Does empty "else" cost any time - c++

Question

Is there any difference in computational cost of
if(something){
return something;
}else{
return somethingElse;
}
and
if(something){
return something;
}
//else (put in comments for readibility purposes)
return somethingElse;
In theory we have command (else) but it doesn't seem it should make an actuall difference.
Edit:
After running code for different set sizes, I found that there actually is a differrence, code without else appears to be about 1.5% more efficient. But it most likely depends on compiler, as stated by many people below. Code I tested it on:
int withoutElse(bool a){
if(a)
return 0;
return 1;
}
int withElse(bool a){
if(a)
return 0;
else
return 1;
}
int main(){
using namespace std;
bool a=true;
clock_t begin,end;
begin= clock();
for(__int64 i=0;i<1000000000;i++){
a=!a;
withElse(a);
}
end = clock();
cout<<end-begin<<endl;
begin= clock();
for(__int64 i=0;i<1000000000;i++){
a=!a;
withoutElse(a);
}
end = clock();
cout<<end-begin<<endl;
return 0;
}
Checked on loops from 1 000 000 to 1 000 000 000, and results were consistently different
Edit 2:
Assembly code (once again, generated using Visual Studio 2010) also shows small difference (appareantly, I'm no good with assemblers :()
?withElse##YAH_N#Z PROC ; withElse, COMDAT
; Line 12
push ebp
mov ebp, esp
sub esp, 192 ; 000000c0H
push ebx
push esi
push edi
lea edi, DWORD PTR [ebp-192]
mov ecx, 48 ; 00000030H
mov eax, -858993460 ; ccccccccH
rep stosd
; Line 13
movzx eax, BYTE PTR _a$[ebp]
test eax, eax
je SHORT $LN2#withElse
; Line 14
xor eax, eax
jmp SHORT $LN3#withElse
; Line 15
jmp SHORT $LN3#withElse
$LN2#withElse:
; Line 16
mov eax, 1
$LN3#withElse:
; Line 17
pop edi
pop esi
pop ebx
mov esp, ebp
pop ebp
ret 0
?withElse##YAH_N#Z ENDP ; withElse
and
?withoutElse##YAH_N#Z PROC ; withoutElse, COMDAT
; Line 4
push ebp
mov ebp, esp
sub esp, 192 ; 000000c0H
push ebx
push esi
push edi
lea edi, DWORD PTR [ebp-192]
mov ecx, 48 ; 00000030H
mov eax, -858993460 ; ccccccccH
rep stosd
; Line 5
movzx eax, BYTE PTR _a$[ebp]
test eax, eax
je SHORT $LN1#withoutEls
; Line 6
xor eax, eax
jmp SHORT $LN2#withoutEls
$LN1#withoutEls:
; Line 7
mov eax, 1
$LN2#withoutEls:
; Line 9
pop edi
pop esi
pop ebx
mov esp, ebp
pop ebp
ret 0
?withoutElse##YAH_N#Z ENDP ; withoutElse

Answer 1

It is generically different, but the compiler may decide to execute the same jump in both cases (it will practically do this always).
The best way to see what a compiler does is reading the assembler. Assuming that you are using gcc you can try with
gcc -g -c -fverbose-asm myfile.c; objdump -d -M intel -S myfile.o > myfile.s
which creates a mix of assembler/c code and makes the job easier at the beginning.
As for your example it is:
CASE1
if(something){
23: 83 7d fc 00 cmp DWORD PTR [ebp-0x4],0x0
27: 74 05 je 2e <main+0x19>
return something;
29: 8b 45 fc mov eax,DWORD PTR [ebp-0x4]
2c: eb 05 jmp 33 <main+0x1e>
}else{
return 0;
2e: b8 00 00 00 00 mov eax,0x0
}
CASE2
if(something){
23: 83 7d fc 00 cmp DWORD PTR [ebp-0x4],0x0
27: 74 05 je 2e <main+0x19>
return something;
29: 8b 45 fc mov eax,DWORD PTR [ebp-0x4]
2c: eb 05 jmp 33 <main+0x1e>
return 0;
2e: b8 00 00 00 00 mov eax,0x0
As you could imagine there are no differences!

Answer 2

It won't compile if you type `return
Think that once the code gets compiled, all ifs, elses and loops are changed to goto's
If (cond) { code A } code B
turns to
if cond is false jump to code b
code A
code B
and
If (cond) { code A } else { code B } code C
turns to
if cond is false jump to code B
code A
ALWAYS jump to code C
code B
code C
Most processors 'guess' whether they're going to jump or not before checking if they actually jump. Depending on the processor, it might affect the performance to fail a guess.
So the answer is YES! (unless there's an ALWAYS jump at the end of first comparison) It will take 2-3 cycles to do the ALWAYS jump which isn't in the first if.