C++ Pass by Reference vs Pass by Value – Cost Comparison

A good way to find out why there are any differences is to check the disassembly. Here are the results I got on my machine with Visual Studio 2012.

With optimization flags, both functions generate the same code:

009D1270 57                   push        edi  
009D1271 FF 15 D4 30 9D 00    call        dword ptr ds:[9D30D4h]  
009D1277 8B F8                mov         edi,eax  
009D1279 FF 15 D4 30 9D 00    call        dword ptr ds:[9D30D4h]  
009D127F 8B 0D 48 30 9D 00    mov         ecx,dword ptr ds:[9D3048h]  
009D1285 2B C7                sub         eax,edi  
009D1287 50                   push        eax  
009D1288 E8 A3 04 00 00       call        std::operator<<<std::char_traits<char> > (09D1730h)  
009D128D 8B C8                mov         ecx,eax  
009D128F FF 15 2C 30 9D 00    call        dword ptr ds:[9D302Ch]  
009D1295 33 C0                xor         eax,eax  
009D1297 5F                   pop         edi  
009D1298 C3                   ret  

This is basically equivalent to:

int main ()
{
    clock_t start, stop ;
    start = clock () ;
    stop = clock () ;
    cout << "time: " << stop - start ;
    return 0 ;
}

Without optimization flags, you will probably get different results.

function (no optimizations):

00114890 55                   push        ebp  
00114891 8B EC                mov         ebp,esp  
00114893 81 EC C0 00 00 00    sub         esp,0C0h  
00114899 53                   push        ebx  
0011489A 56                   push        esi  
0011489B 57                   push        edi  
0011489C 8D BD 40 FF FF FF    lea         edi,[ebp-0C0h]  
001148A2 B9 30 00 00 00       mov         ecx,30h  
001148A7 B8 CC CC CC CC       mov         eax,0CCCCCCCCh  
001148AC F3 AB                rep stos    dword ptr es:[edi]  
001148AE 8B 45 08             mov         eax,dword ptr [ptr]  
001148B1 8B 08                mov         ecx,dword ptr [eax]  
001148B3 6B C9 05             imul        ecx,ecx,5  
001148B6 8B 55 08             mov         edx,dword ptr [ptr]  
001148B9 89 0A                mov         dword ptr [edx],ecx  
001148BB 5F                   pop         edi  
001148BC 5E                   pop         esi  
001148BD 5B                   pop         ebx  
001148BE 8B E5                mov         esp,ebp  
001148C0 5D                   pop         ebp  
001148C1 C3                   ret 

function2 (no optimizations)

00FF4850 55                   push        ebp  
00FF4851 8B EC                mov         ebp,esp  
00FF4853 81 EC C0 00 00 00    sub         esp,0C0h  
00FF4859 53                   push        ebx  
00FF485A 56                   push        esi  
00FF485B 57                   push        edi  
00FF485C 8D BD 40 FF FF FF    lea         edi,[ebp-0C0h]  
00FF4862 B9 30 00 00 00       mov         ecx,30h  
00FF4867 B8 CC CC CC CC       mov         eax,0CCCCCCCCh  
00FF486C F3 AB                rep stos    dword ptr es:[edi]  
00FF486E 8B 45 08             mov         eax,dword ptr [val]  
00FF4871 6B C0 05             imul        eax,eax,5  
00FF4874 89 45 08             mov         dword ptr [val],eax  
00FF4877 5F                   pop         edi  
00FF4878 5E                   pop         esi  
00FF4879 5B                   pop         ebx  
00FF487A 8B E5                mov         esp,ebp  
00FF487C 5D                   pop         ebp  
00FF487D C3                   ret  

Why is pass by value faster (in the no optimization case)?

Well, function() has two extra mov operations. Let's take a look at the first extra mov operation:

001148AE 8B 45 08             mov         eax,dword ptr [ptr]  
001148B1 8B 08                mov         ecx,dword ptr [eax]  
001148B3 6B C9 05             imul        ecx,ecx,5

Here we are dereferencing the pointer. In function2 (), we already have the value, so we avoid this step. We first move the address of the pointer into register eax. Then we move the value of the pointer into register ecx. Finally, we multiply the value by five.

Let's look at the second extra mov operation:

001148B3 6B C9 05             imul        ecx,ecx,5  
001148B6 8B 55 08             mov         edx,dword ptr [ptr]  
001148B9 89 0A                mov         dword ptr [edx],ecx 

Now we are moving backwards. We have just finished multiplying the value by 5, and we need to place the value back into the memory address.

Because function2 () does not have to deal with referencing and dereferencing a pointer, it gets to skip these two extra mov operations.

This answer applies to static functions, or with LTO (link time optimization), but not in general across boundaries where the calling convention has to be followed. (Like between libraries and the main executable. Or without LTO, between source files.)

There is a certain GCC optimization called IPA SRA, that replaces "pass by reference" with "pass by value" automatically: https://gcc.gnu.org/onlinedocs/gcc/Optimize-Options.html (-fipa-sra)

This is most likely done for scalar types (eg. int, double, etc), that does not have non-default copy semantics and can fit into cpu registers.

This makes

void(const int &f)

probably as fast (and space optimized)

void(int f)

So with this optimization enabled, using references for small types should be as fast as passing them by value.

On the other hand passing (for example) std::string by value could not be optimized to by-reference speed, as custom copy semantics are being involved.

From what I understand, using pass by reference for everything should never be slower than manually picking what to pass by value and what to pass by reference.

This is extremely useful especially for templates:

template<class T>
void f(const T&)
{
    // Something
}

is always optimal when the template inlines or at least is private to the call-sites using it. e.g. you don't take the address of the template function and pass it as a callback.