如何解决BLAS库如何直接与升压多阵列一起使用?
给出两个矩阵A
和B
,并给出C
的相乘结果。
#include "boost/multi_array.hpp"
typedef boost::multi_array<double,2> matrix;
int m=5;
int n=6;
int k=7;
matrix A(boost::extents[m][k]);
matrix B(boost::extents[k][n]);
matrix C(boost::extents[m][n]);
如何调用dgemm
库中的blas
函数来计算A
和B
的矩阵乘积?
我知道uBLAS
部分boost
库,armadillo
,MTL 4
,eigen
和其他一些为blas
函数提供方便包装的库。这里的问题是如何直接在多阵列上调用dgemm
。
解决方法
您可以访问连续元素存储。
原型是
void cblas_dgemm( CBLAS_LAYOUT layout,CBLAS_TRANSPOSE TransA,CBLAS_TRANSPOSE TransB,const int M,const int N,const int K,const double alpha,const double *A,const int lda,const double *B,const int ldb,const double beta,double *C,const int ldc )
因此,让我们填写它:
cblas_dgemm(
CBLAS_LAYOUT::CblasRowMajor,CBLAS_TRANSPOSE::CblasNoTrans,m,n,k,1.0,// alpha
A.data(),A.shape()[1],B.data(),B.shape()[1],0.0,// beta
C.data(),C.shape()[1]);
那是使用docs here来获得有关alpha / beta的一些指导:
演示
#include <boost/multi_array.hpp>
typedef boost::multi_array<double,2> matrix;
#include <iostream>
namespace io { // for debug output
auto& dump(std::ostream& os,double v) { return os << v; }
template <typename R> auto& dump(std::ostream& os,R const& r) {
std::string_view sep = "";
os << "{";
for (auto const& el : r) { dump(os << sep,el); sep = ","; }
return os << "}";
}
}
std::ostream& operator<<(std::ostream& os,matrix const& m) { return io::dump(os,m); }
// demo
#include <cblas-netlib.h>
#include <numeric> // iota
int main() {
constexpr auto m=5,n=6,k=7;
matrix A(boost::extents[m][k]);
matrix B(boost::extents[k][n]);
matrix C(boost::extents[m][n]);
std::iota(A.data(),A.data() + A.num_elements(),0);
std::iota(B.data(),B.data() + B.num_elements(),50);
std::iota(C.data(),C.data() + C.num_elements(),100);
std::cout << "A: " << A << "\nB: " << B << "\n";
assert(A.storage_order().all_dims_ascending());
/*
* void cblas_dgemm(
* CBLAS_LAYOUT layout,* CBLAS_TRANSPOSE TransA,* CBLAS_TRANSPOSE TransB,* const int M,* const double alpha,* const double *A,* const double *B,* const double beta,* double *C,const int ldc )
*/
cblas_dgemm(
CBLAS_LAYOUT::CblasRowMajor,// alpha
A.data(),// beta
C.data(),C.shape()[1]);
std::cout << "C:\n" << C << "\n";
}
哪些印刷品:
A: {{0,1,2,3,4,5,6},{7,8,9,10,11,12,13},{14,15,16,17,18,19,20},{21,22,23,24,25,26,27},{28,29,30,31,32,33,34}}
B: {{50,51,52,53,54,55},{56,57,58,59,60,61},{62,63,64,65,66,67},{68,69,70,71,72,73},{74,75,76,77,78,79},{80,81,82,83,84,85},{86,87,88,89,90,91}}
C:
{{1596,1617,1638,1659,1680,1701},{4928,4998,5068,5138,5208,5278},{8260,8379,8498,8617,8736,8855},{11592,11760,11928,12096,12264,12432},{14924,15141,15358,15575,15792,16009}}
此操作以Wolfram Alpha签出:
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