Dmitri,
Perfect! Thanks so much for the response. Your guess about the barrier was
exactly correct. The problem has disappeared.
I was ignorant about the proper way to specify shared variables. Thanks for
the correction. Is the following use of reduction acceptable?
!$OMP PARALLEL
!$OMP& PRIVATE(i,j)
!$OMP& SHARED(Ny,Nx,dv,v,boundary))
!
!$OMP DO
...
...
!$OMP END DO
max_dv = 0.d0
!$OMP BARRIER
!$OMP DO
!$OMP& REDUCTION(MAX:max_dv)
do j=2,(Ny-1)
do i=2,(Nx-1)
v(i,j) = v(i,j) + dv(i,j)
if(dabs(dv(i,j)) .gt. max_dv) then
max_dv = dv(i,j)
endif
end do
end do
!$OMP END DO
On Thu, Nov 27, 2008 at 4:07 AM, Dmitri Chubarov
<[EMAIL PROTECTED]>wrote:
> Nathan, hello,
>
> I gave your code a second look and noticed this:
>
> !$OMP PARALLEL
>>
>
>> !$OMP DO
>>
> ....
>> !$OMP END DO
>>
>> max_dv = 0.d0
>> !$OMP DO
>>
> ....
>
>> !$OMP END DO
>> !$OMP END PARALLEL
>>
>>
> There is a BARRIER missing between max_dv = 0.d0 and the following loop.
> One of the threads in the pool might've been late and get to this statement
> when the rest have already completed the reduction loop.
>
> The barrier is also important to ensure that no thread would use the values
> of dv(i,j) in the reduction loop before they are updated by the main
> computational loop above.
>
> Finally
> if(dv(i,j) .gt. max_dv) then
> max_dv = dv(i,j)
> endif
> Does not look right since it would not handle negative values of dv(i,j)
> correctly. I assume it should read
> as
> max_dv = max(max_dv, dabs(dv(i,j) ))
>
> Best regards,
> Dmitri Chubarov
>
> --
> Junior Researcher
> Siberian Branch of the Russian Academy of Sciences
> Institute of Computational Technologies
>
>
--
- - - - - - - - - - - - - - - - - - - - -
Nathan Moore
Assistant Professor, Physics
Winona State University
AIM: nmoorewsu
- - - - - - - - - - - - - - - - - - - - -
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