[Bug fortran/17229] New: parser confused by arithmetic if inside an if

["bdavis at gcc dot gnu dot org" <gcc-bugzilla at gcc dot gnu dot org>]

reduced example to demonstrate:

$ cat z.f
      INTEGER INCX
      INTEGER INCY
      IF (INCX.EQ.INCY) IF (INCX-1) 5,20,60
 5    CONTINUE
 20   CONTINUE
 60   CONTINUE
      END
$ gfc z.f
 In file z.f:3
 
      IF (INCX.EQ.INCY) IF (INCX-1) 5,20,60
                       1
Error: Unclassifiable statement in IF-clause at (1)
 In file z.f:4
 
 5    CONTINUE
 1
Warning: Label 5 at (1) defined but not used
 In file z.f:5
 
 20   CONTINUE
  1
Warning: Label 20 at (1) defined but not used
 In file z.f:6
 
 60   CONTINUE
  1
Warning: Label 60 at (1) defined but not used
$ gfc --version
GNU Fortran 95 (GCC 3.5.0 20040829 (experimental))
Copyright (C) 2003 Free Software Foundation, Inc.
 
GNU Fortran comes with NO WARRANTY, to the extent permitted by law.
You may redistribute copies of GNU Fortran
under the terms of the GNU General Public License.
For more information about these matters, see the file named COPYING


works under g77:

$ g77 z.f

-- 
           Summary: parser confused by arithmetic if inside an if
           Product: gcc
           Version: 3.5.0
            Status: UNCONFIRMED
          Severity: normal
          Priority: P2
         Component: fortran
        AssignedTo: unassigned at gcc dot gnu dot org
        ReportedBy: bdavis at gcc dot gnu dot org
                CC: gcc-bugs at gcc dot gnu dot org


http://gcc.gnu.org/bugzilla/show_bug.cgi?id=17229