I think that forall statement must be at least as fast as equivalent
do-
-end do construction.
But the next program (variant of LU-decomposition) shows that fragment
containing forall statement is approximately at 2.5(!) times slower then
fragment with do-end do.
program test
implicit none
integer, parameter :: n = 2000
integer i, j
double precision, dimension (n, n) :: a, a1
double precision, dimension (n) :: work
real time_begin, time_end
integer, dimension(1) :: max_loc
intrinsic random_number, maxloc, CPU_TIME
call random_number(a)
a1=a
CALL CPU_TIME(time_begin)
do i = 1, n-1
max_loc = maxloc(abs(a(i:,i)))
j = max_loc(1) + i - 1
if (a(j,i) == 0.0) stop 'Zero pivot'
if (i /= j) then
work(i:n) = a(i,i:n)
a(i,i:n) = a(j,i:n)
a(j,i:n) = work(i:n)
end if
a(i+1:,i) = a(i+1:,i) / a(i,i)
do j = i+1, n
a(i+1:,j) = a(i+1:,j) - a(i,j) * a(i+1:,i)
end do
end do
CALL CPU_TIME(time_end)
print *, 'Time of operation was ', time_end - time_begin, ' seconds'
a=a1
CALL CPU_TIME(time_begin)
do i = 1, n-1
max_loc = maxloc(abs(a(i:,i)))
j = max_loc(1) + i - 1
if (a(j,i) == 0.0) stop 'Zero pivot'
if (i /= j) then
work(i:n) = a(i,i:n)
a(i,i:n) = a(j,i:n)
a(j,i:n) = work(i:n)
end if
a(i+1:,i) = a(i+1:,i) / a(i,i)
forall (j = i+1:n) a(i+1:,j) = a(i+1:,j) - a(i,j) * a(i+1:,i)
end do
CALL CPU_TIME(time_end)
print *, 'Time of operation was ', time_end - time_begin, ' seconds'
end program test
GCC version 4.4.2. Windows Vista SP2, CPU: Intel Core 2 Quad Q6600, RAM: 3 GB
Gfortran O3 file_name.f95
--
Summary: Slow forall
Product: gcc
Version: 4.4.2
Status: UNCONFIRMED
Severity: normal
Priority: P3
Component: fortran
AssignedTo: unassigned at gcc dot gnu dot org
ReportedBy: gretsov at gmail dot com
http://gcc.gnu.org/bugzilla/show_bug.cgi?id=42118
