[Bug c++/93807] -std=c++2a allows to omit out-of-class declaration in template class

[cvs-commit at gcc dot gnu.org]

https://gcc.gnu.org/bugzilla/show_bug.cgi?id=93807

--- Comment #4 from CVS Commits <cvs-commit at gcc dot gnu.org> ---
The master branch has been updated by Marek Polacek <mpolacek@gcc.gnu.org>:

https://gcc.gnu.org/g:edfa7c684d329708dcee733fcb552bed9d931621

commit r10-7894-gedfa7c684d329708dcee733fcb552bed9d931621
Author: Marek Polacek <polacek@redhat.com>
Date:   Tue Apr 21 18:11:33 2020 -0400

    c++: Add test for c++/93807

    This PR was initially accepts-invalid, but I think it's actually valid
    C++20 code.  My reasoning is that in C++20 we no longer require the
    declaration of operator== (#if-defed in the test), because C++20's
    [temp.names]/2 says "A name is also considered to refer to a template
    if it is an unqualified-id followed by a < and name lookup either finds
    one or more functions or finds nothing." so when we're parsing

      constexpr friend bool operator==<T>(T lhs, const Foo& rhs);

    we treat "operator==" as a template name, because name lookup of
    "operator==" found nothing and we have an operator-function-id, which is
    an unqualified-id, and it's followed by a <.  So the declaration isn't
    needed to treat "operator==<T>" as a template-id.

            PR c++/93807
            * g++.dg/cpp2a/fn-template20.C: New test.