What's the real type of int (*T::*)(long);

Question

This is the sample code:

typedef int (T::*t1)(long);
typedef int (*T::*t2)(long);

t1 p1 = NULL;
t2 p2 = NULL;

The type t2 get's extra * (by mistake), but this code compile and run OK. But when t2 is not NULL, it gets error:

struct T
{
    int func(long)   {return 0;}
};
t2 p2 = &T::func;

error: cannot convert ‘int (T::*)(long int)’ to ‘t2’ {aka ‘int (* T::*)(long int)’} in initialization

I'm curious, what's the real type of t2? How can I use t2?

user_817643 · Answer

Eliminate the pointer-to-member bit of syntax (by applying the spiral rule heuristic), and a specific type stands out:

int (* /* T::*t2 */)(long) -> int (*)(long).

So t2 is a pointer to a member of T, which is itself a pointer to a free function.

int bar(long) { return 0; }

struct T
{
    int (*func) (long) = &bar;
};
t2 p2 = &T::func;

What's the real type of int (T::)(long);