嵌套名称说明符的名称查找规则

Question

I've read the following (3.4.3/1):

If a :: scope resolution operator in a nested-name-specifier is not preceded by a decltype-specifier, lookup of the name preceding that :: considers only namespaces, types, and templates whose specializations are types.

What is the lookup rule for nested-name-specifier?

For example:

#include <iostrem>

namespace A
{
    int j=5;
}

int main()
{
    std::cout << A::j //I assume that A will be searched as just *unqualified-name* appeared inside the function which is a member of namespace
}

The second example:

namespace A
{
    namespace B
    {
        int j=5;
    }
}

int main()
{
    std::cout << A::B::j
}

Is it true that in the second example A::B will be looking as qualified name inside the namespace? Ie we can define rules for nested-name-specifier lookup inductively. But I cant find anything like that in the standard. Is it true at all?

Answer 1

Yes, it's inductive, and I'd say it simply follows from the wording. First, let's add full parenthesizing based on the associativity of :: :

(std::cout) << ((A::B)::j)

^{(The above is just to demonstrate how the parser understands precedence, it's not valid code).}

So j is qualified by the name A::B . It's a qualified name, so it's looked up according to 3.4.3.

A::B is itself a qualified name (it conforms to the syntactic form outlined by 5.1.1/8), so it is looked up according to rules for a qualified name.