c++/10146: [3.4 regression] [new parser] template function lookup failure(s)

[Gabriel Dos Reis]

The following reply was made to PR c++/10146; it has been noted by GNATS.

From: Gabriel Dos Reis <gdr@integrable-solutions.net>
To: Richard Guenther <rguenth@tat.physik.uni-tuebingen.de>
Cc: Giovanni Bajo <giovannibajo@libero.it>, gcc-gnats@gcc.gnu.org,
   <gcc-bugs@gcc.gnu.org>
Subject: Re: c++/10146: [3.4 regression] [new parser] template function lookup failure(s)
Date: 19 Mar 2003 13:26:14 +0100

 Richard Guenther <rguenth@tat.physik.uni-tuebingen.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