------- 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
