http://gcc.gnu.org/bugzilla/show_bug.cgi?id=46687
fabien at gcc dot gnu.org changed:
What|Removed |Added
Status|UNCONFIRMED |ASSIGNED
Last reconfirmed|
http://gcc.gnu.org/bugzilla/show_bug.cgi?id=46687
--- Comment #1 from Jonathan Wakely redi at gcc dot gnu.org 2010-11-28
01:43:26 UTC ---
why do you think it's not ambiguous?
C has two bases of type A, so two copies of A::foo()
http://gcc.gnu.org/bugzilla/show_bug.cgi?id=46687
--- Comment #2 from Jonathan Wakely redi at gcc dot gnu.org 2010-11-28
01:57:06 UTC ---
C has two copies of the name A::foo, as B1::foo and B2::foo.
if C only saw A::foo then it would be unambiguous because the same members
would be found, as in
http://gcc.gnu.org/bugzilla/show_bug.cgi?id=46687
--- Comment #3 from Hubert Tong hstong at ca dot ibm.com 2010-11-28 04:23:55
UTC ---
(In reply to comment #2)
However, because you have using declarations in B1 and B2 name lookup finds
B1::foo and B2::foo ... at least by my reading, which
