Date: 1999-10-15.00:00:00

Proposed Resolution (10/99): Change the two referenced bullets to read:

• If T is a class type (including unions), its associated classes are: the class itself; the class of which it is a member, if any; and its direct and indirect base classes. Its associated namespaces are the namespaces in which its associated classes are defined.
• If T is an enumeration type, its associated namespace is the namespace in which it is defined. If it is class member, its associated class is the member's class; else it has no associated class.

(This proposal also addresses Core issue 91.)

Date: 2022-11-20.07:54:16

Section 6.5.4 [basic.lookup.argdep] includes the following:

• If T is a class type, its associated classes are the class itself and its direct and indirect base classes. Its associated namespaces are the namespaces in which its associated classes are defined.
• If T is a union or enumeration type, its associated namespace is the namespace in which it is defined. If it is a class member, its associated class is the member's class; else it has no associated class.

Note that for a union, the enclosing class is an "associated class", but for a class type the enclosing class is not an "associated class". This results in some surprising behavior, as shown in the example below.

    struct A {
        union U {};
        friend void f(U);
    };
    struct B {
        struct S {};
        friend void f(S);
    };
    int main() {
        A::U    u;
        f(u);        // okay: A is an associated class
        B::S    s;
        f(s);        // error: no matching f(), B is not an associated class
    }

Certainly the enclosing class should also be an associated class for nested class types, shouldn't it?

Proposed Resolution (10/99): Change the two referenced bullets to read:

• If T is a class type (including unions), its associated classes are: the class itself; the class of which it is a member, if any; and its direct and indirect base classes. Its associated namespaces are the namespaces in which its associated classes are defined.
• If T is an enumeration type, its associated namespace is the namespace in which it is defined. If it is class member, its associated class is the member's class; else it has no associated class.

(This proposal also addresses Core issue 91.)