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Re: c++/10146: [3.4 regression] [new parser] template function lookup failure(s)
- From: Gabriel Dos Reis <gdr at integrable-solutions dot net>
- To: Richard Guenther <rguenth at tat dot physik dot uni-tuebingen dot de>
- Cc: Giovanni Bajo <giovannibajo at libero dot it>, gcc-gnats at gcc dot gnu dot org, <gcc-bugs at gcc dot gnu dot org>
- Date: 19 Mar 2003 13:26:14 +0100
- Subject: Re: c++/10146: [3.4 regression] [new parser] template function lookup failure(s)
- Organization: Integrable Solutions
- References: <Pine.LNX.4.44.0303191319110.22005-100000@bellatrix.tat.physik.uni-tuebingen.de>
Richard Guenther <rguenth at tat dot physik dot uni-tuebingen dot de> writes:
| On Wed, 19 Mar 2003, Giovanni Bajo wrote:
|
| >
| > http://gcc.gnu.org/cgi-bin/gnatsweb.pl?cmd=view%20audit-trail&database=gcc&p
| > r=10146
| >
| > To sum it up:
| >
| > >Foo<int>().template foo<U>(u); // does not work
| > >Foo<int>().template bar<U>(u); // does not work
| >
| > These should compile.
| >
| > >Foo<int>().foo(u); // does work ??
| > >Foo<int>::foo(u); // does work ??
| > >Foo<int>().bar(u); // does work ??
| >
| > Yes, because the template parameter of the template member function is
| > deduced from the call. What's wrong with them?
|
| I think foo() and bar() needs to be qualified with the template keyword
| due to two-stage namelookup. But I may be wrong (dont have a standard
| to look at).
Two-phase name lookup has nothing to do here. The Foo<int> part is
-not- dependent, therefore there need not be any extra "template"
qualifier.
-- Gaby