time-to-botec

floorn.js (5815B)

---

 /**
 * @license Apache-2.0
 *
 * Copyright (c) 2018 The Stdlib Authors.
 *
 * Licensed under the Apache License, Version 2.0 (the "License");
 * you may not use this file except in compliance with the License.
 * You may obtain a copy of the License at
 *
 *    http://www.apache.org/licenses/LICENSE-2.0
 *
 * Unless required by applicable law or agreed to in writing, software
 * distributed under the License is distributed on an "AS IS" BASIS,
 * WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
 * See the License for the specific language governing permissions and
 * limitations under the License.
 */
 
 'use strict';
 
 // MODULES //
 
 var isnan = require( './../../../../base/assert/is-nan' );
 var isInfinite = require( './../../../../base/assert/is-infinite' );
 var pow = require( './../../../../base/special/pow' );
 var abs = require( './../../../../base/special/abs' );
 var floor = require( './../../../../base/special/floor' );
 var MAX_SAFE_INTEGER = require( '@stdlib/constants/float64/max-safe-integer' );
 var MAX_EXP = require( '@stdlib/constants/float64/max-base10-exponent' );
 var MIN_EXP = require( '@stdlib/constants/float64/min-base10-exponent' );
 var MIN_EXP_SUBNORMAL = require( '@stdlib/constants/float64/min-base10-exponent-subnormal' );
 var NINF = require( '@stdlib/constants/float64/ninf' );
 
 
 // VARIABLES //
 
 var MAX_INT = MAX_SAFE_INTEGER + 1;
 var HUGE = 1.0e+308;
 
 
 // MAIN //
 
 /**
 * Rounds a numeric value to the nearest multiple of \\(10^n\\) toward negative infinity.
 *
 * ## Method
 *
 * 1.  If \\(|x| <= 2^{53}\\) and \\(|n| <= 308\\), we can use the formula
 *
 *     ```tex
 *     \operatorname{floorn}(x,n) = \frac{\operatorname{floor}(x \cdot 10^{-n})}{10^{-n}}
 *     ```
 *
 *     which shifts the decimal to the nearest multiple of \\(10^n\\), performs a standard \\(\mathrm{floor}\\) operation, and then shifts the decimal to its original position.
 *
 *     <!-- <note> -->
 *
 *     If \\(x \cdot 10^{-n}\\) overflows, \\(x\\) lacks a sufficient number of decimal digits to have any effect when rounding. Accordingly, the rounded value is \\(x\\).
 *
 *     <!-- </note> -->
 *
 *     <!-- <note> -->
 *
 *     Note that rescaling \\(x\\) can result in unexpected behavior. For instance, the result of \\(\operatorname{floorn}(-0.2-0.1,-16)\\) is \\(-0.3000000000000001\\) and not \\(-0.3\\). While possibly unexpected, this is not a bug. The behavior stems from the fact that most decimal fractions cannot be exactly represented as floating-point numbers. And further, rescaling can lead to slightly different fractional values, which, in turn, affects the result of \\(\mathrm{floor}\\).
 *
 *     <!-- </note> -->
 *
 * 2.  If \\(n > 308\\), we recognize that the maximum absolute double-precision floating-point number is \\(\approx 1.8\mbox{e}308\\) and, thus, the result of rounding any possible negative finite number \\(x\\) to the nearest \\(10^n\\) is \\(-\infty\\) and any possible positive finite number \\(x\\) is \\(+0\\). To ensure consistent behavior with \\(\operatorname{floor}(x)\\), if \\(x > 0\\), the sign of \\(x\\) is preserved.
 *
 * 3.  If \\(n < -324\\), \\(n\\) exceeds the maximum number of possible decimal places (such as with subnormal numbers), and, thus, the rounded value is \\(x\\).
 *
 * 4.  If \\(x > 2^{53}\\), \\(x\\) is **always** an integer (i.e., \\(x\\) has no decimal digits). If \\(n <= 0\\), the rounded value is \\(x\\).
 *
 * 5.  If \\(n < -308\\), we let \\(m = n + 308\\) and modify the above formula to avoid overflow.
 *
 *     ```tex
 *     \operatorname{floorn}(x,n) = \frac{\biggl(\frac{\operatorname{floor}( (x \cdot 10^{308}) 10^{-m})}{10^{308}}\biggr)}{10^{-m}}
 *     ```
 *
 *     If overflow occurs, the rounded value is \\(x\\).
 *
 *
 * ## Special Cases
 *
 * ```tex
 * \begin{align*}
 * \operatorname{floorn}(\mathrm{NaN}, n) &= \mathrm{NaN} \\
 * \operatorname{floorn}(x, \mathrm{NaN}) &= \mathrm{NaN} \\
 * \operatorname{floorn}(x, \pm\infty) &= \mathrm{NaN} \\
 * \operatorname{floorn}(\pm\infty, n) &= \pm\infty \\
 * \operatorname{floorn}(\pm 0, n) &= \pm 0
 * \end{align*}
 * ```
 *
 *
 * @param {number} x - input value
 * @param {integer} n - integer power of 10
 * @returns {number} rounded value
 *
 * @example
 * // Round a value to 4 decimal places:
 * var v = floorn( 3.141592653589793, -4 );
 * // returns 3.1415
 *
 * @example
 * // If n = 0, `floorn` behaves like `floor`:
 * var v = floorn( 3.141592653589793, 0 );
 * // returns 3.0
 *
 * @example
 * // Round a value to the nearest thousand:
 * var v = floorn( 12368.0, 3 );
 * // returns 12000.0
 */
 function floorn( x, n ) {
 	var s;
 	var y;
 	if (
 		isnan( x ) ||
 		isnan( n ) ||
 		isInfinite( n )
 	) {
 		return NaN;
 	}
 	if (
 		// Handle infinities...
 		isInfinite( x ) ||
 
 		// Handle +-0...
 		x === 0.0 ||
 
 		// If `n` exceeds the maximum number of feasible decimal places (such as with subnormal numbers), nothing to round...
 		n < MIN_EXP_SUBNORMAL ||
 
 		// If `|x|` is large enough, no decimals to round...
 		( abs( x ) > MAX_INT && n <= 0 )
 	) {
 		return x;
 	}
 	// The maximum absolute double is ~1.8e308. Accordingly, any possible positive finite `x` rounded to the nearest >=10^309 is infinity and any negative finite `x` is zero.
 	if ( n > MAX_EXP ) {
 		if ( x >= 0.0 ) {
 			return 0.0; // preserve the sign (same behavior as floor)
 		}
 		return NINF;
 	}
 	// If we overflow, return `x`, as the number of digits to the right of the decimal is too small (i.e., `x` is too large / lacks sufficient fractional precision) for there to be any effect when rounding...
 	if ( n < MIN_EXP ) {
 		s = pow( 10.0, -(n + MAX_EXP) );
 		y = (x*HUGE) * s; // order of operation matters!
 		if ( isInfinite( y ) ) {
 			return x;
 		}
 		return ( floor(y)/HUGE ) / s;
 	}
 	s = pow( 10.0, -n );
 	y = x * s;
 	if ( isInfinite( y ) ) {
 		return x;
 	}
 	return floor( y ) / s;
 }
 
 
 // EXPORTS //
 
 module.exports = floorn;