[Bug c++/46839] New: Parser confuses derived member with std::forward

[Bug c++/46839] New: Parser confuses derived member with std::forward

[foehst at informatik dot uni-kl.de]

http://gcc.gnu.org/bugzilla/show_bug.cgi?id=46839

           Summary: Parser confuses derived member with std::forward
           Product: gcc
           Version: 4.4.3
            Status: UNCONFIRMED
          Severity: normal
          Priority: P3
         Component: c++
        AssignedTo: unassigned@gcc.gnu.org
        ReportedBy: foehst@informatik.uni-kl.de


Created attachment 22673
  --> http://gcc.gnu.org/bugzilla/attachment.cgi?id=22673
Info about gcc build configuration

01: #include <iostream>
02: #include <utility>
03: using namespace std;
04:
05: template <typename T> struct A
06: {
07:   T forward;
08:   A() : forward(0) {}
09: };
10:
11: template <typename T> struct B : public A<T>
12: {
13:   void Function()
14:   {
15:     cout << (this->forward > 1) << endl;
16:     cout << (this->forward < 1) << endl;
17:   }
18: };
19:
20: int main(int argc, char **argv)
21: {
22:   return 0;
23: }

g++ -Wall -std=gnu++0x

: In member function ‘void B<T>::Function()’:
:16: error: parse error in template argument list

Used compiler version: gcc version 4.4.3 (Gentoo 4.4.3-r2 p1.2)
4.3 and 4.5 versions show the same behavior.

It seems, that in line 16 "forward <" is parsed as the beginning of a template
instantiation. To get this effect, the inclusion of <utility>, using namespace
std and using derived members in a template class is necessary. Maybe, using
namespace typically appears after template definition in other h-files, but a
scenario like this might occur. Apart from that: Would
"this->std::forward<...>" be valid C++ code?

Therefore, I think that is a bug in the parsers which neglects the occurrence
of this and starts to interpret forward as the std::forward template.

Regards,
Tobias Föhst

[Bug c++/46839] Parser confuses derived member with std::forward

[foehst at informatik dot uni-kl.de]

http://gcc.gnu.org/bugzilla/show_bug.cgi?id=46839

--- Comment #1 from Tobias Föhst <foehst at informatik dot uni-kl.de> 2010-12-07 17:12:45 UTC ---
I forgot: Line 15 is only there to prove that this->forward with another
successor than '<' can be resolved as the member derived from A<T>.

[Bug c++/46839] Parser confuses derived member with std::forward

[paolo.carlini at oracle dot com]

http://gcc.gnu.org/bugzilla/show_bug.cgi?id=46839

--- Comment #2 from Paolo Carlini <paolo.carlini at oracle dot com> 2010-12-07 17:19:42 UTC ---
Note that Intel compiler, based on the EDG C++ front-end, also rejects this
code in strict mode. Likewise of course Comeau. This is enough to show the
issue:

template<typename T>
  void forward(T);

template <typename T> struct A
{
  T forward;
};

template <typename T> struct B : public A<T>
{
  void Function()
  {
    this->forward < 1;
  }
};

[Bug c++/46839] Parser confuses derived member with std::forward

[pinskia at gcc dot gnu.org]

http://gcc.gnu.org/bugzilla/show_bug.cgi?id=46839

Andrew Pinski <pinskia at gcc dot gnu.org> changed:

           What    |Removed                     |Added
----------------------------------------------------------------------------
             Status|UNCONFIRMED                 |RESOLVED
         Resolution|                            |DUPLICATE

--- Comment #3 from Andrew Pinski <pinskia at gcc dot gnu.org> 2010-12-07 17:24:19 UTC ---
See PR 20308 and such.

*** This bug has been marked as a duplicate of bug 20308 ***