我对以下代码有问题:
#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,互通器互通器不是在函数MPI_messenger的范围内定义的。我想知道我是否以及如何在函数中获取这个相互交流器,而无需将其作为参数传递。
与
任何其他变量相同的方式:全局声明它,或者在main()
和MPI_messenger()
都可见的其他公共作用域中声明它(例如,在与成员相同的类中)。