>
> It's not "a funny form of partial specialization of a function".
> It's a prohibited attempt to specialise a member without first
> specialising the class template. You need to specialise C<B<T>>
> first and declare that it does have the same 'func' member.
>
Hmm, so why don't I need to specialize C<A> first before creating
the specialized C<A>::func implementation?

And as it relates to my motivation, your answer means I'm going
to have to copy the entire implementation of std::allocator and
specialize every member functionf or C<B<T>>?

Thanks,

-stephen diverdi
-stephen.diverdi@gmail.com