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Re: c++/10146: [3.4 regression] [new parser] template function lookupfailure(s)
- From: Richard Guenther <rguenth at tat dot physik dot uni-tuebingen dot de>
- To: Giovanni Bajo <giovannibajo at libero dot it>
- Cc: gcc-gnats at gcc dot gnu dot org, <gcc-bugs at gcc dot gnu dot org>, <nobody at gcc dot gnu dot org>, <gcc-prs at gcc dot gnu dot org>
- Date: Wed, 19 Mar 2003 13:20:42 +0100 (CET)
- Subject: Re: c++/10146: [3.4 regression] [new parser] template function lookupfailure(s)
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).
Richard.
--
Richard Guenther <richard dot guenther at uni-tuebingen dot de>
WWW: http://www.tat.physik.uni-tuebingen.de/~rguenth/