ddot.c (2070B)
1 /** 2 * @license Apache-2.0 3 * 4 * Copyright (c) 2019 The Stdlib Authors. 5 * 6 * Licensed under the Apache License, Version 2.0 (the "License"); 7 * you may not use this file except in compliance with the License. 8 * You may obtain a copy of the License at 9 * 10 * http://www.apache.org/licenses/LICENSE-2.0 11 * 12 * Unless required by applicable law or agreed to in writing, software 13 * distributed under the License is distributed on an "AS IS" BASIS, 14 * WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied. 15 * See the License for the specific language governing permissions and 16 * limitations under the License. 17 */ 18 19 /** 20 * Compute the dot product of two double-precision floating-point vectors. 21 * 22 * @see <a href="http://www.netlib.org/lapack/expolore-html/df/d28/group__single__blas__level1.html">ddot</a> 23 */ 24 #include "stdlib/blas/base/ddot.h" 25 26 /** 27 * Computes the dot product of two double-precision floating-point vectors. 28 * 29 * @param N number of values over which to compute the dot product 30 * @param X first array 31 * @param strideX X stride length 32 * @param Y second array 33 * @param strideY Y stride length 34 * @returns the dot product of X and Y 35 */ 36 double c_ddot( const int N, const double *X, const int strideX, const double *Y, const int strideY ) { 37 double dot; 38 int ix; 39 int iy; 40 int m; 41 int i; 42 43 dot = 0.0; 44 if ( N <= 0 ) { 45 return dot; 46 } 47 // If both strides are equal to `1`, use unrolled loops... 48 if ( strideX == 1 && strideY == 1 ) { 49 m = N % 5; 50 51 // If we have a remainder, do a clean-up loop... 52 if ( m > 0 ) { 53 for ( i = 0; i < m; i++ ) { 54 dot += X[ i ] * Y[ i ]; 55 } 56 } 57 if ( N < 5 ) { 58 return dot; 59 } 60 for ( i = m; i < N; i += 5 ) { 61 dot += ( X[i]*Y[i] ) + ( X[i+1]*Y[i+1] ) + ( X[i+2]*Y[i+2] ) + ( X[i+3]*Y[i+3] ) + ( X[i+4]*Y[i+4] ); 62 } 63 return dot; 64 } 65 if ( strideX < 0 ) { 66 ix = (1-N) * strideX; 67 } else { 68 ix = 0; 69 } 70 if ( strideY < 0 ) { 71 iy = (1-N) * strideY; 72 } else { 73 iy = 0; 74 } 75 for ( i = 0; i < N; i++ ) { 76 dot += X[ ix ] * Y[ iy ]; 77 ix += strideX; 78 iy += strideY; 79 } 80 return dot; 81 } 82