[Bug fortran/55086] New: ICE with FORALL in allocate_temp_for_forall_nest_1

burnus at gcc dot gnu.org gcc-bugzilla@gcc.gnu.org
Fri Oct 26 16:32:00 GMT 2012


http://gcc.gnu.org/bugzilla/show_bug.cgi?id=55086

             Bug #: 55086
           Summary: ICE with FORALL in allocate_temp_for_forall_nest_1
    Classification: Unclassified
           Product: gcc
           Version: 4.8.0
            Status: UNCONFIRMED
          Keywords: ice-on-valid-code
          Severity: normal
          Priority: P3
         Component: fortran
        AssignedTo: unassigned@gcc.gnu.org
        ReportedBy: burnus@gcc.gnu.org


The following program crashes in trans-stmt.c:3234's
allocate_temp_for_forall_nest_1 for:

3234      unit = fold_convert (gfc_array_index_type, TYPE_SIZE_UNIT (type));

as TYPE_SIZE_UNIT(type) == NULL.


ICEs in the such-marked line. ICEs with GCC 4.1 to 4.8.

  implicit none
  character(len=5), pointer :: a(:), b(:)
  character(len=5), pointer :: c, d
  allocate (a(2), b(2), c, d)
  a = [ "abcde", "ABCDE" ]
  call aloct_pointer_copy_4 (b, a)
  print *, b(1)
  print *, b(2)
  if (any (a /= b)) stop 'WRONG'

  call aloct_copy_4 (b, a)
  print *, b(1)
  print *, b(2)
  if (any (a /= b)) stop 'WRONG'

  d = '12345'
  c = "abcde"
  call test2 (d, c)
  print *, d
  if (d /= '1cb15') stop 'WRONG'

  call test2p (d, c)
  print *, d
  if (d /= '1cb15') stop 'WRONG'

contains
 subroutine aloct_pointer_copy_4(o, i)
  character(len=*), pointer :: o(:), i(:)
  integer :: nl1, nu1
  integer :: i1
  nl1 = lbound(i,dim=1)
  nu1 = ubound(i,dim=1)
  forall (i1 = nl1:nu1) o(i1) = i(i1)
 end subroutine aloct_pointer_copy_4
 subroutine aloct_copy_4(o, i)
  character(len=*), pointer :: o(:), i(:)
  integer :: nl1, nu1
  integer :: i1
  nl1 = lbound(i,dim=1)
  nu1 = ubound(i,dim=1)
  forall (i1 = nl1:nu1) o(i1) = i(i1)
 end subroutine aloct_copy_4
 subroutine test2(o, i)
  character(len=*) :: o, i
  integer :: nl1, nu1
  integer :: i1
  nl1 = 2
  nu1 = 4
  forall (i1 = nl1:nu1) o(i1:i1) = i(i1:i1)
  forall (i1 = nl1:nu1) o(i1:i1) = o(nu1+1-i1:nu1+1-i1)
 end subroutine test2
 subroutine test2p(o, i)
  character(len=*), pointer :: o, i
  integer :: nl1, nu1
  integer :: i1
  nl1 = 2
  nu1 = 4
  forall (i1 = nl1:nu1) o(i1:i1) = i(i1:i1)   ! <<<< ICE
  forall (i1 = nl1:nu1) o(i1:i1) = o(nu1+1-i1:nu1+1-i1)
 end subroutine test2p
end



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