CODEWITHSUNDEEP
c++assemblyinline-assembly
06needhamt
06needhamt

Reputation: 1605

Why Does this code not compile when the compiler creates the exact same type of instruction from the c++ code

Consider the following code

int a = 10;
int b = 2;
int c = 0;

int asmdivide(void);

int main(int argc, char* argv[])
{

     c = a / b;
    asmdivide();
    return 0;
}

int asmdivide()
{
    __asm

    {
        push dword ptr[a]
        push dword ptr[b]
        pop ecx
        idiv ecx, dword ptr[a] //causes compile error
        push ecx
        pop dword ptr[c]
    }
    return c;
}

why does this line idiv ecx, ds:dword ptr[a] //causes compile error Cause a compile error when the compiler generates this instruction idiv eax,dword ptr ds:[1308004h]from this line `int b = 2;

The error outputted is 

error C2414: illegal number of operands

Please see the screenshot below for assembly output

Assembly Output

I am using Visual Studio 2013. Any explanation would be valued

Upvotes: 0

Views: 482

Answers (1)

Michael
Michael

Reputation: 58487

It might seem like your code and the disassembly are the same, but that's just due to how the debugger chooses to display the disassembly.

The dividend for IDIV is implicit, and is always EAX when dividing by a 32-bit value (actually EDX:EAX). To quote Intel's manual:

F7 /7    IDIV r/m32   M   Valid   Valid   Signed divide EDX:EAX by r/m32, with result
                                          stored in EAX ←Quotient, EDX ←Remainder.

So your attempt to specify a different register as the dividend will fail.

You can only specify the divisor for IDIV, as in:

cdq                 ; sign-extend eax into edx:eax
idiv dword ptr [a]  ; divide edx:eax by [a]

Upvotes: 5

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