Re: MPI_Win_fence problem.

Thanks,
this is the exact code:
-------------------------------------------------------------------------------
#include <iostream>
#include <windows.h>
#include <math.h>

#undef SEEK_SET
#undef SEEK_END
#undef SEEK_CUR

#include "mpi.h"
#pragma comment(lib, "mpi.lib")
using namespace std;

int main(int argc, char *argv[])
{
int n, myid, numprocs, i;
double PI25DT = 3.141592653589793238462643;
double mypi, pi, h, sum, x;
MPI_Win nwin, piwin;

MPI_Init(&argc,&argv);
MPI_Comm_size(MPI_COMM_WORLD,&numprocs);
MPI_Comm_rank(MPI_COMM_WORLD,&myid);

if (myid == 0) {
MPI_Win_create(&n, sizeof(int), 1, MPI_INFO_NULL,
MPI_COMM_WORLD, &nwin);
MPI_Win_create(&pi, sizeof(double), 1, MPI_INFO_NULL,
MPI_COMM_WORLD, &piwin);
}
else {
MPI_Win_create(MPI_BOTTOM, 0, 1, MPI_INFO_NULL,
MPI_COMM_WORLD, &nwin);
MPI_Win_create(MPI_BOTTOM, 0, 1, MPI_INFO_NULL,
MPI_COMM_WORLD, &piwin);
}
while (1) {
if (myid == 0) {
printf("Enter the number of intervals: (0 quits) ");
scanf("%d",&n);
pi = 0.0;
}

MPI_Win_fence(0, nwin);

if (myid != 0)
MPI_Get(&n, 1, MPI_INT, 0, 0, 1, MPI_INT, nwin);
MPI_Win_fence(0, nwin);
if (n == 0)
break;
else {
h = 1.0 / (double) n;
sum = 0.0;
for (i = myid + 1; i <= n; i += numprocs) {
x = h * ((double)i - 0.5);
sum += (4.0 / (1.0 + x*x));
}
mypi = h * sum;
MPI_Win_fence( 0, piwin);
MPI_Accumulate(&mypi, 1, MPI_DOUBLE, 0, 0, 1, MPI_DOUBLE,
MPI_SUM, piwin);
MPI_Win_fence(0, piwin);
if (myid == 0)
printf("pi is approximately %.16f, Error is %.16f\n",
pi, fabs(pi - PI25DT));
}
}
MPI_Win_free(&nwin);
MPI_Win_free(&piwin);
MPI_Finalize();
return 0;
}
--------------------------------------------
so it is called three times for the slaves and only one time for the
master.

I tried both to skip the 2 fences, and to let node 0 call them to, but
it didn't worked.

How should I adapt the code ? And does it works for you (if you have
mpich 2 installed, at least).

Thanks again.

.

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