cuBLAS中的Level-1函数asum()和axpy()

2.5.3. cublas<t>asum()

cublasStatus_t  cublasSasum(cublasHandle_t handle, int n,
                            const float           *x, int incx, float  *result)
cublasStatus_t  cublasDasum(cublasHandle_t handle, int n,
                            const double          *x, int incx, double *result)
cublasStatus_t cublasScasum(cublasHandle_t handle, int n,
                            const cuComplex       *x, int incx, float  *result)
cublasStatus_t cublasDzasum(cublasHandle_t handle, int n,
                            const cuDoubleComplex *x, int incx, double *result)

This function computes a vector x The sum of the absolute values ​​of the elements of . 因此,结果是 ∑ i = 1 n | I m ( x [ j ] ) | + | R e ( x [ j ] ) | 其中 j = 1 + ( i - 1 ) * incx . 请注意,最后一个等式反映了用于与 Fortran 兼容的基于 1 的索引.

该函数可能返回的错误值及其含义如下所列.

请参考:

sasum, dasum, scasum, dzasum

2.5.4. cublas<t>axpy()

cublasStatus_t cublasSaxpy(cublasHandle_t handle, int n,
                           const float           *alpha,
                           const float           *x, int incx,
                           float                 *y, int incy)
cublasStatus_t cublasDaxpy(cublasHandle_t handle, int n,
                           const double          *alpha,
                           const double          *x, int incx,
                           double                *y, int incy)
cublasStatus_t cublasCaxpy(cublasHandle_t handle, int n,
                           const cuComplex       *alpha,
                           const cuComplex       *x, int incx,
                           cuComplex             *y, int incy)
cublasStatus_t cublasZaxpy(cublasHandle_t handle, int n,
                           const cuDoubleComplex *alpha,
                           const cuDoubleComplex *x, int incx,
                           cuDoubleComplex       *y, int incy)

This function converts the vector x Multiply a scalar and add it to a vector y 中,and overwrite the latest vector with the result. 因此,对 i = 1 , … , n 、 k = 1 + ( i - 1 ) * incx 和 j = 1 + ( i - 1 ) * incy 执行的操作是 y [ j ] = α × x [ k ] + y [ j ]. 请注意,The last two equations reflect that for AND Fortran 兼容的基于 1 的索引.

请参考:

saxpy, daxpy, caxpy, zaxpy