Why does taking a member function pointer value requires class name qualification even from inside of the class?

Question

When returning a member function pointer to a class within one of that class's member functions I still have to specify the class. I cannot simply take the address. For example, this code works fine:

class Foo {
public:
    void func(int param) { cout << param << endl; }
    void (Foo::*getPointer())(int) { return &Foo::func; }
};

But if in getPointer I try to simply do: return &func I get this error:

prog.cpp: In member function 'void (Foo::* Foo::getPointer())(int)':
prog.cpp:8:43: error: ISO C++ forbids taking the address of an unqualified or parenthesized non-static member function to form a pointer to member function. Say '&Foo::func' [-fpermissive]
void (Foo::*getPointer())(int) { return &func; }

Why must I specify the class when that's the context that I am within?

Answer 1

Pointers and pointer to members are distinct types, we can see that from section the draft C++ standard section 3.9.2 [basic.compound] which includes a compound type for pointer as well as pointer to non-static class member and notes:

Static class members are objects or functions, and pointers to them are ordinary pointers to objects or functions

This issue with this is I think well described in this quote in an answer from Johannes from the Annotated C++ Reference Manual(ARM):

Note that the address-of operator must be explicitly used to get a pointer to member; there is no implicit conversion ... Had there been, we would have an ambiguity in the context of a member function ... For example,

void B::f() {
    int B::* p = &B::i;   // ok
    p = B::i;         // error: B::i is an int
    p = &i;           // error: '&i'means '&this->i'
                      // which is an 'int*'

    int *q = &i;      // ok
    q = B::i;         // error: 'B::i is an int
    q = &B::i;        // error: '&B::i' is an 'int B::*'
}

In particular these lines:

int B::* p = &B::i; // OK

and:

p = &i; // error: '&i'means '&this->i' which is an 'int*'

demonstrate the difference between the qualified and the unqualified name.