mpi4py Send/Recv with tag

mpi, python

Solution

The function `Recv` will store a received message inside a variable. You have to supply the rank of the expected sender. Thus you always know who the sender is. A message passing interface does never need to identify someone, that information is always intrinsic to the system.

If you expect multiple message from the same sender you can distinguish these using tags. You need to supply these tags yourself, there is no natural way to obtain these. Just label the messages somehow, number them.

If you have a tag, the `Recv` function will only return when a message has been received which has a fitting source and tag. This is a blocking function call.

In your case, `tag=-1` is equal to the universal constant `MPI.ANY_TAG` (verify via `print MPI.ANY_TAG`) and thus the `Recv` will accept any tag. But it will in no way overwrite its input variable `rnk`. Try `rnk = -2 # EDIT` and you'll see.

You can write your code differently, though this will not change the underlying logic (i.e. you as a programmer must always know the sender) it just hides it, makes it implicit:

#passRandomDraw.py
import numpy
from mpi4py import MPI
comm = MPI.COMM_WORLD
rank = comm.Get_rank()

randNum = numpy.zeros(1)
rnk = -1 # EDIT

if rank == 1:
    randNum = numpy.random.random_sample(1)
    print "Process", rank, "drew the number", randNum[0]
    comm.Send(randNum, dest=0, tag=rank) # EDIT

if rank == 0:
    print "Process", rank, "before receiving has the number", randNum[0]
    print "Sender rank:", rnk
    status = MPI.Status()
    comm.Recv(randNum, source=MPI.ANY_SOURCE, tag=MPI.ANY_TAG, status=status) # EDIT
    rnk = status.Get_source()
    print "Process", rank, "received the number", randNum[0]
    print "Sender rank:", rnk # EDIT

Problem

Original source

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