某些线程的矩阵乘法 MPI 程序调试断言失败

Question

美好的一天。我对运行矩阵相乘的 MPI 程序有一些问题。这是代码（不是我的代码）我从 http://dkl.cs.arizona.edu/teaching/csc522-fall16/examples/hybrid-openmp-mm.c
得到的如果你能帮助我，我将不胜感激我也在寻找类似的问题和解决方案，但没有解决我的问题

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

#define TAG 13

int main(int argc, char* argv[]) {
    double** A, ** B, ** C, * tmp;
    double startTime, endTime;
    int numElements, offset, stripSize, myrank, numnodes, N, i, j, k;
    int numThreads, chunkSize = 10;

    MPI_Init(&argc, &argv);

    MPI_Comm_rank(MPI_COMM_WORLD, &myrank);
    MPI_Comm_size(MPI_COMM_WORLD, &numnodes);

    N = atoi(argv[1]);
    numThreads = atoi(argv[2]);  // difference from MPI: how many threads/rank?

    omp_set_num_threads(numThreads);  // OpenMP call to set threads per rank

    // allocate A, B, and C --- note that you want these to be
    // contiguously allocated.  Workers need less memory allocated.

    if (myrank == 0) {
        tmp = (double*)malloc(sizeof(double) * N * N);
        A = (double**)malloc(sizeof(double*) * N);
        for (i = 0; i < N; i++)
            A[i] = &tmp[i * N];
    }
    else {
        tmp = (double*)malloc(sizeof(double) * N * N / numnodes);
        A = (double**)malloc(sizeof(double*) * N / numnodes);
        for (i = 0; i < N / numnodes; i++)
            A[i] = &tmp[i * N];
    }


    tmp = (double*)malloc(sizeof(double) * N * N);
    B = (double**)malloc(sizeof(double*) * N);
    for (i = 0; i < N; i++)
        B[i] = &tmp[i * N];


    if (myrank == 0) {
        tmp = (double*)malloc(sizeof(double) * N * N);
        C = (double**)malloc(sizeof(double*) * N);
        for (i = 0; i < N; i++)
            C[i] = &tmp[i * N];
    }
    else {
        tmp = (double*)malloc(sizeof(double) * N * N / numnodes);
        C = (double**)malloc(sizeof(double*) * N / numnodes);
        for (i = 0; i < N / numnodes; i++)
            C[i] = &tmp[i * N];
    }

    if (myrank == 0) {
        // initialize A and B
        for (i = 0; i < N; i++) {
            for (j = 0; j < N; j++) {
                A[i][j] = 1.0;
                B[i][j] = 1.0;
            }
        }
    }

    // start timer
    if (myrank == 0) {
        startTime = MPI_Wtime();
    }

    stripSize = N / numnodes;

    // send each node its piece of A -- note could be done via MPI_Scatter
    if (myrank == 0) {
        offset = stripSize;
        numElements = stripSize * N;
        for (i = 1; i < numnodes; i++) {
            MPI_Send(A[offset], numElements, MPI_DOUBLE, i, TAG, MPI_COMM_WORLD);
            offset += stripSize;
        }
    }
    else {  // receive my part of A
        MPI_Recv(A[0], stripSize * N, MPI_DOUBLE, 0, TAG, MPI_COMM_WORLD, MPI_STATUS_IGNORE);
    }

    // everyone gets B
    MPI_Bcast(B[0], N * N, MPI_DOUBLE, 0, MPI_COMM_WORLD);

    // Let each process initialize C to zero 
    for (i = 0; i < stripSize; i++) {
        for (j = 0; j < N; j++) {
            C[i][j] = 0.0;
        }
    }

    // do the work---this is the primary difference from the pure MPI program
#pragma omp parallel for shared(A,B,C,numThreads) private(i,j,k) schedule (static, chunkSize)
    for (i = 0; i < stripSize; i++) {
        for (j = 0; j < N; j++) {
            for (k = 0; k < N; k++) {
                C[i][j] += A[i][k] * B[k][j];
            }
        }
    }

    // master receives from workers  -- note could be done via MPI_Gather
    if (myrank == 0) {
        offset = stripSize;
        numElements = stripSize * N;
        for (i = 1; i < numnodes; i++) {
            MPI_Recv(C[offset], numElements, MPI_DOUBLE, i, TAG, MPI_COMM_WORLD, MPI_STATUS_IGNORE);
            offset += stripSize;
        }
    }
    else { // send my contribution to C
        MPI_Send(C[0], stripSize * N, MPI_DOUBLE, 0, TAG, MPI_COMM_WORLD);
    }

    // stop timer
    if (myrank == 0) {
        endTime = MPI_Wtime();
        printf("Time is %f\n", endTime - startTime);
    }

    // print out matrix here, if I'm the master
    if (myrank == 0 && N < 10) {
        for (i = 0; i < N; i++) {
            for (j = 0; j < N; j++) {
                printf("%f ", C[i][j]);
            }
            printf("\n");
        }
    }

    MPI_Finalize();
    return 0;
}

这是我的问题

Answer 1

您正在 B 上执行 MPI_Bcast，就好像它是 N*N 元素的连续块一样。然而，它不是：它是指向 N 个长度为 N 的单独数组的指针数组。所以要么你需要连续分配 B，要么你需要做 N 广播。

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