指针到成员的混乱

Question

I'm trying to understand the consistency in the error that is thrown in this program:

#include <iostream>

class A{
public:
    void test();
    int x = 10;

};

void A::test(){

    std::cout << x << std::endl; //(1)
    std::cout << A::x << std::endl; //(2)

    int* p = &x;
    //int* q = &A::x; //error: cannot convert 'int A::*' to 'int*' in initialization| //(3)


}

int main(){

    const int A::* a = &A::x; //(4)

    A b;

    b.test();

}

The output is 10 10 . I labelled 4 points of the program, but (3) is my biggest concern:

1. x is fetched normally from inside a member function.
2. x of the object is fetched using the scope operator and an lvalue to the object x is returned.
3. Given A::x returned an int lvalue in (2), why then does &A::x return not int* but instead returns int A::* ? The scope operator even takes precedence before the & operator so A::x should be run first, returning an int lvalue, before the address is taken. ie this should be the same as &(A::x) surely? (Adding parentheses does actually work by the way).
4. A little different here of course, the scope operator referring to a class member but with no object to which is refers.

So why exactly does A::x not return the address of the object x but instead returns the address of the member, ignoring precedence of :: before & ?

Answer 1

Basically, it's just that the syntax &A::x (without variation) has been chosen to mean pointer-to-member.

If you write, for example, &(A::x) , you will get the plain pointer you expect.

More information on pointers-to-members, including a note about this very property, can be found here .

Answer 2

The C++ standard doesn't explicitly specify operator precedence; it's possible to deduce operator precedence implicitly from the grammar rules, but this approach fails to appreciate the occasional special case like this which doesn't fit into a traditional model of operator precedence.

[expr.unary.op]/3:

The result of the unary & operator is a pointer to its operand. The operand shall be an lvalue or a qualified-id . If the operand is a qualified-id naming a non-static or variant member m of some class C with type T , the result has type 'pointer to member of class C of type T ' and is a prvalue designating C::m . Otherwise, if the type of the expression is T , the result has type 'pointer to T ' and is a prvalue that is the address of the designated object or a pointer to the designated function.

/4:

A pointer to member is only formed when an explicit & is used and its operand is a qualified-id not enclosed in parentheses. [ Note: that is, the expression &(qualified-id) , where the qualified-id is enclosed in parentheses, does not form an expression of type 'pointer to member'.

[expr.prim.general]/9:

A nested-name-specifier that denotes a class, optionally followed by the keyword template , and then followed by the name of a member of either that class or one of its base classes, is a qualified-id .

What it all adds up to is that an expression of the form &A::x has the type "pointer to member x of class A " if x is a non-static member of a non-union class A , and operator precedence has no influence on this.