Question

What’s purpose of push rdi and pop rdi when calling function in C++?
VS2010, x64, debug, no optimizations

C++

int calc()
{
    return 8 + 7;
}

Disassembly:

int calc()
{
000000013F0B1020  push        rdi  
    return 8 + 7;
000000013F0B1022  mov         eax,0Fh  
}
000000013F0B1027  pop         rdi  
000000013F0B1028  ret 

Answer 1

There is no purpose to it. This is a common artifact of unoptimized code. The code generator emits the push edi instruction in anticipation of having to perform an addition. The EDI register must be preserved across function calls. But then, later, figures out that the addition can be performed at compile time.

Getting rid of extraneous code like this requires “peephole optimization”. But that optimization isn’t enabled in the Debug build. To know what the real code look like, you have to turn on the optimizer, best done by building the Release build. It in fact will completely eliminate the function, you can prevent it from doing so with:

__declspec(noline) int calc()
{
    return 8 + 7;
}

Which produces in the Release build:

    return 8 + 7;
000007F7038E1000  mov         eax,0Fh  
000007F7038E1005  ret