[Bug fortran/33686] FORALL loop gives wrong result

["dominiq at lps dot ens dot fr" <gcc-bugzilla at gcc dot gnu dot org>]

------- Comment #4 from dominiq at lps dot ens dot fr  2007-10-08 12:05 -------
You can add xlf to the (3, 1, 4, 2) list. I think this is the right answer.
The following code

PROGRAM TST
  IMPLICIT NONE

  INTEGER :: P(4),Q(4),I
  P = (/2,4,1,3/)
  FORALL(I=1:4)
    Q(P(I)) = I
  END FORALL
  PRINT *, Q

  do I=1,4
    Q(P(I)) = I
  END  do
  PRINT *, Q

  do I=4,1,-1
    Q(P(I)) = I
  END  do
  PRINT *, Q

  FORALL(I=1:4)
    P(P(I)) = I
  END FORALL
  PRINT *, P

  do I=1,4
    P(P(I)) = I
  END  do
  PRINT *, P

  do I=4,1,-1
    P(P(I)) = I
  END  do
  PRINT *, P

END PROGRAM TST

gives with gfortran

           3           1           4           2
           3           1           4           2
           3           1           4           2
           3           1           4           3
           3           1           4           3
           2           1           4           3

My understanding of the FORALL construct is that it is equivalent to any of the
first three loops, followed by P=Q, i.e., P is changed only when all the Q's
have been computed.  Comparing the fourth and fifth lines show that P is
changed within the FORALL before all the rhs has been visited and the sixth
line shows that this depends on the order of their computation.

Note that the code is valid only if P is a permutation. Would it contains a
single repetition, say (/2,4,1,2/), it would be invalid because Q(2) depends on
the order (4 for the first do loop, 1 for the second).


-- 


http://gcc.gnu.org/bugzilla/show_bug.cgi?id=33686