Function 指针，具有“模板化”返回类型和 arguments

Question

So I want to have a function which takes in a function pointer, with an "unknown" return type and arguments. The function should return the same type as the function it recieves as a parameter. I thought about something like this, but it isn't working:

template<typename type, typename... param>
type do_sth(type (*func)(param... args))
{
  // ...
  type ret = func(std::forward<param>(args)...);
  // ...
  return ret;
}

I do this because I have to execute some other functions before and after the function itself.

Is something like this even possible? It is definately possible with defines (but without the return value). But because they are not effected by namespaces I want to use a function.

With define statement it would look like this (not returning anything):

#define do_sth(x) /*...*/\
                  x;\
                  /*...*/

Answer 1

You need to pass the function and the arguments separately:

template<typename type, typename... param>
type do_sth(type (*func)(param...), param... args)

Demo

Note that you will run into some troubles with implicit conversions if you do it this way.

You can also accept the function by type directly, but getting the return type is a little more involved:

template<typename F, typename... param>
std::invoke_result_t<F, param...> do_sth(F func, param... args)

std::invoke_result is C++17 (but there's also result_of in C++11). With this version, implicit conversions work just fine.

Demo