Computing the type of a function pointer

问题

Consider the following:

template<typename T>
struct S
{
    typedef M< &T::foo > MT;
}

This would work for:

S<Widget> SW;

where Widget::foo() is some function

How would I modify the definition of struct S to allow the following instead:

S<Widget*> SWP;

回答1:

What you need is the following type transformation.

• given T, return T
• given T *, return T

It so happens that the standard library already has implemented this for us in std::remove_pointer (though it's not hard to do yourself).

With this, you can then write

using object_type = std::remove_pointer_t<T>;
using return_type = /* whatever foo returns */;
using MT = M<object_type, return_type, &object_type::foo>;

Regarding your comment that you also want to work with smart pointers, we have to re-define the type transformation.

• given a smart pointer type smart_ptr<T>, return smart_ptr<T>::element_type, which should be T
• given a pointer type T *, return T
• otherwise, given T, return T itself

For this, we'll have to code our own meta-function. At least, I'm not aware of anything in the standard library that would help here.

We start by defining the primary template (the “otherwise” case).

template <typename T, typename = void>
struct unwrap_obect_type { using type = T; };

The second (anonymous) type parameter that is defaulted to void will be of use later.

For (raw) pointers, we provide the following partial specialization.

template <typename T>
struct unwrap_obect_type<T *, void> { using type = T; };

If we'd stop here, we'd basically get std::remove_pointer. But we'll add an additional partial specialization for smart pointers. Of course, we'll first have to define what a “smart pointer” is. For the purpose of this example, we'll treat every type with a nested typedef named element_type as a smart pointer. Adjust this definition as you see fit.

template <typename T>
struct unwrap_obect_type
<
  T,
  std::conditional_t<false, typename T::element_type, void>
>
{
  using type = typename T::element_type;
};

The second type parameter std::conditional_t<false, typename T::element_type, void> is a convoluted way to simulate std::void_t in C++14. The idea is that we have the following partial type function.

• given a type T with a nested typedef named element_type, return void
• otherwise, trigger a substitution failure

Therefore, if we are dealing with a smart pointer, we'll get a better match than the primary template and otherwise, SFINAE will remove this partial specialization from further consideration.

Here is a working example. T.C. has suggested using std::mem_fn to invoke the member function. This makes the code a lot cleaner than my initial example.

#include <cstddef>
#include <functional>
#include <iostream>
#include <memory>
#include <string>
#include <utility>

template <typename ObjT, typename RetT, RetT (ObjT::*Pmf)() const noexcept>
struct M
{
  template <typename ThingT>
  static RetT
  call(ThingT&& thing) noexcept
  {
    auto wrapper = std::mem_fn(Pmf);
    return wrapper(std::forward<ThingT>(thing));
  }
};

template <typename T, typename = void>
struct unwrap_obect_type { using type = T; };

template <typename T>
struct unwrap_obect_type<T *, void> { using type = T; };

template <typename T>
struct unwrap_obect_type<T, std::conditional_t<false, typename T::element_type, void>> { using type = typename T::element_type; };

template <typename T>
struct S
{

  template <typename ThingT>
  void
  operator()(ThingT&& thing) const noexcept
  {
    using object_type = typename unwrap_obect_type<T>::type;
    using id_caller_type          = M<object_type, int,                &object_type::id>;
    using name_caller_type        = M<object_type, const std::string&, &object_type::name>;
    using name_length_caller_type = M<object_type, std::size_t,        &object_type::name_length>;
    std::cout << "id:          " << id_caller_type::call(thing)          << "\n";
    std::cout << "name:        " << name_caller_type::call(thing)        << "\n";
    std::cout << "name_length: " << name_length_caller_type::call(thing) << "\n";
  }

};

class employee final
{

 private:

  int id_ {};
  std::string name_ {};

 public:

  employee(int id, std::string name) : id_ {id}, name_ {std::move(name)}
  {
  }

  int                  id()          const noexcept { return this->id_; }
  const std::string&   name()        const noexcept { return this->name_; }
  std::size_t          name_length() const noexcept { return this->name_.length(); }

};

int
main()
{
  const auto bob = std::make_shared<employee>(100, "Smart Bob");
  const auto s_object = S<employee> {};
  const auto s_pointer = S<employee *> {};
  const auto s_smart_pointer = S<std::shared_ptr<employee>> {};
  s_object(*bob);
  std::cout << "\n";
  s_pointer(bob.get());
  std::cout << "\n";
  s_smart_pointer(bob);
}

来源：https://stackoverflow.com/questions/34690980/computing-the-type-of-a-function-pointer