[mpich-devel] call PMI before MPI_Init_thread?

[Jeff Hammond]

So MPICH doesn't fail when I try this on my Mac, but it would be good
to know if I have a reasonable expectation of doing this in general.

Thanks,

Jeff

jrhammon-mac01:ENDPOINTS jrhammon$ mpiexec -n 4 ./a.out
PMI says I am 2 of 4
PMI says I am 0 of 4
PMI says I am 1 of 4
PMI says I am 3 of 4
MPI says I am 0 of 4 (provided=1)
MPI says I am 2 of 4 (provided=1)
MPI says I am 3 of 4 (provided=3)
MPI says I am 1 of 4 (provided=3)

jrhammon-mac01:ENDPOINTS jrhammon$ cat pmi.c
#include <stdio.h>
#include <stdlib.h>

/*#include <pmi.h>*/
int PMI_Init( int *spawned );
int PMI_Finalize( void );
int PMI_Get_size( int *size );
int PMI_Get_rank( int *rank );

#include <mpi.h>

int main(int argc, char * argv[])
{
    int has_parent;
    PMI_Init(&has_parent);

    int rank, size;
    PMI_Get_size(&size);
    PMI_Get_rank(&rank);

    printf("PMI says I am %d of %d \n", rank, size);
    fflush(stdout);

    int requested = (rank%2==0) ? MPI_THREAD_FUNNELED : MPI_THREAD_MULTIPLE;
    int provided;
    MPI_Init_thread(&argc, &argv, requested, &provided);

    int wsize, wrank;
    MPI_Comm_size(MPI_COMM_WORLD, &wsize);
    MPI_Comm_rank(MPI_COMM_WORLD, &wrank);

    printf("MPI says I am %d of %d (provided=%d)\n", wrank, wsize, provided);
    fflush(stdout);

    MPI_Finalize();
    //PMI_Finalize(); /* required? */

    return 0;
}

On Tue, Apr 21, 2015 at 9:09 PM, Jeff Hammond <jeff.science at gmail.com> wrote:
> Is it valid in _MPICH_ to make calls to PMI before MPI is initialized?
>  I am interested in PMI_Init, PMI_Get_size and PMI_Get_rank.
>
> The motivating use case is to initialize e.g. half my ranks with
> MPI_THREAD_FUNNELED and the other half with MPI_THREAD_MULTIPLE but
> not use MPMD launching.
>
> I have two reasons for wanting to do it this way.  First, it is
> otherwise unnecessary to use MPMD since I can branch at the top of my
> code and meet the requirements of FUNNELED and MULTIPLE.  Second, I
> might be interested in running my code on a supercomputer with an
> MPICH-based MPI implementation that does not support MPMD launching in
> an unrestricted manner.
>
> If MPICH can handle this, then I will worry about whether or not the
> aforementioned supercomputer can.  It is my assumption that, if MPICH
> cannot handle this, then the MPICH derivative in question cannot
> either.
>
> Thanks,
>
> Jeff
>
> --
> Jeff Hammond
> jeff.science at gmail.com
> http://jeffhammond.github.io/



-- 
Jeff Hammond
jeff.science at gmail.com
http://jeffhammond.github.io/