Additional note (October, 2010):

An explicitly declared move constructor/op= should not suppress the implicitly declared copy constructor/op=; it should cause it to be deleted instead. This should prevent a member function taking a (reference to) an un-reference-related type from being chosen by overload resolution in a defaulted member function.

And we should clarify that member functions taking un-reference-related types are not even considered during overload resolution in a defaulted member function, to avoid requiring their parameter types to be complete.

Moving to always doing overload resolution for determining exception specifications and implicit deletion creates some unfortunate cycles:

    template<typename T> struct A {
       T t;
    };

    template <typename T> struct B {
       typename T::U u;
    };

    template <typename T> struct C {
       C(const T&);
    };

    template <typename T> struct D {
       C<B<T> > v;
    };

    struct E {
       typedef A<D<E> > U;
    };

    extern A<D<E> > a;
    A<D<E> > a2(a);

If declaring the copy constructor for A<D<E>> is part of instantiating the class, then we need to do overload resolution on D<E>, and thus C<B<E>>. We consider C(const B<E>&), and therefore look to see if there's a conversion from C<B<E>> to B<E>, which instantiates B<E>, which fails because it has a field of type A<D<E>> which is already being instantiated.

Even if we wait until A<D<E>> is considered complete before finalizing the copy constructor declaration, declaring the copy constructor for B<E> will want to look at the copy constructor for A<D<E>>, so we still have the cycle.

I think that to avoid this cycle we need to short-circuit consideration of C(const T&) somehow. But I don't see how we can do that without breaking

    struct F {
       F(F&);
    };

    struct G;
    struct G2 {
       G2(const G&);
    };

    struct G {
       G(G&&);
       G(const G2&);
    };

    struct H: F, G { };

    extern H h;
    H h2(h);

Here, since G's move constructor suppresses the implicit copy constructor, the defaulted H copy constructor calls G(const G2&) instead. If the move constructor did not suppress the implicit copy constructor, I believe the implicit copy constructor would always be viable, and therefore a better match than a constructor taking a reference to another type.

So perhaps the answer is to reconsider that suppression and then disqualify any constructor taking (a reference to) a type other than the constructor's class from consideration when looking up a subobject constructor in an implicitly defined constructor. (Or assignment operator, presumably.)

Another possibility would be that when we're looking for a conversion from C<B<E>> to B<E> we could somehow avoid considering, or even declaring, the B<E> copy constructor. But that seems a bit dodgy.