Question

Here is the example :

struct A
{
    A(const int a ):b(a)
    {
    }

    int b;
};

struct B
{
    B() : a(5)
    {
    }

    static void A()
    {
    }

    A a;
};

int main()
{
    B::A();
}

And the compiler error is :

a9.cpp:19: error: ‘A’ does not name a type
a9.cpp: In constructor ‘B::B()’:
a9.cpp:24: error: class ‘B’ does not have any field named ‘a’

I am using gcc 4.3.0 on fedora 9.

Can someone explains why is the compiler complaining?
If possible, with references from the standard.

Thanks

Answer 1

This works:

struct B {
    B() : a(5) { }

    static void A() { }

    ::A a;
};

Since you’ve used A as a member name in B, that member’s definition shadows the A type from the outer namespace. Using :: you can get to that namespace.

This behavior is specified in the (draft) standard as:

3.3.7 (1) “A name can be hidden by an explicit declaration of that same name in a nested declarative region” (the definition of struct B, which is nested in the namespace where struct A is also defined).

Carefully read the introduction to chapter 3, Basic concepts, for further clarification. Especially, this section specifies that

3 (7) Two names are the same if

• they are identifiers composed of the same character sequence; or
• they are the names of overloaded operator functions formed with the same operator; or
• they are the names of user-defined conversion functions formed with the same type.

Note that this last definition does not distinguish between types and class members, so the name hiding (shadowing) rule 3.3.7 (1) applies.