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

[Richard Guenther <rguenth at tat dot physik dot uni-tuebingen dot de>]

On 19 Mar 2003, Gabriel Dos Reis wrote:

> Richard Guenther <rguenth at tat dot physik dot uni-tuebingen dot de> writes:
>
> | On 19 Mar 2003, Gabriel Dos Reis wrote:
> |
> | > Richard Guenther <rguenth at tat dot physik dot uni-tuebingen dot 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 dot guenther at uni-tuebingen dot de>
WWW: http://www.tat.physik.uni-tuebingen.de/~rguenth/