Hướng dẫn precision in javascript float

From the Floating-Point Guide:

What can I do to avoid this problem?

That depends on what kind of calculations you’re doing.

• If you really need your results to add up exactly, especially when you work with money: use a special decimal datatype.
• If you just don’t want to see all those extra decimal places: simply format your result rounded to a fixed number of decimal places when displaying it.
• If you have no decimal datatype available, an alternative is to work with integers, e.g. do money calculations entirely in cents. But this is more work and has some drawbacks.

Note that the first point only applies if you really need specific precise decimal behaviour. Most people don't need that, they're just irritated that their programs don't work correctly with numbers like 1/10 without realizing that they wouldn't even blink at the same error if it occurred with 1/3.

If the first point really applies to you, use BigDecimal for JavaScript, which is not elegant at all, but actually solves the problem rather than providing an imperfect workaround.

Mike

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answered Aug 9, 2010 at 12:30

Michael BorgwardtMichael Borgwardt

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I like Pedro Ladaria's solution and use something similar.

function strip(number) {
    return (parseFloat(number).toPrecision(12));
}

Unlike Pedros solution this will round up 0.999...repeating and is accurate to plus/minus one on the least significant digit.

Note: When dealing with 32 or 64 bit floats, you should use toPrecision(7) and toPrecision(15) for best results. See this question for info as to why.

answered Sep 4, 2010 at 23:00

linux_mikelinux_mike

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For the mathematically inclined: http://docs.oracle.com/cd/E19957-01/806-3568/ncg_goldberg.html

The recommended approach is to use correction factors (multiply by a suitable power of 10 so that the arithmetic happens between integers). For example, in the case of 0.1 * 0.2, the correction factor is 10, and you are performing the calculation:

> var x = 0.1
> var y = 0.2
> var cf = 10
> x * y
0.020000000000000004
> (x * cf) * (y * cf) / (cf * cf)
0.02

A (very quick) solution looks something like:

var _cf = (function() {
  function _shift(x) {
    var parts = x.toString().split('.');
    return (parts.length < 2) ? 1 : Math.pow(10, parts[1].length);
  }
  return function() { 
    return Array.prototype.reduce.call(arguments, function (prev, next) { return prev === undefined || next === undefined ? undefined : Math.max(prev, _shift (next)); }, -Infinity);
  };
})();

Math.a = function () {
  var f = _cf.apply(null, arguments); if(f === undefined) return undefined;
  function cb(x, y, i, o) { return x + f * y; }
  return Array.prototype.reduce.call(arguments, cb, 0) / f;
};

Math.s = function (l,r) { var f = _cf(l,r); return (l * f - r * f) / f; };

Math.m = function () {
  var f = _cf.apply(null, arguments);
  function cb(x, y, i, o) { return (x*f) * (y*f) / (f * f); }
  return Array.prototype.reduce.call(arguments, cb, 1);
};

Math.d = function (l,r) { var f = _cf(l,r); return (l * f) / (r * f); };

In this case:

> Math.m(0.1, 0.2)
0.02

I definitely recommend using a tested library like SinfulJS

answered Sep 20, 2013 at 2:55

SheetJSSheetJS

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Are you only performing multiplication? If so then you can use to your advantage a neat secret about decimal arithmetic. That is that NumberOfDecimals(X) + NumberOfDecimals(Y) = ExpectedNumberOfDecimals. That is to say that if we have 0.123 * 0.12 then we know that there will be 5 decimal places because 0.123 has 3 decimal places and 0.12 has two. Thus if JavaScript gave us a number like 0.014760000002 we can safely round to the 5th decimal place without fear of losing precision.

answered Aug 11, 2010 at 14:33

Nate ZauggNate Zaugg

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Surprisingly, this function has not been posted yet although others have similar variations of it. It is from the MDN web docs for Math.round(). It's concise and allows for varying precision.

function precisionRound(number, precision) {
    var factor = Math.pow(10, precision);
    return Math.round(number * factor) / factor;
}

console.log(precisionRound(1234.5678, 1));
// expected output: 1234.6

console.log(precisionRound(1234.5678, -1));
// expected output: 1230

var inp = document.querySelectorAll('input');
var btn = document.querySelector('button');

btn.onclick = function(){
  inp[2].value = precisionRound( parseFloat(inp[0].value) * parseFloat(inp[1].value) , 5 );
};

//MDN function
function precisionRound(number, precision) {
  var factor = Math.pow(10, precision);
  return Math.round(number * factor) / factor;
}

button{
display: block;
}

Get Product

UPDATE: Aug/20/2019

Just noticed this error. I believe it's due to a floating point precision error with Math.round().

precisionRound(1.005, 2) // produces 1, incorrect, should be 1.01

These conditions work correctly:

precisionRound(0.005, 2) // produces 0.01
precisionRound(1.0005, 3) // produces 1.001
precisionRound(1234.5, 0) // produces 1235
precisionRound(1234.5, -1) // produces 1230

Fix:

function precisionRoundMod(number, precision) {
  var factor = Math.pow(10, precision);
  var n = precision < 0 ? number : 0.01 / factor + number;
  return Math.round( n * factor) / factor;
}

This just adds a digit to the right when rounding decimals. MDN has updated the Math.round() page so maybe someone could provide a better solution.

zhulien

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answered Apr 9, 2018 at 9:30

4

I'm finding BigNumber.js meets my needs.

A JavaScript library for arbitrary-precision decimal and non-decimal arithmetic.

It has good documentation and the author is very diligent responding to feedback.

The same author has 2 other similar libraries:

Big.js

A small, fast JavaScript library for arbitrary-precision decimal arithmetic. The little sister to bignumber.js.

and Decimal.js

An arbitrary-precision Decimal type for JavaScript.

Here's some code using BigNumber:

$(function(){      
  var product = BigNumber(.1).times(.2);  
  $('#product').text(product);

  var sum = BigNumber(.1).plus(.2);  
  $('#sum').text(sum);
});

.1 × .2 = 

.1 + .2 = 

Cerbrus

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answered Dec 2, 2014 at 14:35

Ronnie OverbyRonnie Overby

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You are looking for an sprintf implementation for JavaScript, so that you can write out floats with small errors in them (since they are stored in binary format) in a format that you expect.

Try javascript-sprintf, you would call it like this:

var yourString = sprintf("%.2f", yourNumber);

to print out your number as a float with two decimal places.

You may also use Number.toFixed() for display purposes, if you'd rather not include more files merely for floating point rounding to a given precision.

erik258

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answered Aug 6, 2010 at 8:38

DouglasDouglas

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var times = function (a, b) {
    return Math.round((a * b) * 100)/100;
};

---or---

var fpFix = function (n) {
    return Math.round(n * 100)/100;
};

fpFix(0.1*0.2); // -> 0.02

---also---

var fpArithmetic = function (op, x, y) {
    var n = {
            '*': x * y,
            '-': x - y,
            '+': x + y,
            '/': x / y
        }[op];        

    return Math.round(n * 100)/100;
};

--- as in ---

fpArithmetic('*', 0.1, 0.2);
// 0.02

fpArithmetic('+', 0.1, 0.2);
// 0.3

fpArithmetic('-', 0.1, 0.2);
// -0.1

fpArithmetic('/', 0.2, 0.1);
// 2

answered Aug 8, 2010 at 3:16

shawndumasshawndumas

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This function will determine the needed precision from the multiplication of two floating point numbers and return a result with the appropriate precision. Elegant though it is not.

function multFloats(a,b){
  var atens = Math.pow(10,String(a).length - String(a).indexOf('.') - 1), 
      btens = Math.pow(10,String(b).length - String(b).indexOf('.') - 1); 
  return (a * atens) * (b * btens) / (atens * btens); 
}

answered Aug 8, 2010 at 3:35

GabrielGabriel

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You can use parseFloat() and toFixed() if you want to bypass this issue for a small operation:

a = 0.1;
b = 0.2;

a + b = 0.30000000000000004;

c = parseFloat((a+b).toFixed(2));

c = 0.3;

a = 0.3;
b = 0.2;

a - b = 0.09999999999999998;

c = parseFloat((a-b).toFixed(2));

c = 0.1;

gion_13

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answered Mar 4, 2015 at 13:49

SoftwareddySoftwareddy

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You just have to make up your mind on how many decimal digits you actually want - can't have the cake and eat it too :-)

Numerical errors accumulate with every further operation and if you don't cut it off early it's just going to grow. Numerical libraries which present results that look clean simply cut off the last 2 digits at every step, numerical co-processors also have a "normal" and "full" lenght for the same reason. Cuf-offs are cheap for a processor but very expensive for you in a script (multiplying and dividing and using pov(...)). Good math lib would provide floor(x,n) to do the cut-off for you.

So at the very least you should make global var/constant with pov(10,n) - meaning that you decided on the precision you need :-) Then do:

Math.floor(x*PREC_LIM)/PREC_LIM  // floor - you are cutting off, not rounding

You could also keep doing math and only cut-off at the end - assuming that you are only displaying and not doing if-s with results. If you can do that, then .toFixed(...) might be more efficient.

If you are doing if-s/comparisons and don't want to cut of then you also need a small constant, usually called eps, which is one decimal place higher than max expected error. Say that your cut-off is last two decimals - then your eps has 1 at the 3rd place from the last (3rd least significant) and you can use it to compare whether the result is within eps range of expected (0.02 -eps < 0.1*0.2 < 0.02 +eps).

answered Aug 6, 2010 at 8:30

ZXXZXX

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Notice that for the general purpose use, this behavior is likely to be acceptable.
The problem arises when comparing those floating points values to determine an appropriate action.
With the advent of ES6, a new constant Number.EPSILON is defined to determine the acceptable error margin :
So instead of performing the comparison like this

0.1 + 0.2 === 0.3 // which returns false

you can define a custom compare function, like this :

function epsEqu(x, y) {
    return Math.abs(x - y) < Number.EPSILON;
}
console.log(epsEqu(0.1+0.2, 0.3)); // true

Source : http://2ality.com/2015/04/numbers-math-es6.html#numberepsilon

answered Sep 15, 2017 at 23:44

user10089632user10089632

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The round() function at phpjs.org works nicely: http://phpjs.org/functions/round

num = .01 + .06;  // yields 0.0699999999999
rnum = round(num,12); // yields 0.07

answered Mar 1, 2012 at 19:50

TomTom

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decimal.js, big.js or bignumber.js can be used to avoid floating-point manipulation problems in Javascript:

0.1 * 0.2                                // 0.020000000000000004
x = new Decimal(0.1)
y = x.times(0.2)                          // '0.2'
x.times(0.2).equals(0.2)                  // true

big.js: minimalist; easy-to-use; precision specified in decimal places; precision applied to division only.

bignumber.js: bases 2-64; configuration options; NaN; Infinity; precision specified in decimal places; precision applied to division only; base prefixes.

decimal.js: bases 2-64; configuration options; NaN; Infinity; non-integer powers, exp, ln, log; precision specified in significant digits; precision always applied; random numbers.

link to detailed comparisons

Joe Coder

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answered Aug 6, 2019 at 22:26

Erikas PliaukstaErikas Pliauksta

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The result you've got is correct and fairly consistent across floating point implementations in different languages, processors and operating systems - the only thing that changes is the level of the inaccuracy when the float is actually a double (or higher).

0.1 in binary floating points is like 1/3 in decimal (i.e. 0.3333333333333... forever), there's just no accurate way to handle it.

If you're dealing with floats always expect small rounding errors, so you'll also always have to round the displayed result to something sensible. In return you get very very fast and powerful arithmetic because all the computations are in the native binary of the processor.

Most of the time the solution is not to switch to fixed-point arithmetic, mainly because it's much slower and 99% of the time you just don't need the accuracy. If you're dealing with stuff that does need that level of accuracy (for instance financial transactions) Javascript probably isn't the best tool to use anyway (as you've want to enforce the fixed-point types a static language is probably better).

You're looking for the elegant solution then I'm afraid this is it: floats are quick but have small rounding errors - always round to something sensible when displaying their results.

answered Aug 9, 2010 at 8:57

KeithKeith

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0.6 * 3 it's awesome!)) For me this works fine:

function dec( num )
{
    var p = 100;
    return Math.round( num * p ) / p;
}

Very very simple))

answered Feb 12, 2015 at 9:14

4

To avoid this you should work with integer values instead of floating points. So when you want to have 2 positions precision work with the values * 100, for 3 positions use 1000. When displaying you use a formatter to put in the separator.

Many systems omit working with decimals this way. That is the reason why many systems work with cents (as integer) instead of dollars/euro's (as floating point).

answered Aug 9, 2010 at 12:12

GertjanGertjan

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not elegant but does the job (removes trailing zeros)

var num = 0.1*0.2;
alert(parseFloat(num.toFixed(10))); // shows 0.02

answered Sep 22, 2009 at 12:43

PeterPeter

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Problem

Floating point can't store all decimal values exactly. So when using floating point formats there will always be rounding errors on the input values. The errors on the inputs of course results on errors on the output. In case of a discrete function or operator there can be big differences on the output around the point where the function or operator is discrete.

Input and output for floating point values

So, when using floating point variables, you should always be aware of this. And whatever output you want from a calculation with floating points should always be formatted/conditioned before displaying with this in mind.
When only continuous functions and operators are used, rounding to the desired precision often will do (don't truncate). Standard formatting features used to convert floats to string will usually do this for you.
Because the rounding adds an error which can cause the total error to be more then half of the desired precision, the output should be corrected based on expected precision of inputs and desired precision of output. You should

• Round inputs to the expected precision or make sure no values can be entered with higher precision.
• Add a small value to the outputs before rounding/formatting them which is smaller than or equal to 1/4 of the desired precision and bigger than the maximum expected error caused by rounding errors on input and during calculation. If that is not possible the combination of the precision of the used data type isn't enough to deliver the desired output precision for your calculation.

These 2 things are usually not done and in most cases the differences caused by not doing them are too small to be important for most users, but I already had a project where output wasn't accepted by the users without those corrections.

Discrete functions or operators (like modula)

When discrete operators or functions are involved, extra corrections might be required to make sure the output is as expected. Rounding and adding small corrections before rounding can't solve the problem.
A special check/correction on intermediate calculation results, immediately after applying the discrete function or operator might be required. For a specific case (modula operator), see my answer on question: Why does modulus operator return fractional number in javascript?

Better avoid having the problem

It is often more efficient to avoid these problems by using data types (integer or fixed point formats) for calculations like this which can store the expected input without rounding errors. An example of that is that you should never use floating point values for financial calculations.

answered Jul 27, 2017 at 14:04

Elegant, Predictable, and Reusable

Let's deal with the problem in an elegant way reusable way. The following seven lines will let you access the floating point precision you desire on any number simply by appending .decimal to the end of the number, formula, or built in Math function.

// First extend the native Number object to handle precision. This populates
// the functionality to all math operations.

Object.defineProperty(Number.prototype, "decimal", {
  get: function decimal() {
    Number.precision = "precision" in Number ? Number.precision : 3;
    var f = Math.pow(10, Number.precision);
    return Math.round( this * f ) / f;
  }
});


// Now lets see how it works by adjusting our global precision level and 
// checking our results.

console.log("'1/3 + 1/3 + 1/3 = 1' Right?");
console.log((0.3333 + 0.3333 + 0.3333).decimal == 1); // true

console.log(0.3333.decimal); // 0.333 - A raw 4 digit decimal, trimmed to 3...

Number.precision = 3;
console.log("Precision: 3");
console.log((0.8 + 0.2).decimal); // 1
console.log((0.08 + 0.02).decimal); // 0.1
console.log((0.008 + 0.002).decimal); // 0.01
console.log((0.0008 + 0.0002).decimal); // 0.001

Number.precision = 2;
console.log("Precision: 2");
console.log((0.8 + 0.2).decimal); // 1
console.log((0.08 + 0.02).decimal); // 0.1
console.log((0.008 + 0.002).decimal); // 0.01
console.log((0.0008 + 0.0002).decimal); // 0

Number.precision = 1;
console.log("Precision: 1");
console.log((0.8 + 0.2).decimal); // 1
console.log((0.08 + 0.02).decimal); // 0.1
console.log((0.008 + 0.002).decimal); // 0
console.log((0.0008 + 0.0002).decimal); // 0

Number.precision = 0;
console.log("Precision: 0");
console.log((0.8 + 0.2).decimal); // 1
console.log((0.08 + 0.02).decimal); // 0
console.log((0.008 + 0.002).decimal); // 0
console.log((0.0008 + 0.0002).decimal); // 0

Cheers!

answered Sep 12, 2019 at 6:41

BernestoBernesto

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Solved it by first making both numbers integers, executing the expression and afterwards dividing the result to get the decimal places back:

function evalMathematicalExpression(a, b, op) {
    const smallest = String(a < b ? a : b);
    const factor = smallest.length - smallest.indexOf('.');

    for (let i = 0; i < factor; i++) {
        b *= 10;
        a *= 10;
    }

    a = Math.round(a);
    b = Math.round(b);
    const m = 10 ** factor;
    switch (op) {
        case '+':
            return (a + b) / m;
        case '-':
            return (a - b) / m;
        case '*':
            return (a * b) / (m ** 2);
        case '/':
            return a / b;
    }

    throw `Unknown operator ${op}`;
}

Results for several operations (the excluded numbers are results from eval):

0.1 + 0.002   = 0.102 (0.10200000000000001)
53 + 1000     = 1053 (1053)
0.1 - 0.3     = -0.2 (-0.19999999999999998)
53 - -1000    = 1053 (1053)
0.3 * 0.0003  = 0.00009 (0.00008999999999999999)
100 * 25      = 2500 (2500)
0.9 / 0.03    = 30 (30.000000000000004)
100 / 50      = 2 (2)

answered Sep 29, 2019 at 17:53

SimonSimon

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From my point of view, the idea here is to round the fp number in order to have a nice/short default string representation.

The 53-bit significand precision gives from 15 to 17 significant decimal digits precision (2−53 ≈ 1.11 × 10−16). If a decimal string with at most 15 significant digits is converted to IEEE 754 double-precision representation, and then converted back to a decimal string with the same number of digits, the final result should match the original string. If an IEEE 754 double-precision number is converted to a decimal string with at least 17 significant digits, and then converted back to double-precision representation, the final result must match the original number.
...
With the 52 bits of the fraction (F) significand appearing in the memory format, the total precision is therefore 53 bits (approximately 16 decimal digits, 53 log10(2) ≈ 15.955). The bits are laid out as follows ... wikipedia

(0.1).toPrecision(100) ->
0.1000000000000000055511151231257827021181583404541015625000000000000000000000000000000000000000000000

(0.1+0.2).toPrecision(100) ->
0.3000000000000000444089209850062616169452667236328125000000000000000000000000000000000000000000000000

Then, as far as I understand, we can round the value up to 15 digits to keep a nice string representation.

10**Math.floor(53 * Math.log10(2)) // 1e15

eg.

Math.round((0.2+0.1) * 1e15 ) / 1e15
0.3

(Math.round((0.2+0.1) * 1e15 ) / 1e15).toPrecision(100)
0.2999999999999999888977697537484345957636833190917968750000000000000000000000000000000000000000000000

The function would be:

function roundNumberToHaveANiceDefaultStringRepresentation(num) {

    const integerDigits = Math.floor(Math.log10(Math.abs(num))+1);
    const mult = 10**(15-integerDigits); // also consider integer digits
    return Math.round(num * mult) / mult;
}

answered Nov 29, 2019 at 10:55

Franck FreiburgerFranck Freiburger

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Have a look at Fixed-point arithmetic. It will probably solve your problem, if the range of numbers you want to operate on is small (eg, currency). I would round it off to a few decimal values, which is the simplest solution.

answered Sep 22, 2009 at 7:39

MariusMarius

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You are right, the reason for that is limited precision of floating point numbers. Store your rational numbers as a division of two integer numbers and in most situations you'll be able to store numbers without any precision loss. When it comes to printing, you may want to display the result as fraction. With representation I proposed, it becomes trivial.

Of course that won't help much with irrational numbers. But you may want to optimize your computations in the way they will cause the least problem (e.g. detecting situations like sqrt(3)^2).

answered Aug 7, 2010 at 11:21

skaleeskalee

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I had a nasty rounding error problem with mod 3. Sometimes when I should get 0 I would get .000...01. That's easy enough to handle, just test for <= .01. But then sometimes I would get 2.99999999999998. OUCH!

BigNumbers solved the problem, but introduced another, somewhat ironic, problem. When trying to load 8.5 into BigNumbers I was informed that it was really 8.4999… and had more than 15 significant digits. This meant BigNumbers could not accept it (I believe I mentioned this problem was somewhat ironic).

Simple solution to ironic problem:

x = Math.round(x*100);
// I only need 2 decimal places, if i needed 3 I would use 1,000, etc.
x = x / 100;
xB = new BigNumber(x);

answered Dec 14, 2014 at 17:34

mcgeo52mcgeo52

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    You can use library https://github.com/MikeMcl/decimal.js/. 
    it will   help  lot to give proper solution. 
    javascript console output 95 *722228.630 /100 = 686117.1984999999
    decimal library implementation 
    var firstNumber = new Decimal(95);
    var secondNumber = new Decimal(722228.630);
    var thirdNumber = new Decimal(100);
    var partialOutput = firstNumber.times(secondNumber);
    console.log(partialOutput);
    var output = new Decimal(partialOutput).div(thirdNumber);
    alert(output.valueOf());
    console.log(output.valueOf())== 686117.1985

answered May 8, 2015 at 11:22

Avoid dealing with floating points during the operation using Integers

As stated on the most voted answer until now, you can work with integers, that would mean to multiply all your factors by 10 for each decimal you are working with, and divide the result by the same number used.

For example, if you are working with 2 decimals, you multiply all your factors by 100 before doing the operation, and then divide the result by 100.

Here's an example, Result1 is the usual result, Result2 uses the solution:

var Factor1="1110.7";
var Factor2="2220.2";
var Result1=Number(Factor1)+Number(Factor2);
var Result2=((Number(Factor1)*100)+(Number(Factor2)*100))/100;
var Result3=(Number(parseFloat(Number(Factor1))+parseFloat(Number(Factor2))).toPrecision(2));
document.write("Result1: "+Result1+"
Result2: "+Result2+"
Result3: "+Result3);

The third result is to show what happens when using parseFloat instead, which created a conflict in our case.

answered Sep 21, 2019 at 18:34

DavidTaubmannDavidTaubmann

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I could not find a solution using the built in Number.EPSILON that's meant to help with this kind of problem, so here is my solution:

function round(value, precision) {
  const power = Math.pow(10, precision)
  return Math.round((value*power)+(Number.EPSILON*power)) / power
}

This uses the known smallest difference between 1 and the smallest floating point number greater than one to fix the EPSILON rounding error ending up just one EPSILON below the rounding up threshold.

Maximum precision is 15 for 64bit floating point and 6 for 32bit floating point. Your javascript is likely 64bit.

answered Jun 15, 2021 at 13:15

KarlssonKarlsson

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Try my chiliadic arithmetic library, which you can see here. If you want a later version, I can get you one.

answered Oct 6, 2009 at 0:04

Robert LRobert L

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1