No, it is not right. With the ibarrier usage you're making below, the output
should be similar to the first case (all should leave at earlist at 6.0). The
ibarrier is still a synchronizing point, all processes MUST reach it before
anyone is allowed to leave.
However, if you move the ibarrier on proc < 2 before the sleep, the output you
got become possible.
George
On Jul 6, 2012, at 7:53, Eugene Loh <eugene....@oracle.com> wrote:
> Either there is a problem with MPI_Ibarrier or I don't understand the
> semantics.
>
> The following example is with openmpi-1.9a1r26747. (Thanks for the fix in
> 26757. I tried with that as well with same results.) I get similar results
> for different OSes, compilers, bitness, etc.
>
> % cat ibarrier.c
> #include <stdio.h>
> #include <unistd.h>
> #include <stdlib.h>
> #include <mpi.h>
>
> int main(int argc, char** argv) {
> int i, me;
> double t0, t1, t2;
> MPI_Request req;
>
> MPI_Init(&argc,&argv);
> MPI_Comm_rank(MPI_COMM_WORLD,&me);
>
> MPI_Barrier(MPI_COMM_WORLD);
> MPI_Barrier(MPI_COMM_WORLD);
> MPI_Barrier(MPI_COMM_WORLD);
> t0 = MPI_Wtime(); /* set "time zero" */
>
> if ( me < 2 ) sleep(3); /* two processes delay before hitting
> barrier */
> t1 = MPI_Wtime() - t0;
> MPI_Barrier(MPI_COMM_WORLD);
> t2 = MPI_Wtime() - t0;
> printf("%d entered at %3.1lf and exited at %3.1lf\n", me, t1, t2);
>
> if ( me < 2 ) sleep(3); /* two processes delay before hitting
> barrier */
> t1 = MPI_Wtime() - t0;
> MPI_Ibarrier(MPI_COMM_WORLD, &req);
> MPI_Wait(&req, MPI_STATUS_IGNORE);
> t2 = MPI_Wtime() - t0;
> printf("%d entered at %3.1lf and exited at %3.1lf\n", me, t1, t2);
>
> MPI_Finalize();
> return 0;
> }
> % mpirun -n 4 ./a.out
> 0 entered at 3.0 and exited at 3.0
> 1 entered at 3.0 and exited at 3.0
> 2 entered at 0.0 and exited at 3.0
> 3 entered at 0.0 and exited at 3.0
> 0 entered at 6.0 and exited at 6.0
> 1 entered at 6.0 and exited at 6.0
> 2 entered at 3.0 and exited at 3.0
> 3 entered at 3.0 and exited at 3.0
>
> With the first barrier, no one leaves until the last process has entered.
> With the non-blocking barrier, two processes enter and leave before the two
> laggards arrive at the barrier. Is that right?
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