Local function declaration inside namespace

Question

In such a situation

namespace n {
    void f() {
        void another_function();
    }
}

Should the function another_function

Answer 1

C++11 3.5 (as well as C++03)

7 When a block scope declaration of an entity with linkage is not found to refer to some other declaration, then that entity is a member of the innermost enclosing namespace. However such a declaration does not introduce the member name in its namespace scope.

The declaration in your example declares n::another_function.

Answer 2

According to N3485 7.3.1 [namespace.def]/6, the correct answer is n::another_function.

The enclosing namespaces of a declaration are those namespaces in which the declaration lexically appears, except for a redeclaration of a namespace member outside its original namespace (e.g., a definition as specified in 7.3.1.2). Such a redeclaration has the same enclosing namespaces as the original declaration. [ Example:

namespace Q {
    namespace V {
        void f(); // enclosing namespaces are the global namespace, Q, and Q::V
        class C { void m(); };
    }
    void V::f() { // enclosing namespaces are the global namespace, Q, and Q::V
        extern void h(); // ... so this declares Q::V::h
    }
    void V::C::m() { // enclosing namespaces are the global namespace, Q, and Q::V
    }
}

—end example ]