c++ - When instantiating a template, should members of its incomplete argument types be visible?

Question

In the following example, A has a member typedef Instantiate which causes the instantiation of B<A>.

template<typename T>
struct B
{
    typedef typename T::Before Before; // ok
    typedef typename T::After After; // error: no type named 'After' in 'A<int>'
};

template<typename T>
struct A
{
    typedef int Before;
    typedef typename B<A>::After Instantiate;
    typedef int After;
};

template struct A<int>; // instantiate A<int>

All the compilers I've tried report that, while A::Before is visible, A::After is not. Is this behaviour compliant with the standard? If so, where does the standard specify which names in A should be visible during instantiation of B<A>?

If dependent names are "looked up at the point of the template instantiation", what does this mean in the scenario of a name qualified by a template parameter such as T::After?

EDIT: Note that the same behaviour occurs when A is not a template:

template<typename T>
struct B
{
    typedef typename T::Before Before; // ok
    typedef typename T::After After; // error: no type named 'After' in 'A'
};

struct A
{
    typedef int Before;
    typedef B<A>::After Instantiate;
    typedef int After;
};

.. and G++ accepts the following, but Clang does not:

template<typename T>
struct B
{
    static const int value = 0;
    static const int i = T::value; // clang error: not a constant expression
};

struct A
{
    static const int value = B<A>::value;
};

EDIT: After some reading of the C++03 standard:

[temp.dep.type] A type is dependent if it is a template parameter

Therefore T is dependent.

[temp.res] When looking for the declaration of a name used in a template definition, the usual lookup rules are used for nondependent names. The lookup of names dependent on the template parameters is postponed until the actual template argument is known.

The lookup of T::After is therefore postponed until the argument for T is known.

[temp.inst] Unless a class template specialization has been explicitly instantiated ... the class template specialization is implicitly instantiated when the specialization is referenced in a context that requires a completely-defined object type.

Therefore the declaration of A<int>::Instantiate requires the instantiation of B<A> (because it is used in a nested-name-specifier.)

A<int>::After is not visible at the point of declaration of A<int>::Instantiate, so the behaviour of the compiler makes sense - but I haven't seen anything in C++03 that explicitly describes this behaviour. The closest thing was this somewhat vague paragraph:

[temp.dep.res] In resolving dependent names, names from the following sources are considered:

— Declarations that are visible at the point of definition of the template.

Answer 1

Whether typename T::Before is valid is not explicitly said by the spec. It is subject of a defect report (because the Standard can very reasonably be read to forbid it): http://www.open-std.org/jtc1/sc22/wg21/docs/cwg_active.html#287 .

Whether typename T::After is invalid can also very reasonably be read to be true by the spec, and actually it makes quite a bit of sense (and aforementioned DR still keeps it ill-formed). Because you have an instantiation of a class A<Foo>, which references another class A<Bar> during a period where a member Baz has not yet been declared, and that makes a reference back to A<Foo>::Bar. That is ill-formed in the case of non-templates aswell (try to "forget" for a moment that you are dealing with templates: surely the lookup of B<A>::After is done after the A template was completely parsed, but not after the specific instantiation of it was completely created. And it is the instantiation of it that actually will do the reference!).

struct A {
   typedef int Foo;
   typedef A::Foo Bar; // valid
   typedef A::Baz Lulz; // *not* valid
   typedef int Baz; 
};