How do you get the type of a member function

Question

The question is inspired by a note in the standard [class.mem]

The type of a non-static member function is an ordinary function type, and the type of

Answer 1

The standard explicitly states that you can't do this.

[expr.prim.id]

2  An id-expression that denotes a non-static data member or non-static member function of a class can only be used:

   (2.1) as part of a class member access in which the object expression refers to the member's class⁵⁸ or a class derived from that class, or

   (2.2) to form a pointer to member ([expr.unary.op]), or

   (2.3) if that id-expression denotes a non-static data member and it appears in an unevaluated operand.

[ Example:

struct S {
  int m;
};
int i = sizeof(S::m);           // OK
int j = sizeof(S::m + 42);      // OK

— end example ]

Note my emphasis on can only be used: Those are the only ways that you can use an expression denoting a member function. Any other way you can come up with is ill-formed.

Note that 2.3 is exactly what you want to to - to use S::m (m being a member function) in an unevaluated context (i.e.: decltype), but it specifically (and I would assume deliberately) only applies to data members.

I can think of at least one implication of allowing this (See below). There's probably more.

• Let's assume m is declared as void m(); and it is a member of class S. If decltype(S::m) is valid, then std::add_pointer<decltype(S::m)> should also be valid.
  Considering that member functions have the implicit this parameter, what would this second type be? void (S::*)(), or maybe something like void (*)(S*)? Or even void (*)()? It may be obvious to us that we want void (S::*)(), but knowing that S::m is just a regular function type, why would add_pointer turn it into a pointer-to-member? How could it even differentiate it?