------- Comment #1 from pinskia at gcc dot gnu dot org 2006-12-02 18:13 -------
You want (without template<>):
void A<int>::outer() const { cout << "<int>outer" << endl; }
Because A<int> is fully specialized. If A<int> was not fully specialized then:
template <> void A<int>::outer() const { cout << "<int>outer" << endl; }
Would be correct.
--
pinskia at gcc dot gnu dot org changed:
What |Removed |Added
----------------------------------------------------------------------------
Status|UNCONFIRMED |RESOLVED
Resolution| |INVALID
http://gcc.gnu.org/bugzilla/show_bug.cgi?id=30053
