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

[Gabriel Dos Reis <gdr at integrable-solutions dot net>]

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).

-- Gaby