Extracting function parameter types from templated function

May 11, 2024

Suppose I have a template that expect a function pointer:

template <typename FncPtr>
void f(FncPtr fp)
{
   if constexpr ("is_first_arg_double") { // is something like this possible?
      fp(1.0);
   }
   else if constexpr ("is_first_arg_int") { // is something like this possible?
      fp(25);
   }
   else {
      // ...
   }
}

So my particular question is:
Is it possible to extract the arguments that the type FncPtr expects and act accordingly in the template, as indicated in the code comments above?

For my particular case, I know that it will always only contain one argument, but I’d like to know if it is possible to also do this in a variadic way.

I have not been able to find anything in the reference. Closest I could get was result_of.

>Solution :

if it is possible to also do this in a variadic way.

Yes, it is certainly as shown below:

template<typename Ret, typename... Args> void f(Ret (*fp)(Args...))
{
      if constexpr(std::is_same_v<typename std::tuple_element<0, std::tuple<Args...>>::type, double>)
      {
      }
      else if constexpr(std::is_same_v<typename std::tuple_element<0, std::tuple<Args...>>::type, int>)
      {
      }
}

how about for member function pointers?

For member function pointers you can write:

template<typename Ret, typename C, typename... Args> void f(Ret (C::*fp)(Args...))
{
      if constexpr(std::is_same_v<typename std::tuple_element<0, std::tuple<Args...>>::type, double>)
      {
      }
      else if constexpr(std::is_same_v<typename std::tuple_element<0, std::tuple<Args...>>::type, int>)
      {
      }
}