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* @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( '@stdlib/math/base/assert/is-nan' );
var isInfinite = require( '@stdlib/math/base/assert/is-infinite' );
var pow = require( '@stdlib/math/base/special/pow' );
var abs = require( '@stdlib/math/base/special/abs' );
var round = require( '@stdlib/math/base/special/round' );
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' );
// VARIABLES //
var MAX_INT = MAX_SAFE_INTEGER + 1;
var HUGE = 1.0e+308;
// MAIN //
/**
* Rounds a double-precision floating-point number to the nearest multiple of \\(10^n\\).
*
* ## Method
*
* 1. If \\(|x| <= 2^{53}\\) and \\(|n| <= 308\\), we can use the formula
*
* ```tex
* \operatorname{roundn}(x,n) = \frac{\operatorname{round}(x \cdot 10^{-n})}{10^{-n}}
* ```
*
* which shifts the decimal to the nearest multiple of \\(10^n\\), performs a standard \\(\mathrm{round}\\) 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{roundn}(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{round}\\).
*
* <!-- </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 finite number \\(x\\) to the nearest \\(10^n\\) is \\(0.0\\). To ensure consistent behavior with \\(\operatorname{round}(x)\\), 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{roundn}(x,n) = \frac{\biggl(\frac{\operatorname{round}( (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{roundn}(\mathrm{NaN}, n) &= \mathrm{NaN} \\
* \operatorname{roundn}(x, \mathrm{NaN}) &= \mathrm{NaN} \\
* \operatorname{roundn}(x, \pm\infty) &= \mathrm{NaN} \\
* \operatorname{roundn}(\pm\infty, n) &= \pm\infty \\
* \operatorname{roundn}(\pm 0, n) &= \pm 0
* \end{align*}
* ```
*
* ## Notes
*
* 1. Alternative algorithms:
*
* - Round by [casting][1] \\(x\\) to an exponential string.
* - Native Python implementation [1][2] and [2][3].
*
* [1]: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Math/round
* [2]: https://hg.python.org/releasing/2.7.9/file/tip/Objects/floatobject.c#l1082
* [3]: https://hg.python.org/releasing/2.7.9/file/tip/Objects/floatobject.c#l1226
*
* @param {number} x - input value
* @param {integer} n - integer power of `10`
* @returns {number} rounded value
*
* @example
* // Round a value to 2 decimal places:
* var v = roundn( 3.141592653589793, -2 );
* // returns 3.14
*
* @example
* // If n = 0, `roundn` behaves like `round`:
* var v = roundn( 3.141592653589793, 0 );
* // returns 3.0
*
* @example
* // Round a value to the nearest thousand:
* var v = roundn( 12368.0, 3 );
* // returns 12000.0
*/
function roundn( 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 finite `x` rounded to the nearest >=10^309 is 0.0.
if ( n > MAX_EXP ) {
return 0.0 * x; // preserve the sign (same behavior as round)
}
// 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 ( round(y)/HUGE ) / s;
}
s = pow( 10.0, -n );
y = x * s;
if ( isInfinite( y ) ) {
return x;
}
return round( y ) / s;
}
// EXPORTS //
module.exports = roundn;
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