time-to-botec

Benchmark sampling in different programming languages
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dmidrange.c (1724B)


      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/stats/base/dmidrange.h"
     20 #include "stdlib/math/base/assert/is_nan.h"
     21 #include "stdlib/math/base/assert/is_positive_zero.h"
     22 #include "stdlib/math/base/assert/is_negative_zero.h"
     23 #include <stdint.h>
     24 
     25 /**
     26 * Computes the mid-range of a double-precision floating-point strided array.
     27 *
     28 * @param N       number of indexed elements
     29 * @param X       input array
     30 * @param stride  stride length
     31 * @return        output value
     32 */
     33 double stdlib_strided_dmidrange( const int64_t N, const double *X, const int64_t stride ) {
     34 	double max;
     35 	double min;
     36 	int64_t ix;
     37 	int64_t i;
     38 	double v;
     39 
     40 	if ( N <= 0 ) {
     41 		return 0.0 / 0.0; // NaN
     42 	}
     43 	if ( N == 1 || stride == 0 ) {
     44 		return X[ 0 ];
     45 	}
     46 	if ( stride < 0 ) {
     47 		ix = (1-N) * stride;
     48 	} else {
     49 		ix = 0;
     50 	}
     51 	min = X[ ix ];
     52 	max = min;
     53 	for ( i = 1; i < N; i++ ) {
     54 		ix += stride;
     55 		v = X[ ix ];
     56 		if ( stdlib_base_is_nan( v ) ) {
     57 			return v;
     58 		}
     59 		if ( v < min || ( v == min && stdlib_base_is_negative_zero( v ) ) ) {
     60 			min = v;
     61 		} else if ( v > max || ( v == max && stdlib_base_is_positive_zero( v ) ) ) {
     62 			max = v;
     63 		}
     64 	}
     65 	return ( max+min ) / 2.0;
     66 }