Ranking of implicit conversion sequences [f(int) and f(const int&)]

Question

Consider the following functions:

void f(int) {...}
void f(const int&) {...}

They are different and their definitions compile together successfully. But is there a way to call any of them when they both participate in overload resolution? And if there is no way, why are they not considered the same function like these two:

void g(int) {...}
void g(const int) {...} // error: redefinition of 'void g(int)'

Answer 1

If you want to explicitly call a particular function of an overload set you can cast the function to a function pointer with the signature yo want. That would look like

void f(int) { std::cout << "void f(int) \n"; }
void f(const int&) { std::cout << "void f(const int&)\n"; }

int main () 
{
    auto fi = static_cast<void(*)(int)>(f);
    auto fciref = static_cast<void(*)(const int&)>(f);
    fi(2);
    fciref(2);
}

which outputs

void f(int) 
void f(const int&)

Otherwise you can't call your function as neither is better than the other according to the tie-break rules.