我有一个与 MPI 并行化的应用程序，它被分成许多不同的任务。每个处理器只分配一个任务，分配相同任务的一组处理器分配它自己的通信器。定期地，任务需要同步。目前，同步是通过 MPI_COMM_WORLD 完成的，但这样做的缺点是不能使用集体操作，因为不能保证其他任务会到达该代码块。
作为一个更具体的例子：
task1: equation1_solver, N nodes, communicator: mpi_comm_solver1
task2: equation2_solver, M nodes, communicator: mpi_comm_solver2
task3: file IO , 1 node , communicator: mpi_comm_io
我想 MPI_SUM 在 task1 上的一个数组，并让结果出现在 task3 上。有没有一种有效的方法来做到这一点？（如果这是一个愚蠢的问题，我深表歉意，我在创建和使用自定义 MPI 通信器方面没有太多经验）
查尔斯完全正确；相互沟通者允许您在沟通者之间进行交谈（或者，在这种情况下区分“正常”沟通者，“内部沟通者”，这对我来说并没有太大的改进）。
对于那些刚接触它的人来说，我总是发现这些互通器的使用有点令人困惑。不是有意义的基本想法，而是使用（比如说）MPI_Reduce其中之一的机制。执行归约的任务组指定远程通信器上的根等级，到目前为止一切顺利；但是在远程等级通信器中，不是根的每个人都指定MPI_PROC_NULL为根，而实际的根则指定MPI_ROOT. 为向后兼容所做的事情，嘿？
#include <mpi.h>
#include <stdio.h>
int main(int argc, char **argv)
{
int commnum = 0; /* which of the 3 comms I belong to */
MPI_Comm mycomm; /* Communicator I belong to */
MPI_Comm intercomm; /* inter-communicator */
int cw_rank, cw_size; /* size, rank in MPI_COMM_WORLD */
int rank; /* rank in local communicator */
MPI_Init(&argc, &argv);
MPI_Comm_rank(MPI_COMM_WORLD, &cw_rank);
MPI_Comm_size(MPI_COMM_WORLD, &cw_size);
if (cw_rank == cw_size-1) /* last task is IO task */
commnum = 2;
else {
if (cw_rank < (cw_size-1)/2)
commnum = 0;
else
commnum = 1;
}
printf("Rank %d in comm %d\n", cw_rank, commnum);
/* create the local communicator, mycomm */
MPI_Comm_split(MPI_COMM_WORLD, commnum, cw_rank, &mycomm);
const int lldr_tag = 1;
const int intercomm_tag = 2;
if (commnum == 0) {
/* comm 0 needs to communicate with comm 2. */
/* create an intercommunicator: */
/* rank 0 in our new communicator will be the "local leader"
* of this commuicator for the purpose of the intercommuniator */
int local_leader = 0;
/* Now, since we're not part of the other communicator (and vice
* versa) we have to refer to the "remote leader" in terms of its
* rank in COMM_WORLD. For us, that's easy; the remote leader
* in the IO comm is defined to be cw_size-1, because that's the
* only task in that comm. But for them, it's harder. So we'll
* send that task the id of our local leader. */
/* find out which rank in COMM_WORLD is the local leader */
MPI_Comm_rank(mycomm, &rank);
if (rank == 0)
MPI_Send(&cw_rank, 1, MPI_INT, cw_size-1, 1, MPI_COMM_WORLD);
/* now create the inter-communicator */
MPI_Intercomm_create( mycomm, local_leader,
MPI_COMM_WORLD, cw_size-1,
intercomm_tag, &intercomm);
}
else if (commnum == 2)
{
/* there's only one task in this comm */
int local_leader = 0;
int rmt_ldr;
MPI_Status s;
MPI_Recv(&rmt_ldr, 1, MPI_INT, MPI_ANY_SOURCE, lldr_tag, MPI_COMM_WORLD, &s);
MPI_Intercomm_create( mycomm, local_leader,
MPI_COMM_WORLD, rmt_ldr,
intercomm_tag, &intercomm);
}
/* now let's play with our communicators and make sure they work */
if (commnum == 0) {
int max_of_ranks = 0;
/* try it internally; */
MPI_Reduce(&rank, &max_of_ranks, 1, MPI_INT, MPI_MAX, 0, mycomm);
if (rank == 0) {
printf("Within comm 0: maximum of ranks is %d\n", max_of_ranks);
printf("Within comm 0: sum of ranks should be %d\n", max_of_ranks*(max_of_ranks+1)/2);
}
/* now try summing it to the other comm */
/* the "root" parameter here is the root in the remote group */
MPI_Reduce(&rank, &max_of_ranks, 1, MPI_INT, MPI_SUM, 0, intercomm);
}
if (commnum == 2) {
int sum_of_ranks = -999;
int rootproc;
/* get reduction data from other comm */
if (rank == 0) /* am I the root of this reduce? */
rootproc = MPI_ROOT;
else
rootproc = MPI_PROC_NULL;
MPI_Reduce(&rank, &sum_of_ranks, 1, MPI_INT, MPI_SUM, rootproc, intercomm);
if (rank == 0)
printf("From comm 2: sum of ranks is %d\n", sum_of_ranks);
}
if (commnum == 0 || commnum == 2);
MPI_Comm_free(&intercomm);
MPI_Finalize();
}
您所需要的只是创建一个新的通信器，其中包括要一起通信的两个任务的节点。看看 MPI Groups and Communicators。你可以在网上找到很多例子，例如这里。