No, there is no such call in MPI since it is trivial to implement it using point-to-point communication. Of course you could write one, for example (with some rudimentary support for error handling):
// Just a random tag that is unlikely to be used by the rest of the program
#define TAG_IDEAL_SNDRCV 11223
int Ideal_MPI_SendRecv(void *buf, int count, MPI_Datatype datatype,
int source, int dest, MPI_Comm comm)
{
int rank;
int err;
if (source == dest)
return MPI_SUCCESS;
err = MPI_Comm_rank(comm, &rank);
if (err != MPI_SUCCESS)
return err;
if (rank == source)
err = MPI_Send(buf, count, datatype, dest, TAG_IDEAL_SNDRCV, comm);
else if (rank == dest)
err = MPI_Recv(buf, count, datatype, source, TAG_IDEAL_SNDRCV, comm,
MPI_STATUS_IGNORE);
return err;
}
// Example: transfer 'int buf[10]' from rank 0 to rank 2
Ideal_MPI_SendRecv(buf, 10, MPI_INT, 0, 2, MPI_COMM_WORLD);
You could also add another output argument of type MPI_Status *
and store the status of MPI_Recv
there. It could be useful if both processes have different buffer sizes.
Another option would be, if you have to do that many times within a fixed set of ranks, e.g. always from rank 0 to rank 2, you could simply create a new communicator and broadcast inside it:
int rank;
MPI_Comm buddycomm;
MPI_Comm_rank(MPI_COMM_WORLD, &rank);
MPI_Comm_split(MPI_COMM_WORLD, (!rank || rank == 2) ? 0 : MPI_UNDEFINED, rank,
&buddycomm);
// Transfer 'int buf[10]' from rank 0 to rank 2
MPI_Bcast(buf, 10, MPI_INT, 0, buddycomm);
This, of course, is an overkill since the broadcast is more expensive than the simple combination of MPI_Send
and MPI_Recv
.