Assembly Program 8086 to add two 8-bit BCD number and to add two 16-bit BCD number

Hey !! Let's see program to multiply two 8 bit numbers and another program to add two 16 bit BCD number. First let's take a look at the theory.

Let's go!!

In assembly language 8086 we use mnemonics in order to perform arithmetic operations like ADD/ADC in addition.
So, Let's dive deep into what ADD and ADC is ....Scroll Down for Program↓...

ADD:- It adds number from source to destination without carry. Source can be memory location or register or a number but destination should be register or a memory location(not immediate number). Remember that source and destination should be of same type and both shouldn't be memory locations.

Syntax: ADD <destination> <source>
Operation: destination← destination + source

ADC:- It adds number from source to destination with carry i.e , adds carry flag into result of addition. Source can be memory location or register or a number but destination should be register or a memory location(not immediate number). Remember that source and destination should be of same type and both shouldn't be memory locations.

Syntax: ADC <destination> <source>
Operation: destination← destination + source + carry

DAA:- Usually used after ADD/ADC because DAA works on AL registers. This instruction is used to convert result of addition of 2packed numbers into 1-packed BCD number. in simple language, ADD/ADC adds BCD number in hexadecimal format and then DAA instruction converts this hexadecimal result into BCD result.

Syntax: DAA

Which flags get affected?
Ans- OF , CF, PF , AF, SF , ZF .



Below are 2 programs : Two 8-bit BCD addition and Two 16-bit BCD addition.

Sum of two 8-Bit BCD numbers using ALP 8086 :-

data segment

no1 db 02h

no2 db 03h

ans db ?

data ends

code segment

start : assume  cs :code,  ds :data

mov ax ,data

mov ds , ax

mov al , no1

mov bl ,no2

ADD al, b1

DAA

mov ans ,al

mov ah ,4ch

int 21h

code ends

end start

Sum of two 16-Bit BCD numbers using ALP 8086 :-

data segment

no1 dw 21h

no2 dw 31h

ans dw ?

data ends

code segment

start : assume  cs :code,  ds :data

mov ax ,data

mov ds , ax

mov ax , no1

mov bx , no2

ADD ax , bx

DAA

mov ans ,ax

mov ah ,4ch

int 21h

code ends

end start

Programs:-

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