MPI_Send/Recv 发送动态分配的 int 数组失败

Question

我试图了解发送数组的工作原理。我写了一个简单的程序，但效果不佳，因为它通常会因分段错误而崩溃。我的程序有参数：

mpirun -np 2 sendTest

代码如下：

#include <mpi.h>
#include <iostream> 

using namespace std;

int main(int argc, char *argv[])
{
    int cpuNum;
    int myId;
    MPI_Status mpiStatus;

    MPI_Init(&argc,&argv);
    MPI_Comm_size(MPI_COMM_WORLD,&cpuNum);
    MPI_Comm_rank(MPI_COMM_WORLD,&myId);

    int *myNumbs=(int*)malloc(2*(sizeof(int)));
    int *neighNumbs=(int*)malloc(2*(sizeof(int)));

    if(myId==0){
        myNumbs[0]=0;
        myNumbs[1]=0;

        MPI_Send(&myNumbs, 2, MPI_INT, 1, 0, MPI_COMM_WORLD);
        MPI_Recv(&neighNumbs, 2, MPI_INT, 1, 0, MPI_COMM_WORLD, &mpiStatus);

        cout<<"this is cpu 0, neigh myNumbs=";
        for (int i = 0; i < 2; ++i){
            printf("%d,",neighNumbs[i]);
        }
        cout<<endl;
    }
    else{
        myNumbs[0]=1;
        myNumbs[1]=1;

        MPI_Recv(&neighNumbs, 2, MPI_INT, 0, 0, MPI_COMM_WORLD, &mpiStatus);
        MPI_Send(&myNumbs, 2, MPI_INT, 0, 0, MPI_COMM_WORLD);

        cout<<"this is cpu 1, neigh myNumbs=";
        for (int i = 0; i < 2; ++i){
            printf("%d,",neighNumbs[i]);
        }
        cout<<endl;
    }

    MPI_Finalize();
    return 0;
}

正如我所说，它通常会因以下错误而崩溃：*** Process received signal *** Signal: Segmentation fault (11) Signal code: Address not mapped (1)。但有时一个或另一个进程会接收数据。

Answer 1

您必须传递已分配数组的地址，而不是指向已分配数组的指针的地址。

去掉指针变量前的两个&：试试

        MPI_Send(myNumbs, 2, MPI_INT, 1, 0, MPI_COMM_WORLD);
        MPI_Recv(neighNumbs, 2, MPI_INT, 1, 0, MPI_COMM_WORLD, &mpiStatus);

而不是

        MPI_Send(&myNumbs, 2, MPI_INT, 1, 0, MPI_COMM_WORLD);
        MPI_Recv(&neighNumbs, 2, MPI_INT, 1, 0, MPI_COMM_WORLD, &mpiStatus);

和

        MPI_Recv(neighNumbs, 2, MPI_INT, 0, 0, MPI_COMM_WORLD, &mpiStatus);
        MPI_Send(myNumbs, 2, MPI_INT, 0, 0, MPI_COMM_WORLD);

而不是

        MPI_Recv(&neighNumbs, 2, MPI_INT, 0, 0, MPI_COMM_WORLD, &mpiStatus);
        MPI_Send(&myNumbs, 2, MPI_INT, 0, 0, MPI_COMM_WORLD);