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

[Gabriel Dos Reis]

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. 

-- Gaby