--- Comment #4 from redi at gcc dot gnu dot org 2009-08-13 10:59 ---
(In reply to comment #2)
> Well, if the call is on foo then surely a foo can call; its own methods,
Yes, a foo can call its own methods, but a bar can only call them through a bar
object, not on an object with static t
--- Comment #3 from rguenth at gcc dot gnu dot org 2009-08-13 08:42 ---
EDG rejects it as well:
t.C(12): error #453: protected function "foo::foo(int)" (declared at line 5) is
not accessible through a "foo" pointer or object
new (this) foo(5);
^
compilatio
--- Comment #2 from igodard at pacbell dot net 2009-08-13 02:47 ---
Well, if the call is on foo then surely a foo can call; its own methods,
whereas if the call is on bar then a bar should see the protected methods of
its base class foo. Either way it should be visible.
--
http://gc
--- Comment #1 from pinskia at gcc dot gnu dot org 2009-08-13 02:42 ---
I think this is correct as you are not calling a protected method on this but a
new object that is created from this.
--
http://gcc.gnu.org/bugzilla/show_bug.cgi?id=41050
