[Bug fortran/46244] New: gfc_compare_derived_types is buggy

[janus at gcc dot gnu.org]

http://gcc.gnu.org/bugzilla/show_bug.cgi?id=46244

           Summary: gfc_compare_derived_types is buggy
           Product: gcc
           Version: 4.6.0
            Status: UNCONFIRMED
          Severity: normal
          Priority: P3
         Component: fortran
        AssignedTo: unassigned@gcc.gnu.org
        ReportedBy: janus@gcc.gnu.org


As noted by Dominique in PR 46196 comment #5, there is a problem in
gfc_compare_derived_types (interface.c):

--- ../_clean/gcc/fortran/interface.c    2010-10-27 23:47:20.000000000 +0200
+++ gcc/fortran/interface.c    2010-10-29 10:55:07.000000000 +0200
@@ -445,16 +445,16 @@ gfc_compare_derived_types (gfc_symbol *d
       /* Make sure that link lists do not put this function into an 
     endless recursive loop!  */
       if (!(dt1->ts.type == BT_DERIVED && derived1 == dt1->ts.u.derived)
-        && !(dt1->ts.type == BT_DERIVED && derived1 == dt1->ts.u.derived)
+        && !(dt2->ts.type == BT_DERIVED && derived2 == dt2->ts.u.derived)
        && gfc_compare_types (&dt1->ts, &dt2->ts) == 0)
    return 0;

       else if ((dt1->ts.type == BT_DERIVED && derived1 == dt1->ts.u.derived)
-        && !(dt1->ts.type == BT_DERIVED && derived1 == dt1->ts.u.derived))
+        && !(dt2->ts.type == BT_DERIVED && derived2 == dt2->ts.u.derived))
    return 0;

       else if (!(dt1->ts.type == BT_DERIVED && derived1 == dt1->ts.u.derived)
-        && (dt1->ts.type == BT_DERIVED && derived1 == dt1->ts.u.derived))
+        && (dt2->ts.type == BT_DERIVED && derived2 == dt2->ts.u.derived))
    return 0;

       dt1 = dt1->next;