Calling MPI_Comm_Split function

Question

My question is probably stupid but still I'm gonna ask it to be sure !

Question : Do you expect the two codes below to work the same using MPI_Comm_Split to build 1 sub-communicator ? (for example, let's say I'm running the code with 6 procs with a rank between 0 and 5)

NB : The code is in fortran 90 with intel compiler 2019 and I use Mpich for the Mpi.

CODE 1

call Mpi_Init(ierror)
call Mpi_Comm_Rank(mpi_comm_world,rank,ierror)

if (rank > 2) then
    call Mpi_Comm_Split(mpi_comm_world,0,rank,new_comm,ierror)
else
    call Mpi_Comm_Split(mpi_comm_world,mpi_undefined,rank,new_comm,ierror)
endif

CODE 2

call Mpi_Init(ierror)
call Mpi_Comm_Rank(mpi_comm_world,rank,ierror)

if (rank > 2) then
    color = 0
else
    color = mpi_undefined
endif

call Mpi_Comm_Split(mpi_comm_world,color,rank,new_comm,ierror)

The Mpi_Comm_Split is not called the same way in the 2 codes but to me, it should behave the same but I'm not sure... I read that Mpi_Comm_Split has to be invoked at the same line but how procs can know that the call of Mpi_Comm_Split is done at a one line or another one (it doesn't make anysense to me) ?!

NB : With Mpich and intel fortran, I tested it and both implentation of the communicator splitting works, but I'm afraid of the behavior of different Mpi compilers...

Gilles Gouaillardet Gilles Gouaillardet · Accepted Answer · 2020-02-19T00:03:02

Assuming you declared color correctly, both codes are equivalent.

MPI_Comm_split() is a collective operation, and hence must be invoked by all the ranks of the parent communicator. That does not mandate the call has to be performed by the same line of code.

Calling MPI_Comm_Split function

1 Answers