1

我对以下代码有疑问：

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

#define NUM_SPAWNS 2
// Based on the example from: http://mpi.deino.net/mpi_functions/MPI_Comm_spawn.html
void MPI_messenger(int stuff, int dest)
{
    MPI_Send(&stuff, 1, MPI_INT, dest, 1,intercomm);
}

int main( int argc, char *argv[] )
{
    int my_rank;
    int size;
    int np = NUM_SPAWNS;
    int errcodes[NUM_SPAWNS];
    MPI_Comm parentcomm, intercomm, testcomm;
    MPI_Init( &argc, &argv );
    MPI_Status stat;
    MPI_Comm_get_parent( &parentcomm );
    if (parentcomm == MPI_COMM_NULL)
    {
        MPI_Comm_spawn( "spawn_example4", MPI_ARGV_NULL, np, MPI_INFO_NULL, 0, MPI_COMM_WORLD, &intercomm, errcodes );
        MPI_Comm_rank(MPI_COMM_WORLD, &my_rank);
        MPI_Comm_size(MPI_COMM_WORLD, &size);
        int lol = 10;
        MPI_messenger(lol,0);
        MPI_messenger(lol,1);
    }
    else
    {
        MPI_Comm_rank(MPI_COMM_WORLD, &my_rank);
        MPI_Comm_size(MPI_COMM_WORLD, &size);
        int lol;
        MPI_Recv(&lol, 1, MPI_INT,0,1,parentcomm, &stat);


        std::cout << lol << "\n";
    }
    fflush(stdout);
    MPI_Finalize();
    return 0;
}

当然intercommunicator intercomm没有定义在函数MPI_messenger的范围内。我想知道是否以及如何在函数内部获取此交互器而不将其作为参数传递。

4

1 回答 1

2

与任何其他变量相同的方式：全局声明它，或者在对两者可见的其他公共范围内声明它main()（MPI_messenger()例如，在同一个类中作为成员）。

于 2012-06-07T16:59:12.077 回答