Wrong Visual Studio assembly output? - c

Question

I wrote this classic function : (in 32-bit mode)
void ex(size_t a, size_t b)
{
size_t c;
c = a;
a = b;
b = c;
}
I call it inside the main as follows :
size_t a = 4;
size_t b = 5;
ex(a,b);
What I was expecting from the assembly code generated when entering the function is something like this :
1-Push the values of b and a in the stack : (which was done)
mov eax,dword ptr [b]
push eax
mov ecx,dword ptr [a]
push ecx
2-Use the values of a and b in the stack :
push ebp
mov ebp, esp
sub esp, 4
c = a;
mov eax, dword ptr [ebp+8]
mov dword ptr [ebp-4], eax
and so on for the other variables.
However, this is what I find when debugging :
push ebp
mov ebp,esp
sub esp,0CCh // normal since it's in debug with ZI option
push ebx
push esi
push edi
lea edi,[ebp-0CCh]
mov ecx,33h
mov eax,0CCCCCCCCh
rep stos dword ptr es:[edi]
size_t c;
c = a;
mov eax,dword ptr [a]
mov dword ptr [c],eax
Why is it using the variable a directly instead of calling the value stored in the stack? I don't understand...

Answer 1

The debugger doesn't show the instruction using ebp to access a. The same syntax is permitted when you write inline assembly. Otherwise the reason that dword ptr still appears.
It is easy to get it your preferred way, right click > untick "Show Symbol Names".

Answer 2

Using the assembly output option (right click on file name, properties, ...), I get what you expect from debug assembly output. This could depend on which version of VS you use. For this example, I used VS2005. I have VS2015 on a different system, but didn't try it yet.
_c$ = -8 ; size = 4
_a$ = 8 ; size = 4
_b$ = 12 ; size = 4
_ex PROC ; COMDAT
push ebp
mov ebp, esp
sub esp, 204 ; 000000ccH
push ebx
push esi
push edi
lea edi, DWORD PTR [ebp-204]
mov ecx, 51 ; 00000033H
mov eax, -858993460 ; ccccccccH
rep stosd ;fill with 0cccccccch
mov eax, DWORD PTR _a$[ebp]
mov DWORD PTR _c$[ebp], eax
mov eax, DWORD PTR _b$[ebp]
mov DWORD PTR _a$[ebp], eax
mov eax, DWORD PTR _c$[ebp]
mov DWORD PTR _b$[ebp], eax
pop edi
pop esi
pop ebx
mov esp, ebp
pop ebp
ret 0
_ex ENDP
Note this doesn't work, you need to use pointers for the swap to work.
void ex(size_t *pa, size_t *pb)
{
size_t c;
c = *pa;
*pa = *pb;
*pb = c;
}
which gets translated into:
_c$ = -8 ; size = 4
_pa$ = 8 ; size = 4
_pb$ = 12 ; size = 4
_ex PROC ; COMDAT
push ebp
mov ebp, esp
sub esp, 204 ; 000000ccH
push ebx
push esi
push edi
lea edi, DWORD PTR [ebp-204]
mov ecx, 51 ; 00000033H
mov eax, -858993460 ; ccccccccH
rep stosd
mov eax, DWORD PTR _pa$[ebp]
mov ecx, DWORD PTR [eax]
mov DWORD PTR _c$[ebp], ecx
mov eax, DWORD PTR _pa$[ebp]
mov ecx, DWORD PTR _pb$[ebp]
mov edx, DWORD PTR [ecx]
mov DWORD PTR [eax], edx
mov eax, DWORD PTR _pb$[ebp]
mov ecx, DWORD PTR _c$[ebp]
mov DWORD PTR [eax], ecx
pop edi
pop esi
pop ebx
mov esp, ebp
pop ebp
ret 0
_ex ENDP