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Why do I get Floating point exception when I divide?

 
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Why do I get Floating point exception when I divide?, using idiv x86 nasm
no2pencil
27 Aug, 2008 - 12:12 AM
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CODE

_div: ; use division
  push ebp       ; bring in the arguments.
  mov ebp, esp
  mov eax, [ebp+8]; Argument 1
  mov edx, [ebp+12]; Argument 2
  xor edx, edx; clear out register edx
  mov edx, ebx; Place ebx *argument 2* into ebx
  idiv ebx; Divide EDX:EAX by EBX
  pop eax; Return the quotient
  ret


No matter what arguments I give, I get the dreaded floating point exception error. I can't find a proper explanation w/Google as to why! Any ideas?
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no2pencil
RE: Why Do I Get Floating Point Exception When I Divide?
27 Aug, 2008 - 12:41 AM
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I'm not sure if this is correct, but this is what I did to fix it :

CODE

_div:
  push ebp
  mov ebp, esp
  mov eax, [ebp+8] ; low
  mov edx, 0           ; high
  idiv dword [ebp+12] ; divide edx:eax by ebp+12 (2nd argument)
  mov [ebp+12],eax ; Store the remainder back into ebp
  pop ebp                 ; return the quotient
  ret
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AmitTheInfinity
RE: Why Do I Get Floating Point Exception When I Divide?
27 Aug, 2008 - 12:50 AM
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You copied your argument 2 in edx and then cleared it in next statement using XOR. I think you wanted to copy your second argument in ebx instead of edx, right?

CODE

mov edx, [ebp+12]; Argument 2    You copied it to edx and not ebx
  xor edx, edx; clear out register edx
  mov edx, ebx; Place ebx *argument 2* into ebx    but you never had argument 2 in ebx!!!


May be it was just a typo. Correct it and see whether it works.
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no2pencil
RE: Why Do I Get Floating Point Exception When I Divide?
27 Aug, 2008 - 12:57 AM
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QUOTE(AmitTheInfinity @ 27 Aug, 2008 - 04:50 AM) *

Correct it and see whether it works.

Yeah tongue.gif

Now how can I return the remainder? Shouldn't it be in register edx?
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