I'd like to round at most two decimal places, but only if necessary.

Input:

``````10
1.7777777
9.1
``````

Output:

``````10
1.78
9.1
``````

How can I do this in JavaScript?

## Solution 1

``````Math.round(num * 100) / 100
``````

Or to be more specific and to ensure things like 1.005 round correctly, use Number.EPSILON :

``````Math.round((num + Number.EPSILON) * 100) / 100
``````

## Solution 2

If the value is a text type:

``````parseFloat("123.456").toFixed(2);
``````

If the value is a number:

``````var numb = 123.23454;
numb = numb.toFixed(2);
``````

There is a downside that values like 1.5 will give "1.50" as the output. A fix suggested by @minitech:

``````var numb = 1.5;
numb = +numb.toFixed(2);
// Note the plus sign that drops any "extra" zeroes at the end.
// It changes the result (which is a string) into a number again (think "0 + foo"),
// which means that it uses only as many digits as necessary.
``````

It seems like `Math.round` is a better solution. But it is not! In some cases it will not round correctly:

``````Math.round(1.005 * 100)/100 // Returns 1 instead of expected 1.01!
``````

toFixed() will also not round correctly in some cases (tested in Chrome v.55.0.2883.87)!

Examples:

``````parseFloat("1.555").toFixed(2); // Returns 1.55 instead of 1.56.
parseFloat("1.5550").toFixed(2); // Returns 1.55 instead of 1.56.
// However, it will return correct result if you round 1.5551.
parseFloat("1.5551").toFixed(2); // Returns 1.56 as expected.

1.3555.toFixed(3) // Returns 1.355 instead of expected 1.356.
// However, it will return correct result if you round 1.35551.
1.35551.toFixed(2); // Returns 1.36 as expected.
``````

I guess, this is because 1.555 is actually something like float 1.55499994 behind the scenes.

Solution 1 is to use a script with required rounding algorithm, for example:

``````function roundNumber(num, scale) {
if(!("" + num).includes("e")) {
return +(Math.round(num + "e+" + scale)  + "e-" + scale);
} else {
var arr = ("" + num).split("e");
var sig = ""
if(+arr[1] + scale > 0) {
sig = "+";
}
return +(Math.round(+arr[0] + "e" + sig + (+arr[1] + scale)) + "e-" + scale);
}
}
``````

It is also at Plunker.

Note: This is not a universal solution for everyone. There are several different rounding algorithms. Your implementation can be different, and it depends on your requirements. See also Rounding.

Solution 2 is to avoid front end calculations and pull rounded values from the backend server.

Another possible solution, which is not a bulletproof either.

``````Math.round((num + Number.EPSILON) * 100) / 100
``````

In some cases, when you round a number like 1.3549999999999998, it will return an incorrect result. It should be 1.35, but the result is 1.36.

## Solution 3

You can use

``````function roundToTwo(num) {
return +(Math.round(num + "e+2")  + "e-2");
}
``````

I found this on MDN. Their way avoids the problem with 1.005 that was mentioned.

``````roundToTwo(1.005)
1.01
roundToTwo(10)
10
roundToTwo(1.7777777)
1.78
roundToTwo(9.1)
9.1
roundToTwo(1234.5678)
1234.57
``````

## Solution 4

MarkG's answer is the correct one. Here's a generic extension for any number of decimal places.

``````Number.prototype.round = function(places) {
return +(Math.round(this + "e+" + places)  + "e-" + places);
}
``````

Usage:

``````var n = 1.7777;
n.round(2); // 1.78
``````

Unit test:

``````it.only('should round floats to 2 places', function() {

var cases = [
{ n: 10,      e: 10,    p:2 },
{ n: 1.7777,  e: 1.78,  p:2 },
{ n: 1.005,   e: 1.01,  p:2 },
{ n: 1.005,   e: 1,     p:0 },
{ n: 1.77777, e: 1.8,   p:1 }
]

cases.forEach(function(testCase) {
var r = testCase.n.round(testCase.p);
assert.equal(r, testCase.e, 'didn\'t get right number');
});
})
``````

## Solution 5

You should use:

``````Math.round( num * 100 + Number.EPSILON ) / 100
``````

No one seems to be aware of `Number.EPSILON`.

Also it's worth noting that this is not a JavaScript weirdness like some people stated.

That is simply the way floating point numbers works in a computer. Like 99% of programming languages, JavaScript doesn't have home made floating point numbers; it relies on the CPU/FPU for that. A computer uses binary, and in binary, there isn't any numbers like `0.1`, but a mere binary approximation for that. Why? For the same reason than 1/3 cannot be written in decimal: its value is 0.33333333... with an infinity of threes.

Here come `Number.EPSILON`. That number is the difference between 1 and the next number existing in the double precision floating point numbers. That's it: There is no number between `1` and 1 + `Number.EPSILON`.

EDIT:

As asked in the comments, let's clarify one thing: adding `Number.EPSILON` is relevant only when the value to round is the result of an arithmetic operation, as it can swallow some floating point error delta.

It's not useful when the value comes from a direct source (e.g.: literal, user input or sensor).

EDIT (2019):

Like @maganap and some peoples have pointed out, it's best to add `Number.EPSILON` before multiplying:

``````Math.round( ( num + Number.EPSILON ) * 100 ) / 100
``````

EDIT (december 2019):

Lately, I use a function similar to this one for comparing numbers epsilon-aware:

``````const ESPILON_RATE = 1 + Number.EPSILON ;
const ESPILON_ZERO = Number.MIN_VALUE ;

function epsilonEquals( a , b ) {
if ( Number.isNaN( a ) || Number.isNaN( b ) ) {
return false ;
}
if ( a === 0 || b === 0 ) {
return a <= b + EPSILON_ZERO && b <= a + EPSILON_ZERO ;
}
return a <= b * EPSILON_RATE && b <= a * EPSILON_RATE ;
}
``````

My use-case is an assertion + data validation lib I'm developing for many years.

In fact, in the code I'm using `ESPILON_RATE = 1 + 4 * Number.EPSILON` and `EPSILON_ZERO = 4 * Number.MIN_VALUE` (four times the epsilon), because I want an equality checker loose enough for cumulating floating point error.

So far, it looks perfect for me. I hope it will help.

## Solution 6

In general, decimal rounding is done by scaling: `round(num * p) / p`

Naive implementation

Using the following function with halfway numbers, you will get either the upper rounded value as expected, or the lower rounded value sometimes depending on the input.

This `inconsistency` in rounding may introduce hard to detect bugs in the client code.

``````function naiveRound(num, decimalPlaces = 0) {
var p = Math.pow(10, decimalPlaces);
return Math.round(num * p) / p;
}

console.log( naiveRound(1.245, 2) );  // 1.25 correct (rounded as expected)
console.log( naiveRound(1.255, 2) );  // 1.25 incorrect (should be 1.26)

// testing edge cases
console.log( naiveRound(1.005, 2) );  // 1    incorrect (should be 1.01)
console.log( naiveRound(2.175, 2) );  // 2.17 incorrect (should be 2.18)
console.log( naiveRound(5.015, 2) );  // 5.01 incorrect (should be 5.02)``````

In order to determine whether a rounding operation involves a midpoint value, the Round function multiplies the original value to be rounded by 10 ** n, where n is the desired number of fractional digits in the return value, and then determines whether the remaining fractional portion of the value is greater than or equal to .5. This `"Exact Testing for Equality"` with floating-point values are problematic because of the floating-point format's issues with binary representation and precision. This means that any fractional portion of a number that is slightly less than .5 (because of a loss of precision) will not be rounded upward.

In the previous example, `5.015` is a midpoint value if it is to be rounded to two decimal places, the value of 5.015 * 100 is actually `501.49999999999994`. Because .49999999999994 is less than .5, it is rounded down to 501 and finally the result is 5.01.

Better implementations

Exponential notation

By converting the number to a string in the exponential notation, positive numbers are rounded as expected. But, be aware that negative numbers round differently than positive numbers.

In fact, it performs what is basically equivalent to "round half up" as the rule, you will see that `round(-1.005, 2)` evaluates to `-1` even though `round(1.005, 2)` evaluates to `1.01`. The lodash _.round method uses this technique.

``````/**
* Round half up ('round half towards positive infinity')
* Negative numbers round differently than positive numbers.
*/
function round(num, decimalPlaces = 0) {
num = Math.round(num + "e" + decimalPlaces);
return Number(num + "e" + -decimalPlaces);
}

// test rounding of half
console.log( round(0.5) );  // 1
console.log( round(-0.5) ); // 0

// testing edge cases
console.log( round(1.005, 2) );   // 1.01
console.log( round(2.175, 2) );   // 2.18
console.log( round(5.015, 2) );   // 5.02

console.log( round(-1.005, 2) );  // -1
console.log( round(-2.175, 2) );  // -2.17
console.log( round(-5.015, 2) );  // -5.01``````

If you want the usual behavior when rounding negative numbers, you would need to convert negative numbers to positive before calling Math.round(), and then convert them back to negative numbers before returning.

``````// Round half away from zero
function round(num, decimalPlaces = 0) {
if (num < 0)
return -round(-num, decimalPlaces);

num = Math.round(num + "e" + decimalPlaces);
return Number(num + "e" + -decimalPlaces);
}
``````

Approximate rounding

To correct the rounding problem shown in the previous `naiveRound` example, we can define a custom rounding function that performs a "nearly equal" test to determine whether a fractional value is sufficiently close to a midpoint value to be subject to midpoint rounding.

``````// round half away from zero
function round(num, decimalPlaces = 0) {
if (num < 0)
return -round(-num, decimalPlaces);
var p = Math.pow(10, decimalPlaces);
var n = num * p;
var f = n - Math.floor(n);
var e = Number.EPSILON * n;

// Determine whether this fraction is a midpoint value.
return (f >= .5 - e) ? Math.ceil(n) / p : Math.floor(n) / p;
}

// test rounding of half
console.log( round(0.5) );  // 1
console.log( round(-0.5) ); // -1

// testing edge cases
console.log( round(1.005, 2) );  // 1.01
console.log( round(2.175, 2) );  // 2.18
console.log( round(5.015, 2) );  // 5.02

console.log( round(-1.005, 2) ); // -1.01
console.log( round(-2.175, 2) ); // -2.18
console.log( round(-5.015, 2) ); // -5.02``````

Number.EPSILON

There is a different purely mathematical technique to perform round-to-nearest (using "round half away from zero"), in which epsilon correction is applied before calling the rounding function.

Simply, we add the smallest possible float value (= 1.0 ulp; unit in the last place) to the product before rounding. This moves to the next representable float value, away from zero, thus it will offset the binary round-off error that may occur during the multiplication by `10 ** n`.

``````/**
* Round half away from zero ('commercial' rounding)
* Uses correction to offset floating-point inaccuracies.
* Works symmetrically for positive and negative numbers.
*/
function round(num, decimalPlaces = 0) {
var p = Math.pow(10, decimalPlaces);
var n = (num * p) * (1 + Number.EPSILON);
return Math.round(n) / p;
}

// rounding of half
console.log( round(0.5) );  // 1
console.log( round(-0.5) ); // -1

// testing edge cases
console.log( round(1.005, 2) );  // 1.01
console.log( round(2.175, 2) );  // 2.18
console.log( round(5.015, 2) );  // 5.02

console.log( round(-1.005, 2) ); // -1.01
console.log( round(-2.175, 2) ); // -2.18
console.log( round(-5.015, 2) ); // -5.02``````

After adding 1 ulp, the value of 5.015 * 100 which is `501.49999999999994` will be corrected to `501.50000000000006`, this will rounded up to 502 and finally the result is 5.02.

Note that the size of a unit in last place ("ulp") is determined by (1) the magnitude of the number and (2) the relative machine epsilon (2^-52). Ulps are relatively larger at numbers with bigger magnitudes than they are at numbers with smaller magnitudes.

Double rounding

Here, we use the toPrecision() method to strip the floating-point round-off errors in the intermediate calculations. Simply, we round to 15 significant figures to strip the round-off error at the 16th significant digit. This technique to preround the result to significant digits is also used by PHP 7 round function.

The value of 5.015 * 100 which is `501.49999999999994` will be rounded first to 15 significant digits as `501.500000000000`, then it will rounded up again to 502 and finally the result is 5.02.

``````// Round half away from zero
function round(num, decimalPlaces = 0) {
if (num < 0)
return -round(-num, decimalPlaces);
var p = Math.pow(10, decimalPlaces);
var n = (num * p).toPrecision(15);
return Math.round(n) / p;
}

// rounding of half
console.log( round(0.5) );  // 1
console.log( round(-0.5) ); // -1

// testing edge cases
console.log( round(1.005, 2) );  // 1.01
console.log( round(2.175, 2) );  // 2.18
console.log( round(5.015, 2) );  // 5.02

console.log( round(-1.005, 2) ); // -1.01
console.log( round(-2.175, 2) ); // -2.18
console.log( round(-5.015, 2) ); // -5.02``````

Arbitrary-precision JavaScript library - decimal.js

``````// Round half away from zero
function round(num, decimalPlaces = 0) {
return new Decimal(num).toDecimalPlaces(decimalPlaces).toNumber();
}

// rounding of half
console.log( round(0.5) );  // 1
console.log( round(-0.5) ); // -1

// testing edge cases
console.log( round(1.005, 2) );  // 1.01
console.log( round(2.175, 2) );  // 2.18
console.log( round(5.015, 2) );  // 5.02

console.log( round(-1.005, 2) ); // -1.01
console.log( round(-2.175, 2) ); // -2.18
console.log( round(-5.015, 2) ); // -5.02``````
``<script src="https://cdnjs.cloudflare.com/ajax/libs/decimal.js/10.2.1/decimal.js" integrity="sha512-GKse2KVGCCMVBn4riigHjXE8j5hCxYLPXDw8AvcjUtrt+a9TbZFtIKGdArXwYOlZvdmkhQLWQ46ZE3Q1RIa7uQ==" crossorigin="anonymous"></script>``

Solution 1: string in exponential notation

Inspired by the solution provided by KFish here: https://stackoverflow.com/a/55521592/4208440

A simple drop in solution that provides accurate decimal rounding, flooring, and ceiling to a specific number of decimal places without adding a whole library. It treats floats more like decimals by fixing the binary rounding issues to avoid unexpected results: for example, floor((0.1+0.7)*10) will return the expected result 8.

Numbers are rounded to a specific number of fractional digits. Specifying a negative precision will round to any number of places to the left of the decimal point.

``````// Solution 1
var DecimalPrecision = (function() {
if (Math.trunc === undefined) {
Math.trunc = function(v) {
return v < 0 ? Math.ceil(v) : Math.floor(v);
};
}
var decimalAdjust = function myself(type, num, decimalPlaces) {
if (type === 'round' && num < 0)
return -myself(type, -num, decimalPlaces);
var shift = function(value, exponent) {
value = (value + 'e').split('e');
return +(value[0] + 'e' + (+value[1] + (exponent || 0)));
};
var n = shift(num, +decimalPlaces);
return shift(Math[type](n), -decimalPlaces);
};
return {
// Decimal round (half away from zero)
round: function(num, decimalPlaces) {
},
// Decimal ceil
ceil: function(num, decimalPlaces) {
},
// Decimal floor
floor: function(num, decimalPlaces) {
},
// Decimal trunc
trunc: function(num, decimalPlaces) {
},
// Format using fixed-point notation
toFixed: function(num, decimalPlaces) {
}
};
})();

// test rounding of half
console.log(DecimalPrecision.round(0.5));  // 1
console.log(DecimalPrecision.round(-0.5)); // -1

// testing very small numbers
console.log(DecimalPrecision.ceil(1e-8, 2) === 0.01);
console.log(DecimalPrecision.floor(1e-8, 2) === 0);

// testing simple cases
console.log(DecimalPrecision.round(5.12, 1) === 5.1);
console.log(DecimalPrecision.round(-5.12, 1) === -5.1);
console.log(DecimalPrecision.ceil(5.12, 1) === 5.2);
console.log(DecimalPrecision.ceil(-5.12, 1) === -5.1);
console.log(DecimalPrecision.floor(5.12, 1) === 5.1);
console.log(DecimalPrecision.floor(-5.12, 1) === -5.2);
console.log(DecimalPrecision.trunc(5.12, 1) === 5.1);
console.log(DecimalPrecision.trunc(-5.12, 1) === -5.1);

// testing edge cases for round
console.log(DecimalPrecision.round(1.005, 2) === 1.01);
console.log(DecimalPrecision.round(39.425, 2) === 39.43);
console.log(DecimalPrecision.round(-1.005, 2) === -1.01);
console.log(DecimalPrecision.round(-39.425, 2) === -39.43);

// testing edge cases for ceil
console.log(DecimalPrecision.ceil(9.13, 2) === 9.13);
console.log(DecimalPrecision.ceil(65.18, 2) === 65.18);
console.log(DecimalPrecision.ceil(-2.26, 2) === -2.26);
console.log(DecimalPrecision.ceil(-18.15, 2) === -18.15);

// testing edge cases for floor
console.log(DecimalPrecision.floor(2.26, 2) === 2.26);
console.log(DecimalPrecision.floor(18.15, 2) === 18.15);
console.log(DecimalPrecision.floor(-9.13, 2) === -9.13);
console.log(DecimalPrecision.floor(-65.18, 2) === -65.18);

// testing edge cases for trunc
console.log(DecimalPrecision.trunc(2.26, 2) === 2.26);
console.log(DecimalPrecision.trunc(18.15, 2) === 18.15);
console.log(DecimalPrecision.trunc(-2.26, 2) === -2.26);
console.log(DecimalPrecision.trunc(-18.15, 2) === -18.15);

// testing round to tens and hundreds
console.log(DecimalPrecision.round(1262.48, -1) === 1260);
console.log(DecimalPrecision.round(1262.48, -2) === 1300);

// testing toFixed()
console.log(DecimalPrecision.toFixed(1.005, 2) === "1.01");``````

Solution 2: purely mathematical (Number.EPSILON)

This solution avoids any string conversion / manipulation of any kind for performance reasons.

``````// Solution 2
var DecimalPrecision2 = (function() {
if (Number.EPSILON === undefined) {
Number.EPSILON = Math.pow(2, -52);
}
if (Math.trunc === undefined) {
Math.trunc = function(v) {
return v < 0 ? Math.ceil(v) : Math.floor(v);
};
}
var powers = [
1e0,  1e1,  1e2,  1e3,  1e4,  1e5,  1e6,  1e7,
1e8,  1e9,  1e10, 1e11, 1e12, 1e13, 1e14, 1e15,
1e16, 1e17, 1e18, 1e19, 1e20, 1e21, 1e22
];
var intpow10 = function(power) {
if (power < 0 || power > 22) {
return Math.pow(10, power);
}
return powers[power];
};
var isRound = function(num, decimalPlaces) {
//return decimalPlaces >= 0 &&
//    +num.toFixed(decimalPlaces) === num;
var p = intpow10(decimalPlaces);
return Math.round(num * p) / p === num;
};
var decimalAdjust = function(type, num, decimalPlaces) {
if (type !== 'round' && isRound(num, decimalPlaces || 0))
return num;
var p = intpow10(decimalPlaces || 0);
var n = (num * p) * (1 + Number.EPSILON);
return Math[type](n) / p;
};
return {
// Decimal round (half away from zero)
round: function(num, decimalPlaces) {
},
// Decimal ceil
ceil: function(num, decimalPlaces) {
},
// Decimal floor
floor: function(num, decimalPlaces) {
},
// Decimal trunc
trunc: function(num, decimalPlaces) {
},
// Format using fixed-point notation
toFixed: function(num, decimalPlaces) {
}
};
})();

// test rounding of half
console.log(DecimalPrecision2.round(0.5));  // 1
console.log(DecimalPrecision2.round(-0.5)); // -1

// testing very small numbers
console.log(DecimalPrecision2.ceil(1e-8, 2) === 0.01);
console.log(DecimalPrecision2.floor(1e-8, 2) === 0);

// testing simple cases
console.log(DecimalPrecision2.round(5.12, 1) === 5.1);
console.log(DecimalPrecision2.round(-5.12, 1) === -5.1);
console.log(DecimalPrecision2.ceil(5.12, 1) === 5.2);
console.log(DecimalPrecision2.ceil(-5.12, 1) === -5.1);
console.log(DecimalPrecision2.floor(5.12, 1) === 5.1);
console.log(DecimalPrecision2.floor(-5.12, 1) === -5.2);
console.log(DecimalPrecision2.trunc(5.12, 1) === 5.1);
console.log(DecimalPrecision2.trunc(-5.12, 1) === -5.1);

// testing edge cases for round
console.log(DecimalPrecision2.round(1.005, 2) === 1.01);
console.log(DecimalPrecision2.round(39.425, 2) === 39.43);
console.log(DecimalPrecision2.round(-1.005, 2) === -1.01);
console.log(DecimalPrecision2.round(-39.425, 2) === -39.43);

// testing edge cases for ceil
console.log(DecimalPrecision2.ceil(9.13, 2) === 9.13);
console.log(DecimalPrecision2.ceil(65.18, 2) === 65.18);
console.log(DecimalPrecision2.ceil(-2.26, 2) === -2.26);
console.log(DecimalPrecision2.ceil(-18.15, 2) === -18.15);

// testing edge cases for floor
console.log(DecimalPrecision2.floor(2.26, 2) === 2.26);
console.log(DecimalPrecision2.floor(18.15, 2) === 18.15);
console.log(DecimalPrecision2.floor(-9.13, 2) === -9.13);
console.log(DecimalPrecision2.floor(-65.18, 2) === -65.18);

// testing edge cases for trunc
console.log(DecimalPrecision2.trunc(2.26, 2) === 2.26);
console.log(DecimalPrecision2.trunc(18.15, 2) === 18.15);
console.log(DecimalPrecision2.trunc(-2.26, 2) === -2.26);
console.log(DecimalPrecision2.trunc(-18.15, 2) === -18.15);

// testing round to tens and hundreds
console.log(DecimalPrecision2.round(1262.48, -1) === 1260);
console.log(DecimalPrecision2.round(1262.48, -2) === 1300);

// testing toFixed()
console.log(DecimalPrecision2.toFixed(1.005, 2) === "1.01");``````

Solution 3: double rounding

This solution uses the toPrecision() method to strip the floating-point round-off errors.

``````// Solution 3
var DecimalPrecision3 = (function() {
if (Math.trunc === undefined) {
Math.trunc = function(v) {
return v < 0 ? Math.ceil(v) : Math.floor(v);
};
}
var powers = [
1e0,  1e1,  1e2,  1e3,  1e4,  1e5,  1e6,  1e7,
1e8,  1e9,  1e10, 1e11, 1e12, 1e13, 1e14, 1e15,
1e16, 1e17, 1e18, 1e19, 1e20, 1e21, 1e22
];
var intpow10 = function(power) {
if (power < 0 || power > 22) {
return Math.pow(10, power);
}
return powers[power];
};
// Eliminate binary floating-point inaccuracies.
var stripError = function(num) {
if (Number.isInteger(num))
return num;
return parseFloat(num.toPrecision(15));
};
var decimalAdjust = function myself(type, num, decimalPlaces) {
if (type === 'round' && num < 0)
return -myself(type, -num, decimalPlaces);
var p = intpow10(decimalPlaces || 0);
var n = stripError(num * p);
return Math[type](n) / p;
};
return {
// Decimal round (half away from zero)
round: function(num, decimalPlaces) {
},
// Decimal ceil
ceil: function(num, decimalPlaces) {
},
// Decimal floor
floor: function(num, decimalPlaces) {
},
// Decimal trunc
trunc: function(num, decimalPlaces) {
},
// Format using fixed-point notation
toFixed: function(num, decimalPlaces) {
}
};
})();

// test rounding of half
console.log(DecimalPrecision3.round(0.5));  // 1
console.log(DecimalPrecision3.round(-0.5)); // -1

// testing very small numbers
console.log(DecimalPrecision3.ceil(1e-8, 2) === 0.01);
console.log(DecimalPrecision3.floor(1e-8, 2) === 0);

// testing simple cases
console.log(DecimalPrecision3.round(5.12, 1) === 5.1);
console.log(DecimalPrecision3.round(-5.12, 1) === -5.1);
console.log(DecimalPrecision3.ceil(5.12, 1) === 5.2);
console.log(DecimalPrecision3.ceil(-5.12, 1) === -5.1);
console.log(DecimalPrecision3.floor(5.12, 1) === 5.1);
console.log(DecimalPrecision3.floor(-5.12, 1) === -5.2);
console.log(DecimalPrecision3.trunc(5.12, 1) === 5.1);
console.log(DecimalPrecision3.trunc(-5.12, 1) === -5.1);

// testing edge cases for round
console.log(DecimalPrecision3.round(1.005, 2) === 1.01);
console.log(DecimalPrecision3.round(39.425, 2) === 39.43);
console.log(DecimalPrecision3.round(-1.005, 2) === -1.01);
console.log(DecimalPrecision3.round(-39.425, 2) === -39.43);

// testing edge cases for ceil
console.log(DecimalPrecision3.ceil(9.13, 2) === 9.13);
console.log(DecimalPrecision3.ceil(65.18, 2) === 65.18);
console.log(DecimalPrecision3.ceil(-2.26, 2) === -2.26);
console.log(DecimalPrecision3.ceil(-18.15, 2) === -18.15);

// testing edge cases for floor
console.log(DecimalPrecision3.floor(2.26, 2) === 2.26);
console.log(DecimalPrecision3.floor(18.15, 2) === 18.15);
console.log(DecimalPrecision3.floor(-9.13, 2) === -9.13);
console.log(DecimalPrecision3.floor(-65.18, 2) === -65.18);

// testing edge cases for trunc
console.log(DecimalPrecision3.trunc(2.26, 2) === 2.26);
console.log(DecimalPrecision3.trunc(18.15, 2) === 18.15);
console.log(DecimalPrecision3.trunc(-2.26, 2) === -2.26);
console.log(DecimalPrecision3.trunc(-18.15, 2) === -18.15);

// testing round to tens and hundreds
console.log(DecimalPrecision3.round(1262.48, -1) === 1260);
console.log(DecimalPrecision3.round(1262.48, -2) === 1300);

// testing toFixed()
console.log(DecimalPrecision3.toFixed(1.005, 2) === "1.01");``````

Solution 4: double rounding v2

This solution is just like Solution 3, however it uses a custom `toPrecision()` function.

``````// Solution 4
var DecimalPrecision4 = (function() {
if (Math.trunc === undefined) {
Math.trunc = function(v) {
return v < 0 ? Math.ceil(v) : Math.floor(v);
};
}
var powers = [
1e0,  1e1,  1e2,  1e3,  1e4,  1e5,  1e6,  1e7,
1e8,  1e9,  1e10, 1e11, 1e12, 1e13, 1e14, 1e15,
1e16, 1e17, 1e18, 1e19, 1e20, 1e21, 1e22
];
var intpow10 = function(power) {
if (power < 0 || power > 22) {
return Math.pow(10, power);
}
return powers[power];
};
var toPrecision = function(num, significantDigits) {
// Return early for ±0, NaN and Infinity.
if (!num || !Number.isFinite(num))
return num;
// Compute shift of the decimal point (sf - leftSidedDigits).
var shift = significantDigits - 1 - Math.floor(Math.log10(Math.abs(num)));
// Return if rounding to the same or higher precision.
var decimalPlaces = 0;
for (var p = 1; num != Math.round(num * p) / p; p *= 10) decimalPlaces++;
if (shift >= decimalPlaces)
return num;
// Round to "shift" fractional digits
var scale = intpow10(Math.abs(shift));
return shift > 0 ?
Math.round(num * scale) / scale :
Math.round(num / scale) * scale;
};
// Eliminate binary floating-point inaccuracies.
var stripError = function(num) {
if (Number.isInteger(num))
return num;
};
var decimalAdjust = function myself(type, num, decimalPlaces) {
if (type === 'round' && num < 0)
return -myself(type, -num, decimalPlaces);
var p = intpow10(decimalPlaces || 0);
var n = stripError(num * p);
return Math[type](n) / p;
};
return {
// Decimal round (half away from zero)
round: function(num, decimalPlaces) {
},
// Decimal ceil
ceil: function(num, decimalPlaces) {
},
// Decimal floor
floor: function(num, decimalPlaces) {
},
// Decimal trunc
trunc: function(num, decimalPlaces) {
},
// Format using fixed-point notation
toFixed: function(num, decimalPlaces) {
}
};
})();

// test rounding of half
console.log(DecimalPrecision4.round(0.5));  // 1
console.log(DecimalPrecision4.round(-0.5)); // -1

// testing very small numbers
console.log(DecimalPrecision4.ceil(1e-8, 2) === 0.01);
console.log(DecimalPrecision4.floor(1e-8, 2) === 0);

// testing simple cases
console.log(DecimalPrecision4.round(5.12, 1) === 5.1);
console.log(DecimalPrecision4.round(-5.12, 1) === -5.1);
console.log(DecimalPrecision4.ceil(5.12, 1) === 5.2);
console.log(DecimalPrecision4.ceil(-5.12, 1) === -5.1);
console.log(DecimalPrecision4.floor(5.12, 1) === 5.1);
console.log(DecimalPrecision4.floor(-5.12, 1) === -5.2);
console.log(DecimalPrecision4.trunc(5.12, 1) === 5.1);
console.log(DecimalPrecision4.trunc(-5.12, 1) === -5.1);

// testing edge cases for round
console.log(DecimalPrecision4.round(1.005, 2) === 1.01);
console.log(DecimalPrecision4.round(39.425, 2) === 39.43);
console.log(DecimalPrecision4.round(-1.005, 2) === -1.01);
console.log(DecimalPrecision4.round(-39.425, 2) === -39.43);

// testing edge cases for ceil
console.log(DecimalPrecision4.ceil(9.13, 2) === 9.13);
console.log(DecimalPrecision4.ceil(65.18, 2) === 65.18);
console.log(DecimalPrecision4.ceil(-2.26, 2) === -2.26);
console.log(DecimalPrecision4.ceil(-18.15, 2) === -18.15);

// testing edge cases for floor
console.log(DecimalPrecision4.floor(2.26, 2) === 2.26);
console.log(DecimalPrecision4.floor(18.15, 2) === 18.15);
console.log(DecimalPrecision4.floor(-9.13, 2) === -9.13);
console.log(DecimalPrecision4.floor(-65.18, 2) === -65.18);

// testing edge cases for trunc
console.log(DecimalPrecision4.trunc(2.26, 2) === 2.26);
console.log(DecimalPrecision4.trunc(18.15, 2) === 18.15);
console.log(DecimalPrecision4.trunc(-2.26, 2) === -2.26);
console.log(DecimalPrecision4.trunc(-18.15, 2) === -18.15);

// testing round to tens and hundreds
console.log(DecimalPrecision4.round(1262.48, -1) === 1260);
console.log(DecimalPrecision4.round(1262.48, -2) === 1300);

// testing toFixed()
console.log(DecimalPrecision4.toFixed(1.005, 2) === "1.01");``````

## Benchmarks

http://jsbench.github.io/#31ec3a8b3d22bd840f8e6822e681a3ac

Here is a benchmark comparing the operations per second in the solutions above on Chrome 85.0.4183.83. Obviously all browsers differ, so your mileage may vary.

(Note: More is better)

Thanks @Mike for adding a screenshot of the benchmark.

## Solution 7

This question is complicated.

Suppose we have a function, `roundTo2DP(num)`, that takes a float as an argument and returns a value rounded to 2 decimal places. What should each of these expressions evaluate to?

• `roundTo2DP(0.014999999999999999)`
• `roundTo2DP(0.0150000000000000001)`
• `roundTo2DP(0.015)`

The 'obvious' answer is that the first example should round to 0.01 (because it's closer to 0.01 than to 0.02) while the other two should round to 0.02 (because 0.0150000000000000001 is closer to 0.02 than to 0.01, and because 0.015 is exactly halfway between them and there is a mathematical convention that such numbers get rounded up).

The catch, which you may have guessed, is that `roundTo2DP` cannot possibly be implemented to give those obvious answers, because all three numbers passed to it are the same number. IEEE 754 binary floating point numbers (the kind used by JavaScript) can't exactly represent most non-integer numbers, and so all three numeric literals above get rounded to a nearby valid floating point number. This number, as it happens, is exactly

0.01499999999999999944488848768742172978818416595458984375

which is closer to 0.01 than to 0.02.

You can see that all three numbers are the same at your browser console, Node shell, or other JavaScript interpreter. Just compare them:

``````> 0.014999999999999999 === 0.0150000000000000001
true``````

So when I write `m = 0.0150000000000000001`, the exact value of `m` that I end up with is closer to `0.01` than it is to `0.02`. And yet, if I convert `m` to a String...

``````> var m = 0.0150000000000000001;
> console.log(String(m));
0.015
> var m = 0.014999999999999999;
> console.log(String(m));
0.015``````

... I get 0.015, which should round to 0.02, and which is noticeably not the 56-decimal-place number I earlier said that all of these numbers were exactly equal to. So what dark magic is this?

The answer can be found in the ECMAScript specification, in section 7.1.12.1: ToString applied to the Number type. Here the rules for converting some Number m to a String are laid down. The key part is point 5, in which an integer s is generated whose digits will be used in the String representation of m:

let n, k, and s be integers such that k 1, 10k-1 s < 10k, the Number value for s × 10n-k is m, and k is as small as possible. Note that k is the number of digits in the decimal representation of s, that s is not divisible by 10, and that the least significant digit of s is not necessarily uniquely determined by these criteria.

The key part here is the requirement that "k is as small as possible". What that requirement amounts to is a requirement that, given a Number `m`, the value of `String(m)` must have the least possible number of digits while still satisfying the requirement that `Number(String(m)) === m`. Since we already know that `0.015 === 0.0150000000000000001`, it's now clear why `String(0.0150000000000000001) === '0.015'` must be true.

Of course, none of this discussion has directly answered what `roundTo2DP(m)` should return. If `m`'s exact value is 0.01499999999999999944488848768742172978818416595458984375, but its String representation is '0.015', then what is the correct answer - mathematically, practically, philosophically, or whatever - when we round it to two decimal places?

There is no single correct answer to this. It depends upon your use case. You probably want to respect the String representation and round upwards when:

• The value being represented is inherently discrete, e.g. an amount of currency in a 3-decimal-place currency like dinars. In this case, the true value of a Number like 0.015 is 0.015, and the 0.0149999999... representation that it gets in binary floating point is a rounding error. (Of course, many will argue, reasonably, that you should use a decimal library for handling such values and never represent them as binary floating point Numbers in the first place.)
• The value was typed by a user. In this case, again, the exact decimal number entered is more 'true' than the nearest binary floating point representation.

On the other hand, you probably want to respect the binary floating point value and round downwards when your value is from an inherently continuous scale - for instance, if it's a reading from a sensor.

These two approaches require different code. To respect the String representation of the Number, we can (with quite a bit of reasonably subtle code) implement our own rounding that acts directly on the String representation, digit by digit, using the same algorithm you would've used in school when you were taught how to round numbers. Below is an example which respects the OP's requirement of representing the number to 2 decimal places "only when necessary" by stripping trailing zeroes after the decimal point; you may, of course, need to tweak it to your precise needs.

``````/**
* Converts num to a decimal string (if it isn't one already) and then rounds it
* to at most dp decimal places.
*
* For explanation of why you'd want to perform rounding operations on a String
* rather than a Number, see http://stackoverflow.com/a/38676273/1709587
*
* @param {(number|string)} num
* @param {number} dp
* @return {string}
*/
function roundStringNumberWithoutTrailingZeroes (num, dp) {
if (arguments.length != 2) throw new Error("2 arguments required");

num = String(num);
if (num.indexOf('e+') != -1) {
// Can't round numbers this large because their string representation
// contains an exponent, like 9.99e+37
throw new Error("num too large");
}
if (num.indexOf('.') == -1) {
// Nothing to do
return num;
}

var parts = num.split('.'),
beforePoint = parts[0],
afterPoint = parts[1],
shouldRoundUp = afterPoint[dp] >= 5,
finalNumber;

afterPoint = afterPoint.slice(0, dp);
if (!shouldRoundUp) {
finalNumber = beforePoint + '.' + afterPoint;
} else if (/^9+\$/.test(afterPoint)) {
// If we need to round up a number like 1.9999, increment the integer
// before the decimal point and discard the fractional part.
finalNumber = Number(beforePoint)+1;
} else {
// Starting from the last digit, increment digits until we find one
// that is not 9, then stop
var i = dp-1;
while (true) {
if (afterPoint[i] == '9') {
afterPoint = afterPoint.substr(0, i) +
'0' +
afterPoint.substr(i+1);
i--;
} else {
afterPoint = afterPoint.substr(0, i) +
(Number(afterPoint[i]) + 1) +
afterPoint.substr(i+1);
break;
}
}

finalNumber = beforePoint + '.' + afterPoint;
}

// Remove trailing zeroes from fractional part before returning
return finalNumber.replace(/0+\$/, '')
}
``````

Example usage:

``````> roundStringNumberWithoutTrailingZeroes(1.6, 2)
'1.6'
> roundStringNumberWithoutTrailingZeroes(10000, 2)
'10000'
> roundStringNumberWithoutTrailingZeroes(0.015, 2)
'0.02'
> roundStringNumberWithoutTrailingZeroes('0.015000', 2)
'0.02'
> roundStringNumberWithoutTrailingZeroes(1, 1)
'1'
> roundStringNumberWithoutTrailingZeroes('0.015', 2)
'0.02'
> roundStringNumberWithoutTrailingZeroes(0.01499999999999999944488848768742172978818416595458984375, 2)
'0.02'
> roundStringNumberWithoutTrailingZeroes('0.01499999999999999944488848768742172978818416595458984375', 2)
'0.01'``````

The function above is probably what you want to use to avoid users ever witnessing numbers that they have entered being rounded wrongly.

(As an alternative, you could also try the round10 library which provides a similarly-behaving function with a wildly different implementation.)

But what if you have the second kind of Number - a value taken from a continuous scale, where there's no reason to think that approximate decimal representations with fewer decimal places are more accurate than those with more? In that case, we don't want to respect the String representation, because that representation (as explained in the spec) is already sort-of-rounded; we don't want to make the mistake of saying "0.014999999...375 rounds up to 0.015, which rounds up to 0.02, so 0.014999999...375 rounds up to 0.02".

Here we can simply use the built-in `toFixed` method. Note that by calling `Number()` on the String returned by `toFixed`, we get a Number whose String representation has no trailing zeroes (thanks to the way JavaScript computes the String representation of a Number, discussed earlier in this answer).

``````/**
* Takes a float and rounds it to at most dp decimal places. For example
*
*     roundFloatNumberWithoutTrailingZeroes(1.2345, 3)
*
* returns 1.234
*
* Note that since this treats the value passed to it as a floating point
* number, it will have counterintuitive results in some cases. For instance,
*
*     roundFloatNumberWithoutTrailingZeroes(0.015, 2)
*
* gives 0.01 where 0.02 might be expected. For an explanation of why, see
* http://stackoverflow.com/a/38676273/1709587. You may want to consider using the
*
* @param {number} num
* @param {number} dp
* @return {number}
*/
function roundFloatNumberWithoutTrailingZeroes (num, dp) {
var numToFixedDp = Number(num).toFixed(dp);
return Number(numToFixedDp);
}
``````

## Solution 8

Consider `.toFixed()` and `.toPrecision()`:

http://www.javascriptkit.com/javatutors/formatnumber.shtml

## Solution 9

One can use `.toFixed(NumberOfDecimalPlaces)`.

``````var str = 10.234.toFixed(2); // => '10.23'
var number = Number(str); // => 10.23
``````

## Solution 10

None of the answers found here is correct. stinkycheeseman asked to round up, but you all rounded the number.

To round up, use this:

``````Math.ceil(num * 100)/100;
``````

## Solution 11

Here is a simple way to do it:

``````Math.round(value * 100) / 100
``````

You might want to go ahead and make a separate function to do it for you though:

``````function roundToTwo(value) {
return(Math.round(value * 100) / 100);
}
``````

Then you would simply pass in the value.

You could enhance it to round to any arbitrary number of decimals by adding a second parameter.

``````function myRound(value, places) {
var multiplier = Math.pow(10, places);

return (Math.round(value * multiplier) / multiplier);
}
``````

## Solution 12

A precise rounding method. Source: Mozilla

``````(function(){

/**
* Decimal adjustment of a number.
*
* @param   {String}    type    The type of adjustment.
* @param   {Number}    value   The number.
* @param   {Integer}   exp     The exponent (the 10 logarithm of the adjustment base).
* @returns {Number}            The adjusted value.
*/
// If the exp is undefined or zero...
if (typeof exp === 'undefined' || +exp === 0) {
return Math[type](value);
}
value = +value;
exp = +exp;
// If the value is not a number or the exp is not an integer...
if (isNaN(value) || !(typeof exp === 'number' && exp % 1 === 0)) {
return NaN;
}
// Shift
value = value.toString().split('e');
value = Math[type](+(value[0] + 'e' + (value[1] ? (+value[1] - exp) : -exp)));
// Shift back
value = value.toString().split('e');
return +(value[0] + 'e' + (value[1] ? (+value[1] + exp) : exp));
}

// Decimal round
if (!Math.round10) {
Math.round10 = function(value, exp) {
};
}
// Decimal floor
if (!Math.floor10) {
Math.floor10 = function(value, exp) {
};
}
// Decimal ceil
if (!Math.ceil10) {
Math.ceil10 = function(value, exp) {
};
}
})();
``````

Examples:

``````// Round
Math.round10(55.55, -1); // 55.6
Math.round10(55.549, -1); // 55.5
Math.round10(55, 1); // 60
Math.round10(54.9, 1); // 50
Math.round10(-55.55, -1); // -55.5
Math.round10(-55.551, -1); // -55.6
Math.round10(-55, 1); // -50
Math.round10(-55.1, 1); // -60
Math.round10(1.005, -2); // 1.01 -- compare this with Math.round(1.005*100)/100 above
// Floor
Math.floor10(55.59, -1); // 55.5
Math.floor10(59, 1); // 50
Math.floor10(-55.51, -1); // -55.6
Math.floor10(-51, 1); // -60
// Ceil
Math.ceil10(55.51, -1); // 55.6
Math.ceil10(51, 1); // 60
Math.ceil10(-55.59, -1); // -55.5
Math.ceil10(-59, 1); // -50
``````

## Solution 13

``````var result = Math.round(input*100)/100;
``````

For more information, you can have a look at Math.round(num) vs num.toFixed(0) and browser inconsistencies

## Solution 14

For me Math.round() was not giving correct answer. I found toFixed(2) works better. Below are examples of both:

``````console.log(Math.round(43000 / 80000) * 100); // wrong answer

console.log(((43000 / 80000) * 100).toFixed(2)); // correct answer``````

## Solution 15

Use this function `Number(x).toFixed(2);`

## Solution 16

``````+(10).toFixed(2); // = 10
+(10.12345).toFixed(2); // = 10.12

(10).toFixed(2); // = 10.00
(10.12345).toFixed(2); // = 10.12
``````

## Solution 17

Try this lightweight solution:

``````function round(x, digits){
return parseFloat(x.toFixed(digits))
}

round(1.222,  2);
// 1.22
round(1.222, 10);
// 1.222
``````

## Solution 18

There are a couple of ways to do that. For people like me, Lodash's variant

``````function round(number, precision) {
var pair = (number + 'e').split('e')
var value = Math.round(pair[0] + 'e' + (+pair[1] + precision))
pair = (value + 'e').split('e')
return +(pair[0] + 'e' + (+pair[1] - precision))
}
``````

Usage:

``````round(0.015, 2) // 0.02
round(1.005, 2) // 1.01
``````

If your project uses jQuery or Lodash, you can also find the proper `round` method in the libraries.

## Solution 19

If you are using the Lodash library, you can use the round method of Lodash like following.

``````_.round(number, precision)
``````

For example:

``````_.round(1.7777777, 2) = 1.78
``````

## Solution 20

2017
Just use native code `.toFixed()`

``````number = 1.2345;
number.toFixed(2) // "1.23"
``````

If you need to be strict and add digits just if needed it can use `replace`

``````number = 1; // "1"
number.toFixed(5).replace(/\.?0*\$/g,'');
``````

## Solution 21

Since ES6 there is a 'proper' way (without overriding statics and creating workarounds) to do this by using toPrecision

``````var x = 1.49999999999;
console.log(x.toPrecision(4));
console.log(x.toPrecision(3));
console.log(x.toPrecision(2));

var y = Math.PI;
console.log(y.toPrecision(6));
console.log(y.toPrecision(5));
console.log(y.toPrecision(4));

var z = 222.987654
console.log(z.toPrecision(6));
console.log(z.toPrecision(5));
console.log(z.toPrecision(4));``````

then you can just `parseFloat` and zeroes will 'go away'.

``````console.log(parseFloat((1.4999).toPrecision(3)));
console.log(parseFloat((1.005).toPrecision(3)));
console.log(parseFloat((1.0051).toPrecision(3)));``````

It doesn't solve the '1.005 rounding problem' though - since it is intrinsic to how float fractions are being processed.

``console.log(1.005 - 0.005);``

If you are open to libraries you can use bignumber.js

``````console.log(1.005 - 0.005);
console.log(new BigNumber(1.005).minus(0.005));

console.log(new BigNumber(1.005).round(4));
console.log(new BigNumber(1.005).round(3));
console.log(new BigNumber(1.005).round(2));
console.log(new BigNumber(1.005).round(1));``````
``<script src="https://cdnjs.cloudflare.com/ajax/libs/bignumber.js/2.3.0/bignumber.min.js"></script>``

## Solution 22

The easiest approach would be to use toFixed and then strip trailing zeros using the Number function:

``````const number = 15.5;
Number(number.toFixed(2)); // 15.5
``````
``````const number = 1.7777777;
Number(number.toFixed(2)); // 1.78
``````

## Solution 23

MarkG and Lavamantis offered a much better solution than the one that has been accepted. It's a shame they don't get more upvotes!

Here is the function I use to solve the floating point decimals issues also based on MDN. It is even more generic (but less concise) than Lavamantis's solution:

``````function round(value, exp) {
if (typeof exp === 'undefined' || +exp === 0)
return Math.round(value);

value = +value;
exp  = +exp;

if (isNaN(value) || !(typeof exp === 'number' && exp % 1 === 0))
return NaN;

// Shift
value = value.toString().split('e');
value = Math.round(+(value[0] + 'e' + (value[1] ? (+value[1] + exp) : exp)));

// Shift back
value = value.toString().split('e');
return +(value[0] + 'e' + (value[1] ? (+value[1] - exp) : -exp));
}
``````

Use it with:

``````round(10.8034, 2);      // Returns 10.8
round(1.275, 2);        // Returns 1.28
round(1.27499, 2);      // Returns 1.27
round(1.2345678e+2, 2); // Returns 123.46
``````

Compared to Lavamantis's solution, we can do...

``````round(1234.5678, -2); // Returns 1200
round("123.45");      // Returns 123
``````

## Solution 24

One way to achieve such a rounding only if necessary is to use Number.prototype.toLocaleString():

``````myNumber.toLocaleString('en', {maximumFractionDigits:2, useGrouping:false})
``````

This will provide exactly the output you expect, but as strings. You can still convert those back to numbers if that's not the data type you expect.

## Solution 25

It may work for you,

``````Math.round(num * 100)/100;
``````

to know the difference between toFixed and round. You can have a look at Math.round(num) vs num.toFixed(0) and browser inconsistencies.

## Solution 26

This is the simplest, more elegant solution (and I am the best of the world;):

``````function roundToX(num, X) {
return +(Math.round(num + "e+"+X)  + "e-"+X);
}
//roundToX(66.66666666,2) => 66.67
//roundToX(10,2) => 10
//roundToX(10.904,2) => 10.9
``````

Modern syntax alternative with fallback values

``````const roundToX = (num = 0, X = 20) => +(Math.round(num + `e\${X}`)  + `e-\${X}`)
``````

## Solution 27

``````var roundUpto = function(number, upto){
return Number(number.toFixed(upto));
}
roundUpto(0.1464676, 2);
``````

`toFixed(2)`: Here 2 is the number of digits up to which we want to round this number.

## Solution 28

See @AmrAli's answer for a more thorough run through and performance breakdown of all the various adaptations of this solution.

``````var DecimalPrecision = (function(){
if (Number.EPSILON === undefined) {
Number.EPSILON = Math.pow(2, -52);
}
if(Number.isInteger === undefined){
Number.isInteger = function(value) {
return typeof value === 'number' &&
isFinite(value) &&
Math.floor(value) === value;
};
}
this.isRound = function(n,p){
let l = n.toString().split('.')[1].length;
return (p >= l);
}
this.round = function(n, p=2){
if(Number.isInteger(n) || this.isRound(n,p))
return n;
let r = 0.5 * Number.EPSILON * n;
let o = 1; while(p-- > 0) o *= 10;
if(n<0)
o *= -1;
return Math.round((n + r) * o) / o;
}
this.ceil = function(n, p=2){
if(Number.isInteger(n) || this.isRound(n,p))
return n;
let r = 0.5 * Number.EPSILON * n;
let o = 1; while(p-- > 0) o *= 10;

return Math.ceil((n + r) * o) / o;
}
this.floor = function(n, p=2){
if(Number.isInteger(n) || this.isRound(n,p))
return n;
let r = 0.5 * Number.EPSILON * n;
let o = 1; while(p-- > 0) o *= 10;

return Math.floor((n + r) * o) / o;
}
return this;
})();
console.log(DecimalPrecision.round(1.005));
console.log(DecimalPrecision.ceil(1.005));
console.log(DecimalPrecision.floor(1.005));
console.log(DecimalPrecision.round(1.0049999));
console.log(DecimalPrecision.ceil(1.0049999));
console.log(DecimalPrecision.floor(1.0049999));
console.log(DecimalPrecision.round(2.175495134384,7));
console.log(DecimalPrecision.round(2.1753543549,8));
console.log(DecimalPrecision.round(2.1755465135353,4));
console.log(DecimalPrecision.ceil(17,4));
console.log(DecimalPrecision.ceil(17.1,4));
console.log(DecimalPrecision.ceil(17.1,15));``````

## Solution 29

Easiest way:

`+num.toFixed(2)`

It converts it to a string, and then back into an integer / float.

## Solution 30

Another simple solution (without writing any function) may to use toFixed() and then convert to float again:

For example:

``````var objNumber = 1201203.1256546456;
objNumber = parseFloat(objNumber.toFixed(2))
``````