Division + Modulo in assembly x86

Question

I am trying to get the remainder and quotient of a number divided by 10.

I have this code block right here, is this correct? I can't seem to have an output from the variables when I try to print them

mov al, [dividend]
mov ah, 0
mov byte[dividend], 10
mov byte[quotient], al
mov byte[remainder], ah

Given that I have these variables:

dividend resb 1
quotient resb 1
remainder resb 1

I'm using Ubuntu 13 x86. Thanks!

I don't see any arithmetic operation being performed there... — m0skit0, Jun 26 '13 at 14:55
Im new to assembly, im not sure what im doing. what should I add to perform the div operation? — Kevin Lloyd Bernal, Jun 26 '13 at 15:00
@BurnzZ a `div` operation. Or maybe `idiv`. Should it be signed or unsigned? — harold, Jun 26 '13 at 15:01
it would be unsigned. on which line should I insert the div operation? — Kevin Lloyd Bernal, Jun 26 '13 at 15:03
Depends on what you want to divide. I suggest you first explain in your question **exactly** what you want to do: the more details the better, and examples are welcome. — m0skit0, Jun 26 '13 at 15:24
possible duplicate of [Assembly Language - How to Do Modulo?](http://stackoverflow.com/questions/8021772/assembly-language-how-to-do-modulo) — Ciro Santilli OurBigBook.com, Jun 01 '15 at 10:21

score 2 · Answer 1 · answered Jun 27 '13 at 15:24

2

You will need to use the div instruction, like so:

mov al,[dividend]
mov ah,0
mov cl,10            ; divisor is stored here
div cl               ; division is performed here
mov [quotient],al
mov [remainder],ah

answered Jun 27 '13 at 15:24

Drew McGowen

11,471
1
31
57

Division + Modulo in assembly x86

1 Answers1