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

[Richard Guenther]

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

From: Richard Guenther <rguenth@tat.physik.uni-tuebingen.de>
To: Gabriel Dos Reis <gdr@integrable-solutions.net>
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: Wed, 19 Mar 2003 13:47:40 +0100 (CET)

 On 19 Mar 2003, Gabriel Dos Reis wrote:
 
 > Richard Guenther <rguenth@tat.physik.uni-tuebingen.de> writes:
 >
 > | On 19 Mar 2003, Gabriel Dos Reis wrote:
 > |
 > | > 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.
 > |
 > | Ah, ok - only for Foo<T>:: it would be dependent, yes?
 >
 > Yes.  (Any dependent expression would do also).
 
 If I exchange Foo<int> with Foo<T> only
   Foo<T>().foo<U>(u);
 and
   Foo<T>().bar<U>(u);
 do not work (as expected), but f.i. Foo<T>::foo<U>(u); (while
 Foo<int>::foo<U>(u) is rejected) is ok?
 
 Well, I'll leave it now up to Mark to check&fix all relevant cases
 after 3.3 is out.
 
 Richard.
 
 --
 Richard Guenther <richard.guenther@uni-tuebingen.de>
 WWW: http://www.tat.physik.uni-tuebingen.de/~rguenth/