[Bug c++/18541] New: "using" fails to make name visible without qualification

[Bug c++/18541] New: "using" fails to make name visible without qualification

[igodard at pacbell dot net]

#include <ostream>
template<typename ch = char, typename tr = std::char_traits<ch> >
class buffer : public std::basic_streambuf<ch, tr> {
    virtual
    int_type    overflow(int_type c = tr::eof());

    virtual
    typename int_type   overflow1(typename int_type c = tr::eof());

    using std::basic_streambuf<ch, tr>::int_type;

    virtual
    int_type    overflow2(int_type c = tr::eof());

    virtual
    typename int_type   overflow3(typename int_type c = tr::eof());

    virtual
    std::basic_streambuf<ch, tr>::int_type  overflow4(
        std::basic_streambuf<ch, tr>::int_type c = tr::eof());

    virtual
    typename std::basic_streambuf<ch, tr>::int_type overflow5(
        typename std::basic_streambuf<ch, tr>::int_type c = tr::eof());
    };
int main() {
    buffer<> b;
    }

gets you:

~/ootbc/common/test/src$ g++ foo.cc
foo.cc:5: error: `int_type' does not name a type
foo.cc:5: error: (perhaps `typename std::basic_streambuf<_CharT,
_Traits>::int_type' was intended)
foo.cc:8: error: expected nested-name-specifier before "int_type"
foo.cc:8: error: `int_type' does not name a type
foo.cc:8: error: (perhaps `typename std::basic_streambuf<_CharT,
_Traits>::int_type' was intended)
foo.cc:13: error: `int_type' does not name a type
foo.cc:13: error: (perhaps `typename std::basic_streambuf<_CharT,
_Traits>::int_type' was intended)
foo.cc:16: error: expected nested-name-specifier before "int_type"
foo.cc:16: error: `int_type' does not name a type
foo.cc:16: error: (perhaps `typename std::basic_streambuf<_CharT,
_Traits>::int_type' was intended)
foo.cc:19: error: ISO C++ forbids declaration of `int_type' with no type
foo.cc:19: error: cannot declare member `std::basic_streambuf<_CharT,
_Traits>::int_type' within `buffer<ch, tr>'
foo.cc:19: error: expected `;' before "overflow4"


As suggested in the error for "overflow" (good message BTW), the form used in 
"overflow5" works. But a "using" for the inherited name "int_type" does not make 
the name visible either directly (overflow2) or when tagged with "typename" 
(overflow 3). But the "using" doesn't complain either. If I can "using" it I 
should be able to "use" it :-)

Ivan

-- 
           Summary: "using" fails to make name visible without qualification
           Product: gcc
           Version: 3.4.0
            Status: UNCONFIRMED
          Severity: normal
          Priority: P2
         Component: c++
        AssignedTo: unassigned at gcc dot gnu dot org
        ReportedBy: igodard at pacbell dot net
                CC: gcc-bugs at gcc dot gnu dot org


http://gcc.gnu.org/bugzilla/show_bug.cgi?id=18541

[Bug c++/18541] "using" fails to make name visible without qualification

[pinskia at gcc dot gnu dot org]

------- Additional Comments From pinskia at gcc dot gnu dot org  2004-11-18 12:25 -------
Yes this is a bug in GCC, see PR 14258.

*** This bug has been marked as a duplicate of 14258 ***

-- 
           What    |Removed                     |Added
----------------------------------------------------------------------------
             Status|UNCONFIRMED                 |RESOLVED
         Resolution|                            |DUPLICATE


http://gcc.gnu.org/bugzilla/show_bug.cgi?id=18541