assembly x86 division fpoint exception

Question

I am trying to divide two single byte numbers, and then trying to get the quotient and remainder afterwards (placing them in single byte variables).

Here's my code so far:

;divide 8-bit number by the number 10
mov ax, [numb2]
mov cl, 10
div cl

;get quotient and remainder 
mov byte[quotient], al
mov byte[remainder], ah

The quotient is stored in the al, and the remainder is stored in the ah, right?

After running it, I get a "Floating point exception (core dumped)" from the console.

What's wrong with my code?

edit: the quotient, remainder, and numb2 variables are 8-bits

using Ubuntu x86 -- NASM

^ I'm afraid the solution from the link did not work. please help — Kevin Lloyd Bernal, Jun 26 '13 at 16:52
http://stackoverflow.com/questions/10343155/x86-assembly-handling-the-idiv-instruction — Mysticial, Jun 26 '13 at 16:54
@BurnzZ: This is just a shorter version of the question you asked the other day. Did you try what I suggested in my answer for that question? — Michael, Jun 26 '13 at 16:57
@ChrisStratton: He needs to clear `AH` (by using `MOVZX` instead of `MOV`) since he's dividing `AX` by `CL`, and `numb2` is one byte. — Michael, Jun 26 '13 at 17:01
@Michael - I tried using the movzx but the output prints trash chars — Kevin Lloyd Bernal, Jun 26 '13 at 17:01
@Chris - yep, I tried zeroing the dx reg, as said on this doc (page10): http://www.csie.ntu.edu.tw/~acpang/course/asm_2004/slides/chapt_07_PartIISolve.pdf — Kevin Lloyd Bernal, Jun 26 '13 at 17:01
@BurnzZ: If the rest of your code looks like in your previous question then garbage output is to be expected. You're doing the int->ascii conversion of `numb2` _before_ the division, not _after_ like you should. — Michael, Jun 26 '13 at 17:04
@Michael - I think it was the char to num conversion, that's the sub operator right? — Kevin Lloyd Bernal, Jun 26 '13 at 17:16

score 2 · Answer 1 · answered Jan 27 '15 at 17:07

2

;divide 8-bit number by the number 10
mov ax, [numb2]
mov cl, 10
xor ah,ah ;; add this line  this function allows to clear ah
div cl

;get quotient and remainder 
mov byte[quotient], al
mov byte[remainder], ah

answered Jan 27 '15 at 17:07

Mahesh Uligade

597
8
17

score 1 · Answer 2 · 2013-06-26T18:50:39.890

1

You can't move an 8-bit value into a 16-bit register with "mov". The CPU will pull in 16-bits starting at the memory offset 'numb2'. Whatever it's pulling in is too large to fit in al after the div. You should use:

mov al,byte ptr [numb2]  ;tasm/masm

OR

mov al,byte [numb2]      ;nasm

xor ah,ah
mov cl,10
div cl

Per comments: use "byte ptr" or "byte" depending on your assembler. But it's always good practice to specify the object's size. Otherwise, the assembler has to infer the object's size based upon the registers used.

edited Jun 26 '13 at 18:50

answered Jun 26 '13 at 17:03

it has this on the 1st line, error: comma, colon or end of line expected – Kevin Lloyd Bernal Jun 26 '13 at 17:09
`byte ptr` is TASM/MASM syntax. If you use NASM it would simply be `byte`. – Michael Jun 26 '13 at 17:17
True. I was just trying to point out that specifying is good practice. This error would never have happened had he specified the object's size. – Jun 26 '13 at 18:49

score 0 · Answer 3 · answered Oct 15 '18 at 18:56

0

I solved that problem by using the extended version of the registers (EAX,EBX,ECX,EDX). The remainder is stored in EDX and the quotient in EAX

answered Oct 15 '18 at 18:56

Esteban Cruz

1

assembly x86 division fpoint exception

3 Answers3