CODEWITHSUNDEEP
algorithmassemblyadditionx86-16proc
BananaBuisness
BananaBuisness

Reputation: 369

Adding two numbers in base 10 on Assembler

How can I add 2 numbers that their value is on base 16 and make the result on "base 10" on assembler. For example:

"5h+5h=10h" - I know it's wrong, I just want it to be visually 10h

And not:

5h+5h=Ah

CODE:

MOV AX,5h
MOV BX,5h
ADD AX,BX

result: ax=Ah - Not the result that i want...

result: ax=10h - The result that i want.

I tried to figure it out with google but didn't find anything that can help me...

Upvotes: 0

Views: 721

Answers (2)

BananaBuisness
BananaBuisness

Reputation: 369

Ok so i figure it out with @Fifoernik 's help so the problem is that if i want to do it with 16bit (for example 99h+1h) values i need to do it like this using DAA operan and CF flag

    pop ax
    pop bx
    add al,bl
    daa ; dec id values
    mov cl,al
    mov al,ah
    jc Carry; do i have carry 
    add al,bh
    daa ; do the magic thing
    JMP finito
    Carry:
    add al,1 ; add the carring...
    add al,bh
    daa ;can some one tell me what exactly daa does?
    finito:
    mov ch,al
    push cx
    ret

daa working only on AL so you'ill need to use the carry flag to add the carry like:

AH AL
 1 <----- carry
00 99 <-- DAA take care of the 99 and make it 0 when its A0h
00 01+
-- --
01 00 ---> result 100h

Upvotes: 1

Fifoernik
Fifoernik

Reputation: 9899

Here is the code you are looking for

MOV AX,5h
MOV BX,5h
ADD AX,BX
DAA

Now AX contains 10h

Upvotes: 4

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