Question

Maybe I have missed something about functions or something else?
I use VS 2010.
Here is my code:

#include <iostream>

void swap (int& a, int& b)
{
     __asm push a
     __asm push b
     __asm pop a
     __asm pop b
     std::cout << "Inline function check: a=" << a << " b=" << b << "\n\n";
}

void main()
{
    int arg1=1, arg2=9000;
    std::cout << "Before swap we have: a=" << arg1 << " b=" << arg2 << "\n\n";
    swap(arg1, arg2);
    std::cout << "After swap we have: a=" << arg1 << " b=" << arg2;
    std::cin.get();
}

Answer 1

When you pass by reference with Visual Studio, the compiler passes the address of the referenced variable. Your swap is actually swapping the addresses that were passed to the swap function; it’s not actually swapping the contents.

In order to swap the contents, you would need to dereference the pointers. Unfortunately, the compiler won’t allow you to do this directly, so you’ll have to use registers.

Here is a short example. This is just a demonstration; swapping the two variables can be done with far fewer instructions if you access the stack directly, instead of using a and b. It’s important to note that memory can only be modified by using a register.

void swap(int& a, int& b) {
__asm {
    // Preserve our registers
    push eax 
    push ebx 
    push ecx 
    push edx 
    // Store the address of a in eax and b in ebx
    mov eax, a
    mov ebx, b
    // Store the value of a in ecx and b in edx
    mov ecx, [eax]
    mov edx, [ebx]
    // Swap the values pointed to by a and b
    mov [eax], edx
    mov [ebx], ecx
    // Restore the registers
    pop edx
    pop ecx
    pop ebx
    pop eax
}
}