http://gcc.gnu.org/bugzilla/show_bug.cgi?id=47628
--- Comment #7 from Jonathan Wakely <redi at gcc dot gnu.org> 2011-02-07 17:17:54 UTC --- Here's a reduced form of the code, which works with GCC 4.5 and earlier, but is ambiguous with 4.6 #include <map> struct Key { Key() { } Key(const Key&) { } template<typename T> Key(const T&) { } bool operator<(const Key&) const; }; typedef std::map<Key, int> Map; void f() { Map m; (void) m[Key()]; Map::iterator i = m.begin(); m.erase(i); } I'm not sure if the code is valid or not (Paolo?) In C++0x mode the ambiguity is in the user code, because non-const iterator can be converted to const_iterator or key_type (I believe this behaviour is required by the standard) In C++98 mode it's in the library code when calling _M_t.erase - this is a regression. Paolo, should _Rb_tree::erase take a non-const iterator in c++98 mode? That's what map::erase passes it.