[Bug c++/52065] Declaring template friend function of template class breaks template type inference.

[rguenth at gcc dot gnu.org]

http://gcc.gnu.org/bugzilla/show_bug.cgi?id=52065

--- Comment #3 from Richard Guenther <rguenth at gcc dot gnu.org> 2012-01-31 
12:55:53 UTC ---
    template <typename T2> friend void f(A<T>, T2);

would be a partial specialization of f which is not allowed.  I don't think
it is possible to do what you are trying to do.