使用某些线程进行矩阵乘法的 MPI 程序的调试断言失败

Posted 2023-03-27

【中文标题】使用某些线程进行矩阵乘法的 MPI 程序的调试断言失败

【英文标题】：Debug Assertion Failed For MPI program for matrix multiplication with some threads

【发布时间】：2021-12-26 23:21:59

【问题描述】：

美好的一天。我在运行乘以矩阵的 MPI 程序时遇到了一些问题。 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 获得 如果您能帮助我，我将不胜感激 我也在寻找类似的问题和解决方案，但它没有解决我的问题

#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

【问题讨论】：

程序需要命令行参数。你提供了吗？

hmmm，真的，通过命令行运行程序并提供参数是可能且必要的，而不是像往常一样在工作室中启动项目。我现在试试

你能通过命令行演示如何操作吗？非常非常感谢

你有两行使用给程序的参数：N = atoi(argv[1]) 和numThreads = atoi(argv[2]); 如果你不提供这些参数，程序将有未定义的行为。对我来说，如果我不提供它们，它就会崩溃。顺便说一句，程序泄漏。不要在 C++ 程序中使用malloc。

@Sviatoslavch Visual Studio 调试选项有一个地方可以添加命令行选项。此外，如果程序不检查您是否指定了命令参数（检查argc 的值），我也不会很有信心。一个好的程序会检测到您没有指定参数，并会在命令行上输出“用法”或某种错误消息，说明您缺少命令行参数。程序不应该只是假设您已将这些命令参数放入其中而崩溃。

【参考方案1】：

您正在对B 执行MPI_Bcast，就好像它是N*N 元素的连续块一样。然而，它不是：它是一个指向N 的指针数组，长度为N 的单独数组。所以要么你需要连续分配B，要么你需要做N广播。