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

[Richard Guenther]

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.guenther@uni-tuebingen.de>
WWW: http://www.tat.physik.uni-tuebingen.de/~rguenth/