Get the class type of a static class function

Question

I have a function pointer to a static class function Foo::bar(), and want to get the type of the class (Foo). Now, I know that I could get the class type if bar were a member function of Foo, instead of a static function, with something like the following type trait:

template<class T> struct class_of; template<class T, class R> struct class_of<R T::*> { using type = T; };

However, this doesn't work for static functions. What I'd like to do is the following: class_of<Foo::bar>::type == Foo

It seems to me that the compiler knows all relevant information, so how can this be done?

score 1 · Accepted Answer · answered Dec 03 '17 at 15:37

A bare function pointer to a static member function is of the same type as a function pointer to a non-member function.

Maybe you can use a wrapper around your function pointer to include the class information:

#include <iostream>

struct Foo {
  template<class Arg>
  static void bar(Arg arg) {
    std::cout << "called with " << arg << std::endl;
  }
};

template<class T, class Ret, class... Args>
struct Wrapper {
  using F = Ret(*)(Args...);

  F f_;

  constexpr Wrapper(F f) noexcept : f_{f} {}

  template<class... RealArgs>
  constexpr Ret operator()(RealArgs&&... args) const {
    return f_(std::forward<RealArgs>(args)...);
  }
};

template<class T, class Ret, class... Args>
constexpr Wrapper<T, Ret, Args...> make_wrapper(Ret(*f)(Args...)) {
  return Wrapper<T, Ret, Args...>(f);
}

template<class T>
void inspect(const T&) {
  std::cout << __PRETTY_FUNCTION__ << std::endl;
}

int main() {
  constexpr auto foobar_int = make_wrapper<Foo>(Foo::bar<int>);
  inspect(foobar_int);
  foobar_int(4);

  constexpr auto foobar_double = make_wrapper<Foo>(Foo::bar<double>);
  inspect(foobar_double);
  foobar_double(3.8);

  return 0;
}

Get the class type of a static class function

1 Answers1