Strange C++ rule for member function pointers? [duplicate]

Answer 1

This is just a personal opinion. If &(qualified-id) is allowed as &(unary-expression), qualified-id has to be an expression, and an expression is expected to have a type (even if it is incomplete). However, C++ didn't have a type which denotes a member, had only a pointer to member. For example, the following code cannot be compiled.

struct A { int i; };

template< class T > void f( T* );

int main() {
  (void) typeid( A::i );
  f( &A::i );
}

In order to make &(qualified-id) be valid, the compiler has to hold a member type internally. However, if we abandon &(qualified-id) notation, the compiler doesn't need to handle member type. As member type was always handled in the form of a pointer to it, I guess the standard gave priority to simplify the compiler's type system a little.

Answer 2

Imagine this code:

struct B { int data; };
struct C { int data; };

struct A : B, C {
  void f() {
    // error: converting "int B::*" to "int*" ?
    int *bData = &B::data;

    // OK: a normal pointer
    int *bData = &(B::data);
  }
};

Without the trick with the parentheses, you would not be able to take a pointer directly to B's data member (you would need base-class casts and games with this - not nice).

---

From the 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::*'
}

The IS just kept this pre-Standard concept and explicitly mentioned that parentheses make it so that you don't get a pointer to member.