指向函数成员的已删除指针的类型

Question

The type of:

std::remove_pointer<int(*)(int)>::type

is int(int) . This code:

#include <iostream>
#include <type_traits>

using namespace std;

int main()
{
   cout << boolalpha;
   cout << is_same<remove_pointer<int(*)(int)>, int(int)>::value;
   cout << endl;
}

prints 'true'. But, what is the (written) type of a "function member"?

#include <iostream>
#include <type_trais>

using namespace std;

struct A {};

int main()
{
    cout << boolalpha;
    cout << is_same<remove_pointer<int(A::*)(int)>, int(int)>::value;
    cout << endl;
}

returns false . And something like int A::(int) throws a compilation error (invalid type).

Answer 1

A pointer-to-member is not a pointer.

remove_pointer won't change the type.

• Demo: http://ideone.com/OiGGpf

Answer 2

It goes like this: A non-member type is either an object or a function type:

T = int;                 // object type
T = double(char, bool)   // function type

For non-static class members, you can only have pointer-to-member types. Those are of the form:

class Foo;

PM = U Foo::*;           // pointer to member type

The question is what U is:

U = int                 =>  PM = int Foo::*                  // pointer-to-member-object
U = double(char, bool)  =>  PM = double (Foo::*)(char, bool) // pointer-to-member-function

A "pointer-to-member" is not a pointer, and so you cannot "remove the pointer" from it. At best you can get the underlying type, ie go from PM = U Foo::* to U . To my knowledge, no such trait exists, but it is easily concocted:

template <typename> struct remove_member_pointer;

template <typename U, typename F> struct remove_member_pointer<U F::*>
{
    typedef U member_type; 
    typedef F class_type;
};