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* @license Apache-2.0
*
* Copyright (c) 2025 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 isnanf = require( '@stdlib/math/base/assert/is-nanf' );
var isInfinite = require( '@stdlib/math/base/assert/is-infinitef' );
var pow = require( '@stdlib/math/base/special/pow' );
var absf = require( '@stdlib/math/base/special/absf' );
var floorf = require( '@stdlib/math/base/special/floorf' );
var MAX_SAFE_INTEGER = require( '@stdlib/constants/float32/max-safe-integer' );
var MAX_EXP = require( '@stdlib/constants/float32/max-base10-exponent' );
var MIN_EXP = require( '@stdlib/constants/float32/min-base10-exponent' );
var MIN_EXP_SUBNORMAL = require( '@stdlib/constants/float32/min-base10-exponent-subnormal' );
var NINF = require( '@stdlib/constants/float32/ninf' );
var float64ToFloat32 = require( '@stdlib/number/float64/base/to-float32' );
// VARIABLES //
var MAX_INT = MAX_SAFE_INTEGER + 1;
var HUGE = 1.0e+38;
// MAIN //
/**
* Rounds a single-precision floating-point number to the nearest multiple of `10^n` toward negative infinity.
*
* ## Method
*
* 1. If \\(|x| <= 2^{24}\\) and \\(|n| <= 38\\), we can use the formula
*
* ```tex
* \operatorname{floornf}(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{floornf}(-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 > 38\\), we recognize that the maximum absolute single-precision floating-point number is \\(\approx 3.4\mbox{e}38\\) 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 < -45\\), \\(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^{24}\\), \\(x\\) is **always** an integer (i.e., \\(x\\) has no decimal digits). If \\(n <= 0\\), the rounded value is \\(x\\).
*
* 5. If \\(n < -38\\), we let \\(m = n + 38\\) and modify the above formula to avoid overflow.
*
* ```tex
* \operatorname{floornf}(x,n) = \frac{\biggl(\frac{\operatorname{floor}( (x \cdot 10^{38}) 10^{-m})}{10^{38}}\biggr)}{10^{-m}}
* ```
*
* If overflow occurs, the rounded value is \\(x\\).
*
* ## Special Cases
*
* ```tex
* \begin{align*}
* \operatorname{floornf}(\mathrm{NaN}, n) &= \mathrm{NaN} \\
* \operatorname{floornf}(x, \mathrm{NaN}) &= \mathrm{NaN} \\
* \operatorname{floornf}(x, \pm\infty) &= \mathrm{NaN} \\
* \operatorname{floornf}(\pm\infty, n) &= \pm\infty \\
* \operatorname{floornf}(\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 3 decimal places:
* var v = floornf( 3.14159, -3 );
* // returns 3.141
*
* @example
* // Round a value to the nearest thousand:
* var v = floornf( 12368.0, 3 );
* // returns 12000.0
*
* @example
* // If n = 0, `floornf` behaves like `floor`:
* var v = floornf( 3.14159, 0 );
* // returns 3.0
*/
function floornf( x, n ) {
var s;
var y;
if (
isnanf( x ) ||
isnanf( n ) ||
isInfinite( n )
) {
return NaN;
}
x = float64ToFloat32( x );
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...
( absf( x ) > MAX_INT && n <= 0 )
) {
return x;
}
// The maximum absolute single is ~1.8e38. Accordingly, any possible positive finite `x` rounded to the nearest >=10^39 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( float64ToFloat32( 10.0 ), -(n + MAX_EXP) );
y = float64ToFloat32( float64ToFloat32( x * HUGE ) * float64ToFloat32( s ) ); // eslint-disable-line max-len
if ( isInfinite( y ) ) {
return x;
}
return float64ToFloat32( float64ToFloat32( floorf( y ) / HUGE ) / s );
}
s = pow( float64ToFloat32( 10.0 ), -n );
y = float64ToFloat32( x * float64ToFloat32( s ) );
if ( isInfinite( y ) ) {
return x;
}
return floorf( y ) / s;
}
// EXPORTS //
module.exports = floornf;
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