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Regarding MPI_Comm_create, the MPI standard says

MPI_COMM_CREATE(comm, group, newcomm)

... The function is collective and must be called by all processes in the group of comm.

I took this to mean that, for instance, if the comm argument is MPI_COMM_WORLD, then all processes must call MPI_COMM_WORLD.

However, I wrote a variation on a code available on the internet demonstrating the use of MPI_Comm_create. It is below. You can see that there are two spots where MPI_Comm_create is called, and not by all processes. And yet the code runs just fine.

Did I get lucky? Did I stumble onto some implementation-dependent feature? Am I misunderstanding the MPI standard? Is the idea that the two calls together result in everyone calling MPI_Comm_create so "at the end of the day" it's OK? Thanks. Here's the code:

#include <stdio.h>
#include <stdlib.h>
#include "mpi.h"


int main(int argc, char **argv) {

   MPI_Comm even_comm, odd_comm;
   MPI_Group even_group, odd_group, world_group;
   int id, even_id, odd_id;
   int *even_ranks, *odd_ranks;
   int num_ranks, num_even_ranks, num_odd_ranks;
   int err_mpi, i, j;
   int even_sum, odd_sum;


   err_mpi = MPI_Init(&argc, &argv);
   MPI_Comm_size(MPI_COMM_WORLD, &num_ranks);
   MPI_Comm_rank(MPI_COMM_WORLD, &id);

   MPI_Comm_group(MPI_COMM_WORLD, &world_group);

   num_even_ranks = (num_ranks+1)/2;
   even_ranks = (int *)malloc(num_even_ranks*sizeof(int));
   j=0;
   for (i=0; i<num_ranks; i++){
      if (i%2 == 0) {
         even_ranks[j] = i;
         j++;
      }
   } 

   num_odd_ranks = num_ranks/2;
   odd_ranks = (int *)malloc(num_odd_ranks*sizeof(int));
   j=0;
   for (i=0; i<num_ranks; i++){
      if (i%2 == 1) {
         odd_ranks[j] = i;
         j++;
      }
   } 

   if (id%2 == 0){
      MPI_Group_incl(world_group, num_even_ranks, even_ranks, &even_group);
      // RIGHT HERE, all procs are NOT calling!
      MPI_Comm_create(MPI_COMM_WORLD, even_group, &even_comm);
      MPI_Comm_rank(even_comm, &even_id);
      odd_id = -1;
   } else {
      MPI_Group_incl(world_group, num_odd_ranks, odd_ranks, &odd_group);
      // RIGHT HERE, all procs are NOT calling!
      MPI_Comm_create(MPI_COMM_WORLD, odd_group, &odd_comm);
      MPI_Comm_rank(odd_comm, &odd_id);
      even_id = -1;
   }

   // Just to have something to do, we'll some up the ranks of
   // the various procs in each communicator.       
   if (even_id != -1) MPI_Reduce(&id, &even_sum, 1, MPI_INT, MPI_SUM, 0, even_comm);
   if (odd_id != -1) MPI_Reduce(&id, &odd_sum, 1, MPI_INT, MPI_SUM, 0, odd_comm);

   if (odd_id == 0) printf("odd sum: %d\n", odd_sum);
   if (even_id == 0) printf("even sum: %d\n", even_sum);

   MPI_Finalize();

}
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1 回答 1

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虽然 comm_create 是从两行不同的代码调用的，但重要的是 COMM_WORLD 中的所有进程都在同时调用 comm_create。它们不是来自同一行代码的事实是不相关的——事实上，MPI 库甚至不知道从哪里调用 comm_create。

一个更简单的例子是从两个分支调用 Barrier；无论执行哪一行，所有进程都在执行相同的屏障，因此代码将按预期工作。

您可以轻松地重写要从同一行调用的代码：只需使用名为“num_ranks”、“mycomm”、“mygroup”和“myid”的变量和一个名为“ranks”的数组，并将它们设置为等于偶数或奇数变量取决于等级。然后所有进程都可以调用：

MPI_Group_incl(world_group, num_ranks, ranks, &mygroup);
MPI_Comm_create(MPI_COMM_WORLD, mygroup, &mycomm);
MPI_Comm_rank(mycomm, &myid);

如果你真的想要你可以在之后重新分配这些，例如

if (id%2 == 0) even_comm = mycomm;
于 2016-08-09T13:29:29.203 回答