c++ - What happens in memory when calling a function with literal values? -

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suppose have arbitrary function:

void somefunc(int, double, char);

and call somefunc(8, 2.4, 'a');, happens? how 8, 2.4, , 'a' memory, moved memory, , passed function? type of optimizations compiler have situations these? if mix , match parameters, such somefunc(myintvar, 2.4, somechar);?

what happens if function declared inline?

it makes no difference values literal or not (unless function inlined , compiler can optimize stuff out).

usually, parameters put registers or function parameter stack. regardless of whether explicit values or variables.

without optimizations, parameter gets pushed onto parameter stack. in first case, value of x taken first , put register eax, pushed parameter stack. foo prints x.

    foo(x); 00361a75  mov         eax,dword ptr [x]  00361a78  push        eax   00361a79  call        get_4 (3612b7h)  00361a7e  add         esp,4      foo(3); 00361a81  push        3     00361a83  call        get_4 (3612b7h)  00361a88  add         esp,4

with optimizations function visible compiler (in sample) , call skipped altogether:

    foo(x); 01011000  mov         ecx,dword ptr [__imp_std::cout (101203ch)]  01011006  push        3     01011008  call        dword ptr [__imp_std::basic_ostream<char,std::char_traits<char> >::operator<< (1012038h)]      foo(3); 0101100e  mov         ecx,dword ptr [__imp_std::cout (101203ch)]  01011014  push        3     01011016  call        dword ptr [__imp_std::basic_ostream<char,std::char_traits<char> >::operator<< (1012038h)]

foo defined as:

void foo(int x) {    std::cout << x; }

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