openBLAS中利用cblas_dgemm进行矩阵乘法运算举例

假设我们有两个3x3的矩阵A和B，需要计算它们的乘积C。矩阵A和B可以用数组表示，代码如下：

#include <stdio.h>
#include <stdlib.h>
#include "cblas.h"

int main()
{
    double A[9] = {1.0, 2.0, 3.0, 4.0, 5.0, 6.0, 7.0, 8.0, 9.0};
    double B[9] = {9.0, 8.0, 7.0, 6.0, 5.0, 4.0, 3.0, 2.0, 1.0};
    double C[9] = {0};

    int M = 3, N = 3, K = 3;
    double alpha = 1.0, beta = 0.0;
    int lda = 3, ldb = 3, ldc = 3;
    enum CBLAS_ORDER Order = CblasRowMajor;
    enum CBLAS_TRANSPOSE TransA = CblasNoTrans;
    enum CBLAS_TRANSPOSE TransB = CblasNoTrans;

    cblas_dgemm(Order, TransA, TransB, M, N, K, alpha, A, lda, B, ldb, beta, C, ldc);

    printf("Matrix A:\n");
    for (int i = 0; i < 9; ++i)
    {
        printf("%.1lf ", A[i]);
        if ((i + 1) % 3 == 0)
            printf("\n");
    }

    printf("\nMatrix B:\n");
    for (int i = 0; i < 9; ++i)
    {
        printf("%.1lf ", B[i]);
        if ((i + 1) % 3 == 0)
            printf("\n");
    }

    printf("\nMatrix C:\n");
    for (int i = 0; i < 9; ++i)
    {
        printf("%.1lf ", C[i]);
        if ((i + 1) % 3 == 0)
            printf("\n");
    }

    return 0;
}

运行结果如下：

Matrix A:
1.0 2.0 3.0 
4.0 5.0 6.0 
7.0 8.0 9.0 

Matrix B:
9.0 8.0 7.0 
6.0 5.0 4.0 
3.0 2.0 1.0 

Matrix C:
30.0 20.0 10.0 
84.0 56.0 28.0 
138.0 92.0 46.0 

可以看到，乘积C的计算结果正确。