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 could be wrong
It does find B1::foo and B2::foo, but then again, B1::foo and B2::foo refer to
the same functions. In the N3126 wording, the declaration sets for looking up
B1::foo and B2::foo are the same. I believe the case presented is valid under
both the C++03 and the N3126 wording.
>>>
C++03 subclause 10.2 [class.member.lookup] paragraph 2:
First, every declaration for the name in the class and in each of its base
class sub-objects is considered.
...
Each of these declarations that was introduced by a using-declaration is
considered to be from each sub-object of C that is of the type containing the
declaration designated by the using-declaration. If the resulting set of
declarations are not all from sub-objects of the same type, or the set has a
nonstatic member and includes members from distinct sub-objects, there is an
ambiguity and the program is ill-formed. Otherwise that set is the result of
the lookup.
<<<
My understanding is that the resulting set of declarations are all from
subobjects of the same type (the two subobjects, C::B1::A and C::B2::A) and the
set has no nonstatic members (all functions foo() are static member functions).
>From this paragraph, we have a set of declarations as the result of lookup:
{ ::A::foo(void), ::A::foo(char) }
Overload resolution then takes place.
