Extracting the return type from an overloaded function

Question

I want to extract the return type of a function. Problem is, there are other functions with the same name but different signature, and I can not get C++ to select the approp

Answer 1

Okay, after a few attempts I managed to work around the std::declval method suggested by Mankarse. I used a variadic class template to fixate the parameters, and used the template deduction of functions to get the return value from a function pointer. Its current syntax is typeof(ResultOf::get(function)), unfortunately it is still far from the desired resultof(function) form. Will edit this answer if I find a way to further simplify it.

#include 
#include 

using namespace std;

template 
class ResultOf
{
    public:
        template 
        static R get (R (*) (Args...));
        template 
        static R get (R (C::*) (Args...));
};

class NoDefaultConstructor
{
    public:
        NoDefaultConstructor (int) {}
};

int f ();
double f (int x);
bool f (NoDefaultConstructor);
int f (int x, int y);


int main (int argc, char* argv[])
{
    if(argc||argv){}

    cout << typeid(typeof(ResultOf<>::get(f))).name() << endl;
    cout << typeid(typeof(ResultOf::get(f))).name() << endl;
    cout << typeid(typeof(ResultOf::get(f))).name() << endl;
    cout << typeid(typeof(ResultOf::get(f))).name() << endl;

    typeof(ResultOf::get(f)) d = 1.1;
    cout << d << endl;
}

Edit:

Managed to solve it with variadic macros, the syntax is now resultof(f, param1, param2, etc). Without them I couldn't pass the commas between the parameter types to the template. Tried with the syntax resultof(f, (param1, param2, etc)) to no avail.

#include 

using namespace std;

template 
class Param
{
    public:
        template 
        static R Func (R (*) (Args...));
        template 
        static R Func (R (C::*) (Args...));
};

#define resultof(f, ...) typeof(Param<__VA_ARGS__>::Func(f))

int f ();
double f (int x);
int f (int x, int y);

int main (int argc, char* argv[])
{
    resultof(f, int) d = 1.1;
    cout << d << endl;
}