Why is __fastcall assebmler code larger than __stdcall one in MS C++? - c++

Question

I have disassembled two different variations of Swap function (simple value-swap between two pointers).
1). __fastcall http://pastebin.com/ux5LMktz
2). __stdcall (function without explicit calling convention modifier will have a __stdcall by default, because of MS C++ compiler for Windows) http://pastebin.com/eGR6VUjX
As I know, __fastcall is implemented differently, depending on the compiler, but basically it puts the first two arguments (left to right) into ECX and EDX register. And there could be stack use, but if the arguments are too long.
But as for the link at 1-st option, you can see, that value is pushed into the ECX registry, and there is no real difference between two variations of swap function.
And __fastcall variant does use:
00AA261F pop ecx
00AA2620 mov dword ptr [ebp-14h],edx
00AA2623 mov dword ptr [ebp-8],ecx
Which are not used in __stdcall version.
So it doesn't look like more optimized (as __fasctcall must be , by its definition).
I'm a newbie in ASM language and calling convention, so I ask you for a piece of advice. Maybe __fastcall is faster exactly in my sample, but I don't see it, do I?
Thanks!

Answer 1

Try turning on optimization, then comparing the results. Your fastcall version has many redundant operations because it's not optimized.
Here's output of VS 2010 with /Ox.
fastcall:
; _firstValue$ = ecx
; _secondValue$ = edx
?CallMe1##YIXPAH0#Z PROC ; CallMe1
mov eax, DWORD PTR [ecx]
push esi
mov esi, DWORD PTR [edx]
cmp eax, esi
je SHORT $LN1#CallMe1
mov DWORD PTR [ecx], esi
mov DWORD PTR [edx], eax
$LN1#CallMe1:
pop esi
ret 0
?CallMe1##YIXPAH0#Z ENDP ; CallMe1
stdcall:
_firstValue$ = 8 ; size = 4
_secondValue$ = 12 ; size = 4
?CallMe2##YGXPAH0#Z PROC ; CallMe2
mov edx, DWORD PTR _firstValue$[esp-4]
mov eax, DWORD PTR [edx]
push esi
mov esi, DWORD PTR _secondValue$[esp]
mov ecx, DWORD PTR [esi]
cmp eax, ecx
je SHORT $LN1#CallMe2
mov DWORD PTR [edx], ecx
mov DWORD PTR [esi], eax
$LN1#CallMe2:
pop esi
ret 8
?CallMe2##YGXPAH0#Z ENDP ; CallMe2
cdecl (what you mistakenly call stdcall in your example):
_firstValue$ = 8 ; size = 4
_secondValue$ = 12 ; size = 4
?CallMe3##YAXPAH0#Z PROC ; CallMe3
mov edx, DWORD PTR _firstValue$[esp-4]
mov eax, DWORD PTR [edx]
push esi
mov esi, DWORD PTR _secondValue$[esp]
mov ecx, DWORD PTR [esi]
cmp eax, ecx
je SHORT $LN1#CallMe3
mov DWORD PTR [edx], ecx
mov DWORD PTR [esi], eax
$LN1#CallMe3:
pop esi
ret 0
?CallMe3##YAXPAH0#Z ENDP ; CallMe3