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From: Richard Guenther <rguenth@tat.physik.uni-tuebingen.de> To: nobody@gcc.gnu.org Cc: gcc-prs@gcc.gnu.org, Subject: Re: c++/10146: [3.4 regression] [new parser] template function lookup failure(s) Date: Wed, 19 Mar 2003 12:56:00 -0000 [thread overview] Message-ID: <20030319125600.11409.qmail@sources.redhat.com> (raw) 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/
next reply other threads:[~2003-03-19 12:56 UTC|newest] Thread overview: 9+ messages / expand[flat|nested] mbox.gz Atom feed top 2003-03-19 12:56 Richard Guenther [this message] -- strict thread matches above, loose matches on Subject: below -- 2003-03-20 2:05 bangerth 2003-03-19 12:56 Giovanni Bajo 2003-03-19 12:46 Gabriel Dos Reis 2003-03-19 12:36 Gabriel Dos Reis 2003-03-19 12:36 Richard Guenther 2003-03-19 12:26 Richard Guenther 2003-03-19 12:26 Giovanni Bajo 2003-03-19 8:56 rguenth
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