What is the type of a constant method pointer?

Question

Given a class

class C {
public:
    int f (const int& n) const { return 2*n; }
    int g (const int& n) const { return 3*n; }
};

We can define a function pointer p to C::f like this.

int (C::*p) (const int&) const (&C::f);

The definition of p may be split up using a typedef:

typedef int (C::*Cfp_t) (const int&) const;
Cfp_t p (&C::f);

To make sure p doesn't change (as by p = &C::g; for instance) we can do:

const Cfp_t p (&C::f);

Now, what is the type of p in this case? And how do we accomplish the last definition of p without using a typedef? I am aware that typeid (p).name () cannot distinguish the outermost const as it yields

int (__thiscall C::*)(int const &)const

Answer 1

The type of the variable p is int (C::*const) (const int&) const, you can define it without a typedef as:

int (C::*const p) (const int&) const = &C::f;

Your rule of thumb is: to make the object/type that you're defining const, put the const keyword next to the name of the object/type. So you could also do:

typedef int (C::*const Cfp_t) (const int&) const;
Cfp_t p(&C::f);
p = &C::f; // error: assignment to const variable