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使用blas做矩阵乘法

时间:2014-07-27 23:32:19      阅读:425      评论:0      收藏:0      [点我收藏+]

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#define min(x,y) (((x) < (y)) ? (x) : (y))

#include <stdio.h>
#include <stdlib.h>
#include <cublas_v2.h>
#include <iostream>
#include <vector>
//extern "C"
//{
   #include <cblas.h>
//}

using namespace std;
int main()
{

    const enum CBLAS_ORDER Order=CblasRowMajor;
    const enum CBLAS_TRANSPOSE TransA=CblasNoTrans;
    const enum CBLAS_TRANSPOSE TransB=CblasNoTrans;
    const int M=4;//A的行数,C的行数
    const int N=2;//B的列数,C的列数
    const int K=3;//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
    const float A[M*K]={1,2,3,4,5,6,7,8,9,8,7,6};
    const float B[K*N]={5,4,3,2,1,0};
    float C[M*N];
   
    cblas_sgemm(Order, TransA, TransB, M, N, K, alpha, A, lda, B, ldb, beta, C, ldc);
     
    for(int i=0;i<M;i++)
    {
       for(int j=0;j<N;j++)
       {
           cout<<C[i*N+j]<<"\n";
       }
       cout<<endl;
    }
   
    return EXIT_SUCCESS;

 
}

g++ testblas.c++ -lopenblas  -o testout

g++ testblas.c++ -lopenblas_piledriverp-r0.2.9 -o testout   本地编译openblas版本

注意library放在引用library的函数的后面

cblas_sgemm

Multiplies two matrices (single-precision).

void cblas_sgemm (
const enum CBLAS_ORDER Order,  // Specifies row-major (C) or column-major (Fortran) data ordering.
//typedef enum CBLAS_ORDER     {CblasRowMajor=101, CblasColMajor=102} CBLAS_ORDER;

const enum CBLAS_TRANSPOSE TransA,//Specifies whether to transpose matrix A.
const enum CBLAS_TRANSPOSE TransB,
const int M,   //Number of rows in matrices A and C.
const int N,//Number of rows in matrices A and C.
const int K,  //Number of columns in matrix A; number of rows in matrix B
const float alpha, //Scaling factor for the product of matrices A and B
const float *A, 
const int lda, //The size of the first dimention of matrix A; if you are passing a matrix A[m][n], the value should be m.
const float *B,  
const int ldb,  //The size of the first dimention of matrix B; if you are passing a matrix B[m][n], the value should be m.
const float beta,  //Scaling factor for matrix C.


float *C,
const int ldc    //The size of the first dimention of matrix C; if you are passing a matrix C[m][n], the value should be m.
);

Thus, it calculates either
C←αAB + βC
or
C←αBA + βC
with optional use of transposed forms of A, B, or both.


 

typedef enum CBLAS_ORDER     {CblasRowMajor=101, CblasColMajor=102} CBLAS_ORDER;
typedef enum CBLAS_TRANSPOSE {CblasNoTrans=111, CblasTrans=112, CblasConjTrans=113, CblasConjNoTrans=114} CBLAS_TRANSPOSE;

$C=A*B$

$C^T=(A*B)^T=B^T*A^T$  把A和B的顺序颠倒,可以直接得到转制矩阵乘法的结果,不用作其他变换,(结果C也是转制)。

 

 

cblas_sgemv
Multiplies a matrix by a vector (single precision).
void cblas_sgemv (
const enum CBLAS_ORDER Order,
const enum CBLAS_TRANSPOSE TransA,
const int M,
const int N,
const float alpha,
const float *A,
const int lda,
const float *X,
const int incX,
const float beta,
float *Y,
const int incY
);

Y←αAX + βY

 

STL版本

cblas_daxpy
Computes a constant times a vector plus a vector (double-precision).  

On return, the contents of vector Y are replaced with the result. The value computed is (alpha * X[i]) +
Y[i].

#include <OpenBlas/cblas.h>
#include <OpenBlas/common.h>
#include <iostream>
#include <vector>

int main()
{
    blasint n = 10;
    blasint in_x =1;
    blasint in_y =1;

    std::vector<double> x(n);
    std::vector<double> y(n);

    double alpha = 10;

    std::fill(x.begin(),x.end(),1.0);
    std::fill(y.begin(),y.end(),2.0);

    cblas_daxpy( n, alpha, &x[0], in_x, &y[0], in_y);

    //Print y 
    for(int j=0;j<n;j++)
        std::cout << y[j] << "\t";

    std::cout << std::endl;
}

 

使用blas做矩阵乘法,布布扣,bubuko.com

使用blas做矩阵乘法

标签:style   blog   color   使用   os   io   for   div   

原文地址:http://www.cnblogs.com/huashiyiqike/p/3871927.html

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