cblas_sgemm(CblasRowMajor, CblasNoTrans, CblasNoTrans, M, N, K, 0.0f, U, K, V, N, 1.0f, M, N); 单核情况下,其于部分代码不变,只修改贝塔参数,设置为0.0f时性能为11Gflops.设置为1.0f时性能为40Gflops。这两个都通过了正确性校验,硬件为鲲鹏920.我找不到问题的原因,各位大佬有什么办法吗暂无...
我使用的语句是：cblas_sgemm(CblasRowMajor, CblasNoTrans, CblasNoTrans, M, N, K, 1.0f, U, ...
cblas_sgemm()为openBLAS库实现矩阵乘法的函数。直接上例子代码: floata[3]={2,3,4};floatb[3]={1,0,1};floatc[1]={0};cblas_sgemm(CblasRowMajor,CblasNoTrans,CblasTrans,1,1,3,1.0,a,3,b,3,0.0,c,1);cout<<c[0]<<endl; CblasRowMajor:矩阵的读取方式之一,表示行主序。若a[6]={1,2,...
cblas_sgemm cblas.h BLAS(Basic Linear Algebra Subprograms)库,是用Fortran语言实现的向量和矩阵运算库,是许多数值计算软件库的核心, 但也有一些其它的包装, 如cblas是C语言, 也有C++的包装, boost/ublas 是C++ template class的实现; 另外还有一些特别的实现, 如intel MKL, AMD core math library blas就是做向...
void cblas_sgemm(const CBLAS_LAYOUT Layout, const CBLAS_TRANSPOSE transa, const CBLAS_TRANSPOSE transb, const MKL_INT m, const MKL_INT n, const MKL_INT k, const float alpha, const float *a, const MKL_INT lda, const float *b, const MKL_INT ldb, const float beta, float *c, const ...
func cblas_sgemm( _ __Order: CBLAS_ORDER, _ __TransA: CBLAS_TRANSPOSE, _ __TransB: CBLAS_TRANSPOSE, _ __M: Int32, _ __N: Int32, _ __K: Int32, _ __alpha: Float, _ __A: UnsafePointer<Float>!, _ __lda: Int32, _ __B: UnsafePointer<Float>!, _ __ldb: Int32,...
cblas_sgemm cblas.h 2018-08-01 15:51 − BLAS(Basic Linear Algebra Subprograms)库,是用Fortran语言实现的向量和矩阵运算库,是许多数值计算软件库的核心, 但也有一些其它的包装, 如cblas是C语言, 也有C++的包装, boost/ublas 是C++ template class的实现; 另外... 有梦就要去实现他 0 907 BLAS dge...
{ B = rand()%10000/1000.0; } while (true) { double t0 = cvGetTickCount(); cblas_sgemm(CblasRowMajor, CblasNoTrans, CblasNoTrans, 169, 1024, 64, 1.0, A, 64, B, 1024, .0, C, 1024); double t1 = cvGetTickCount()-t0; cout<<"consume time:"<<t1/cvGetTickFrequency()/1000.0<<...
cblas_sgemm 源码讲解 发布于 2018-08-02 05:29 JDK源码 源码阅读 源代码 写下你的评论... 打开知乎App 在「我的页」右上角打开扫一扫 其他扫码方式:微信 下载知乎App 开通机构号 无障碍模式 验证码登录 密码登录 中国+86 其他方式登录 未注册手机验证后自动登录,注册即代表同意《知乎协议》《隐私保护指引》...
Hi Xianyi, We tried to run a matrix multiplication with cblas_sgemm or cblas_dgemm on android. We tried with A = [1 3 4 6], B = [3 5 9 1], and C = A * B. We initialized C with all zero. The result of C did not end up with A * B, but rema...