Question

我刚刚尝试使用MPI，并从[LLNL MPI教程] [1]的第二个代码示例中复制并运行了此代码。

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

int main(int argc, char ** argv) {
    int num_tasks, rank, next, prev, buf[2], tag1 = 1, tag2 = 2;
    MPI_Request reqs[4];
    MPI_Status status[2];

    MPI_Init(&argc, &argv);
    MPI_Comm_size(MPI_COMM_WORLD, &num_tasks);
    MPI_Comm_rank(MPI_COMM_WORLD, &rank);

    prev = rank - 1;
    next = rank + 1;
    if (rank == 0) prev = num_tasks - 1;
    if (rank == (num_tasks - 1)) next = 0;

    MPI_Irecv(&buf[0], 1, MPI_INT, prev, tag1, MPI_COMM_WORLD,
                    &reqs[0]);
    MPI_Irecv(&buf[1], 1, MPI_INT, next, tag2, MPI_COMM_WORLD,
                    &reqs[1]);
    MPI_Isend(&rank, 1, MPI_INT, prev, tag2, MPI_COMM_WORLD, &reqs[2]);
    MPI_Isend(&rank, 1, MPI_INT, next, tag1, MPI_COMM_WORLD, &reqs[3]);

    MPI_Waitall(4, reqs, status);
    printf("Task %d received %d from %d and %d from %d\n",
                    rank, buf[0], prev, buf[1], next);

    MPI_Finalize();
     return EXIT_SUCCESS;
}

我本来期望这样的输出（例如，4个任务）：

$ mpiexec -n 4 ./m3
Task 0 received 3 from 3 and 1 from 1
Task 1 received 0 from 0 and 2 from 2
Task 2 received 1 from 1 and 3 from 3
Task 3 received 2 from 2 and 0 from 0

然而，相反，我明白了：

$ mpiexec -n 4 ./m3
Task 0 received 0 from 3 and 1 from 1
Task 1 received 0 from 0 and 2 from 2
Task 3 received 0 from 2 and 0 from 0
Task 2 received 0 from 1 and 3 from 3

也就是说，进入缓冲区buf [0]的消息（带有标签== 1）总是得到值0.此外，如果我改变代码以便我将缓冲区声明为buf [3]而不是buf [2] ]，并用buf [2]替换buf [0]的每个实例，然后我得到我预期的输出（即上面给出的第一个输出集）。由于某种原因，这看起来好像是用0覆盖buf [0]中的值。但是我看不出那可能是什么。顺便说一下，据我所知，我的代码（没有修改）与教程中的代码完全匹配，除了我的printf。

谢谢！

Answer 1

状态数组的大小必须为4而不是2.在您的情况下，MPI_Waitall在写入状态时会破坏内存。

MPI Irecv无法正确接收缓冲区的第一个元素？

1 个答案: