time-to-botec

Benchmark sampling in different programming languages
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ssumors.c (1694B)


      1 /**
      2 * @license Apache-2.0
      3 *
      4 * Copyright (c) 2020 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 #include "stdlib/blas/ext/base/ssumors.h"
     20 #include <stdint.h>
     21 
     22 /**
     23 * Computes the sum of single-precision floating-point strided array elements using ordinary recursive summation.
     24 *
     25 * @param N       number of indexed elements
     26 * @param X       input array
     27 * @param stride  stride length
     28 * @return        output value
     29 */
     30 float stdlib_strided_ssumors( const int64_t N, const float *X, const int64_t stride ) {
     31 	int64_t ix;
     32 	int64_t m;
     33 	int64_t i;
     34 	float sum;
     35 
     36 	sum = 0.0f;
     37 	if ( N <= 0 ) {
     38 		return sum;
     39 	}
     40 	if ( N == 1 || stride == 0 ) {
     41 		return X[ 0 ];
     42 	}
     43 	// If the stride is equal to `1`, use unrolled loops...
     44 	if ( stride == 1 ) {
     45 		m = N % 6;
     46 
     47 		// If we have a remainder, run a clean-up loop...
     48 		if ( m > 0 ) {
     49 			for ( i = 0; i < m; i++ ) {
     50 				sum += X[ i ];
     51 			}
     52 		}
     53 		if ( N < 6 ) {
     54 			return sum;
     55 		}
     56 		for ( i = m; i < N; i += 6 ) {
     57 			sum += X[i] + X[i+1] + X[i+2] + X[i+3] + X[i+4] + X[i+5];
     58 		}
     59 		return sum;
     60 	}
     61 	if ( stride < 0 ) {
     62 		ix = (1-N) * stride;
     63 	} else {
     64 		ix = 0;
     65 	}
     66 	for ( i = 0; i < N; i++ ) {
     67 		sum += X[ ix ];
     68 		ix += stride;
     69 	}
     70 	return sum;
     71 }