Question

Compare the codes:

    const char x = 'a';
    std::cout<< x;
00C31000  mov         eax,dword ptr [__imp_std::cout (0C32054h)]  
00C31005  push        eax  
00C31006  call        std::operator<<<std::char_traits<char> > (0C310B0h)  
00C3100B  add         esp,4  

and

    const int x = 'a';
    std::cout<< x;
00271000  mov         ecx,dword ptr [__imp_std::cout (272048h)]  
00271006  push        61h  
00271008  call        dword ptr [__imp_std::basic_ostream<char,std::char_traits<char> >::operator<< (272044h)]  

and

    const char* x = "a";
    std::cout<< x;
00071000  mov         eax,dword ptr [__imp_std::cout (72058h)]  
00071005  push        eax  
00071006  call        std::operator<<<std::char_traits<char> > (710B0h)  
0007100B  add         esp,4 

It seems that the const int version is better optimized than the const char* and the (even more surprising)const char version. The question – Why is there a difference in the generated code?

Answer 1

Some overloads of operator<< (including for int, but not char or const char*) are members of std::ostream; some are non-member functions taking std::ostream& as their first parameter.

Microsoft’s compiler uses different calling conventions for member and non-member functions. I’m guessing that you’re building for 32-bit Windows. In that case member functions will use the thiscall convention, where this is passed in register ecx and the remaining arguments are passed on the stack; and non-member functions use the cdecl convention, where all arguments are passed on the stack.