Why does the disassembly of this function show a two-operand idiv?

December 6, 2024

I wrote the following C program:

#include "stdio.h"

__declspec(noinline) void DivideTest(int num, int denom)
{
    int quo = num / denom;
    int rem = num % denom;
    printf("Quotient: %d\nRemainder: %d\n", quo, rem);
}

int main(int argc, char* argv[])
{
    //Use volatile variables to prevent result from being hardcoded.
    volatile int num = 20;
    volatile int denom = 3;
    DivideTest(num, denom);

    return 0;
}

As expected, the output is as follows:

Quotient: 6
Remainder: 2

However, when I compile in release mode (i.e. with optimizations enabled), debug the program in Visual Studio, and look at the disassembly, it shows DivideTest using a two-operand idiv:

_DivideTest:
push esi
mov esi,edx
mov eax,ecx
cdq
idiv eax,esi

Disassembling using dumpbin produces the same result. But every source I’ve found (
example) says that idiv can only take one operand.

When I try to assemble code using a two-operand idiv, it fails, as I would expect based on the documentation. Why does the disassembly show a two-operand idiv?

>Solution :

The disassembly is technically wrong. It just shows how idiv works. The idiv instruction always uses eax (in fact edx:eax pair) to get the dividend from and to store the result (quotient and remainder).