## The Question :

*659 people think this question is useful*

I have the following dummy test script:

function test() {
var x = 0.1 * 0.2;
document.write(x);
}
test();

This will print the result `0.020000000000000004`

while it should just print `0.02`

(if you use your calculator). As far as I understood this is due to errors in the floating point multiplication precision.

Does anyone have a good solution so that in such case I get the correct result `0.02`

? I know there are functions like `toFixed`

or rounding would be another possibility, but I’d like to really have the whole number printed without any cutting and rounding. Just wanted to know if one of you has some nice, elegant solution.

Of course, otherwise I’ll round to some 10 digits or so.

*The Question Comments :*

## The Answer 1

*494 people think this answer is useful*

From the Floating-Point Guide:

**What can I do to avoid this problem?**

That depends on what kind of
calculations youβre 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 donβt 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.

## The Answer 2

*142 people think this answer is useful*

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.

## The Answer 3

*83 people think this answer is useful*

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

## The Answer 4

*51 people think this answer is useful*

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.

## The Answer 5

*30 people think this answer is useful*

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.

## The Answer 6

*29 people think this answer is useful*

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);
});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<!-- 1.4.1 is not the current version, but works for this example. -->
<script src="http://cdn.bootcss.com/bignumber.js/1.4.1/bignumber.min.js"></script>
.1 &times; .2 = <span id="product"></span><br>
.1 &plus; .2 = <span id="sum"></span><br>

## The Answer 7

*23 people think this answer is useful*

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

## The Answer 8

*22 people think this answer is useful*

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;
}

<input type='text' value='0.1'>
<input type='text' value='0.2'>
<button>Get Product</button>
<input type='text'>

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.

## The Answer 9

*20 people think this answer is useful*

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);
}

## The Answer 10

*13 people think this answer is useful*

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 π

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 π 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).

## The Answer 11

*13 people think this answer is useful*

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;

## The Answer 12

*11 people think this answer is useful*

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

## The Answer 13

*9 people think this answer is useful*

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))

## The Answer 14

*9 people think this answer is useful*

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

## The Answer 15

*8 people think this answer is useful*

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.

## The Answer 16

*8 people think this answer is useful*

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).

## The Answer 17

*7 people think this answer is useful*

**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.

## The Answer 18

*4 people think this answer is useful*

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.

## The Answer 19

*4 people think this answer is useful*

You can’t represent most decimal fractions exactly with binary floating point types (which is what ECMAScript uses to represent floating point values). So there isn’t an elegant solution unless you use arbitrary precision arithmetic types or a decimal based floating point type. For example, the Calculator app that ships with Windows now uses arbitrary precision arithmetic to solve this problem.

## The Answer 20

*4 people think this answer is useful*

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 (2β53 β 1.11 Γ 10β16).
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) β 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;
}

## The Answer 21

*3 people think this answer is useful*

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

## The Answer 22

*3 people think this answer is useful*

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)`

.

## The Answer 23

*3 people think this answer is useful*

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 8.4999β¦ 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);

## The Answer 24

*3 people think this answer is useful*

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

## The Answer 25

*3 people think this answer is useful*

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

## The Answer 26

*2 people think this answer is useful*

Use Number(1.234443).toFixed(2); it will print 1.23

function test(){
var x = 0.1 * 0.2;
document.write(Number(x).toFixed(2));
}
test();

## The Answer 27

*2 people think this answer is useful*

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)

## The Answer 28

*1 people think this answer is useful*

Use

var x = 0.1*0.2;
x =Math.round(x*Math.pow(10,2))/Math.pow(10,2);

## The Answer 29

*1 people think this answer is useful*

not elegant but does the job (removes trailing zeros)

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

## The Answer 30

*1 people think this answer is useful*

This works for me:

function round_up( value, precision ) {
var pow = Math.pow ( 10, precision );
return ( Math.ceil ( pow * value ) + Math.ceil ( pow * value - Math.ceil ( pow * value ) ) ) / pow;
}
round_up(341.536, 2); // 341.54