Openblas加速二维矩阵卷积操作

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Openblas可以加速矩阵卷积的操作，越复杂的卷积，越适合Openblas的加速，以下程序经博主测试准确无误。

测试软件：Openblas，VS2015

Openblas下载地址：http://download.csdn.net/detail/samylee/9870678

//cblas加速二维矩阵卷积操作//注意：默认pad=0，stride=1//作者：samylee#include <cblas.h>  #include <iostream>using namespace std;int main() {//定义被卷积矩阵const int Map = 8;const float A[Map * Map] = { 1,2,3,4,5,6,7,8,1,2,3,4,5,6,7,8,1,2,3,4,5,6,7,8,1,2,3,4,5,6,7,8,1,2,3,4,5,6,7,8,1,2,3,4,5,6,7,8,1,2,3,4,5,6,7,8,1,2,3,4,5,6,7,8 };//定义卷积核const int Kernel = 3;const float B[Kernel * Kernel] = {1,1,1,1,1,1,1,1,1 };//计算卷积输出矩阵宽高const int outM = Map - Kernel + 1;//定义被卷积矩阵宽高const int convAw = Kernel*Kernel;const int convAh = outM*outM;//转换被卷积矩阵float A_convert[convAh*convAw] = { 0 };for (int i = 0; i < outM; i++){for (int j = 0; j < outM; j++){int wh = i * outM * convAw + j * convAw;int col1 = i * Map + j;A_convert[wh] = A[col1];A_convert[wh + 1] = A[col1 + 1];A_convert[wh + 2] = A[col1 + 2];int col2 = (i + 1) * Map + j;A_convert[wh + 3] = A[col2];A_convert[wh + 4] = A[col2 + 1];A_convert[wh + 5] = A[col2 + 2];int col3 = (i + 2) * Map + j;A_convert[wh + 6] = A[col3];A_convert[wh + 7] = A[col3 + 1];A_convert[wh + 8] = A[col3 + 2];}}//定义cblas初始值const enum CBLAS_ORDER Order = CblasRowMajor;const enum CBLAS_TRANSPOSE TransA = CblasNoTrans;const enum CBLAS_TRANSPOSE TransB = CblasNoTrans;const int M = convAh;//A的行数，C的行数const int N = 1;//B的列数，C的列数const int K = convAw;//A的列数，B的行数const float alpha = 1;const float beta = 0;const int lda = K;//A的列const int ldb = N;//B的列const int ldc = N;//C的列//定义卷积输出矩阵float C[M*N];//cblas计算输出矩阵cblas_sgemm(Order, TransA, TransB, M, N, K, alpha, A_convert, lda, B, ldb, beta, C, ldc);//输出验证cout << "A is:" << endl;for (int i = 0; i < Map; i++){for (int j = 0; j < Map; j++){cout << A[i*Map + j] << " ";}cout << endl;}cout << endl;cout << "B is:" << endl;for (int i = 0; i < Kernel; i++){for (int j = 0; j < Kernel; j++){cout << B[i*Kernel + j] << " ";}cout << endl;}cout << endl;cout << "C is:" << endl;for (int i = 0; i < outM; i++){for (int j = 0; j < outM; j++){cout << C[i*outM + j] << " ";}cout << endl;}cout << endl;system("pause");return EXIT_SUCCESS;}

结果如下：

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