Error with address of parenthesized member function

Question

I found something interesting. The error message says it all. What is the reason behind not allowing parentheses while taking the address of a non-static member function? I

Answer 1

From the error message, it looks like you're not allowed to take the address of a parenthesized expression. It's suggesting that you rewrite

fPtr = &(myfoo::foo);  // main.cpp:14

to

fPtr = &myfoo::foo;

This is due to a portion of the spec (§5.3.1/3) that reads

A pointer to member is only formed when an explicit & is used and its operand is a qualified-id not enclosed in parentheses [...]

(my emphasis). I'm not sure why this is a rule (and I didn't actually know this until now), but this seems to be what the compiler is complaining about.

Hope this helps!

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.