[Nek5000-users] How to get another element information that in the different processor?

[nek5000-users at lists.mcs.anl.gov]

Dear Hu,


What information are you needing from another element.


The usual approach, which appears relevant to your situation, is to recognize that communication is generally reciprocal:  If element e needs data from element e', then e' also likely needs the same date from e.


These operations are most readily handled through the gather-scatter operator, gs_op(), which can exchange data among participating processors.  (All processors must participate, but some might not have data to exchange.)


Are you trying to effect local averages? or, what?


Many such operations can also be effected through existing spectral element operators, etc.


hth,


Paul


________________________________
From: Nek5000-users <nek5000-users-bounces at lists.mcs.anl.gov> on behalf of nek5000-users at lists.mcs.anl.gov <nek5000-users at lists.mcs.anl.gov>
Sent: Monday, July 24, 2017 9:06:50 PM
To: nek5000-users at lists.mcs.anl.gov
Subject: [Nek5000-users] How to get another element information that in the different processor?

Hi all,
I want to get another element information when I handle one element, and the two elements maybe not in the same processor. How to get another element information that in the different processor?

Below is some of my code:
1)for element e, get global element number;
2)get the correspond element number at the inlet;
3)call the ym1 value at the inlet element
But I don't know how to get ym1(i,j,k,e_ele) in the e_nid processor.

 do e=1,nelv
        eg=lglel(e)
        call get_exyz(ex,ey,ez,eg,nelx,nely,nelz)
        e_in=(ez-1)*nelx*nely+(ey-1)*nelx+1
        e_nid=gllnid(e_in)
        e_ele=gllel(e_in)
c       if (nid.eq.e_nid) then
           do k=1,nz1
            do j=1,ny1
             do i=1,nx1
                temp=-ym1(i,j,k,e)*u_t/vis/A_plus
                cdyn(i,j,k,e)=cs0*(1-exp(temp))
             enddo
            enddo
           enddo
c       endif
      enddo

Best wishes!
Hu
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