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.

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.
·           The multiplication of two 8 bit numbers may result into a 16 bit number. So result is stored in AX 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 multiply the byte in AL with the byte in BL.
·           Using MUL instruction, multiply the contents of two registers.
·           The multiplication of two 8 bit numbers may result into a 16 bit number. So result is stored in AX register.

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.

8086 Program to unpack the packed BCD number

13:01


ØØ       Explanation :
·            A digit BCD number is available in register AL. We have to unpack this BCD number i.e. we have to separate the BCD digits. e.g : If the number = 92 H then in unpack form the two digits will 02 H and 09 H. i.e. we have to mask the lower nibble, first and rotate four times to the right to get the MSB digit. Then to get the LSB digit mask the upper nibble. Display the result. Masking lower nibble means ANDing the number with OF0 to get MSB.


ØØ       Explanation :
·            A digit BCD number is available in register AL. We have to unpack this BCD number i.e. we have to separate the BCD digits. e.g : If the number = 92 H then in unpack form the two digits will 02 H and 09 H. i.e. we have to mask the lower nibble, first and rotate four times to the right to get the MSB digit. Then to get the LSB digit mask the upper nibble. Display the result. Masking lower nibble means ANDing the number with OF0 to get MSB.

8086 Program to pack the two unpacked BCD numbers

11:34


ØØ       Explanation :
·            We have two digits available in the AL and BL registers in unpacked BCD form. The digit in BL is the LSB, whereas digit in AL is MSB. To pack the two numbers, means to combine the two numbers.
·            To make the digit in AL as MSB we rotate it 4 times to the left. i.e. if digit = 09 H, after rotating it by a 4 times it becomes 90 H. Now add the two numbers. Result will be packed BCD. Display the result.


ØØ       Explanation :
·            We have two digits available in the AL and BL registers in unpacked BCD form. The digit in BL is the LSB, whereas digit in AL is MSB. To pack the two numbers, means to combine the two numbers.
·            To make the digit in AL as MSB we rotate it 4 times to the left. i.e. if digit = 09 H, after rotating it by a 4 times it becomes 90 H. Now add the two numbers. Result will be packed BCD. Display the result.

8086 Program to mask upper nibble

11:32


ØØ       Explanation :
·            Let the 8 bit number be in the AL register. We have to mask the upper nibble ie we have to separate the upper nibble. In the result only LSB number should be present.


ØØ       Explanation :
·            Let the 8 bit number be in the AL register. We have to mask the upper nibble ie we have to separate the upper nibble. In the result only LSB number should be present.

8086 Program to mask lower nibble

11:30


Explanation :
·           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 :


Explanation :
·           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 :

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.


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.

8086 program To find 1’s complement of a number.

11:23



ØØ       Explanation :
·            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.



ØØ       Explanation :
·            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.

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