What is JavaScript's highest integer value that a number can go to without losing precision?

Question

Is this defined by the language? Is there a defined maximum? Is it different in different browsers?

Answer 1

JavaScript has two number types: Number and BigInt.

The most frequently-used number type, Number, is a 64-bit floating point IEEE 754 number.

The largest exact integral value of this type is Number.MAX_SAFE_INTEGER, which is:

• 2⁵³-1, or
• +/- 9,007,199,254,740,991, or
• nine quadrillion seven trillion one hundred ninety-nine billion two hundred fifty-four million seven hundred forty thousand nine hundred ninety-one

To put this in perspective: one quadrillion bytes is a petabyte (or one thousand terabytes).

"Safe" in this context refers to the ability to represent integers exactly and to correctly compare them.

From the spec:

Note that all the positive and negative integers whose magnitude is no greater than 2⁵³ are representable in the Number type (indeed, the integer 0 has two representations, +0 and -0).

To safely use integers larger than this, you need to use BigInt, which has no upper bound.

Note that the bitwise operators and shift operators operate on 32-bit integers, so in that case, the max safe integer is 2³¹-1, or 2,147,483,647.

const log = console.log
var x = 9007199254740992
var y = -x
log(x == x + 1) // true !
log(y == y - 1) // also true !

// Arithmetic operators work, but bitwise/shifts only operate on int32:
log(x / 2)      // 4503599627370496
log(x >> 1)     // 0
log(x | 1)      // 1

---

Technical note on the subject of the number 9,007,199,254,740,992: There is an exact IEEE-754 representation of this value, and you can assign and read this value from a variable, so for very carefully chosen applications in the domain of integers less than or equal to this value, you could treat this as a maximum value.

In the general case, you must treat this IEEE-754 value as inexact, because it is ambiguous whether it is encoding the logical value 9,007,199,254,740,992 or 9,007,199,254,740,993.

Answer 2

>= ES6:

Number.MIN_SAFE_INTEGER;
Number.MAX_SAFE_INTEGER;

<= ES5

From the reference:

Number.MAX_VALUE;
Number.MIN_VALUE;

console.log('MIN_VALUE', Number.MIN_VALUE);
console.log('MAX_VALUE', Number.MAX_VALUE);

console.log('MIN_SAFE_INTEGER', Number.MIN_SAFE_INTEGER); //ES6
console.log('MAX_SAFE_INTEGER', Number.MAX_SAFE_INTEGER); //ES6

Answer 3

It is 2⁵³ == 9 007 199 254 740 992. This is because Numbers are stored as floating-point in a 52-bit mantissa.

The min value is -2⁵³.

This makes some fun things happening

Math.pow(2, 53) == Math.pow(2, 53) + 1
>> true

And can also be dangerous :)

var MAX_INT = Math.pow(2, 53); // 9 007 199 254 740 992
for (var i = MAX_INT; i < MAX_INT + 2; ++i) {
    // infinite loop
}

_{Further reading: http://blog.vjeux.com/2010/javascript/javascript-max_int-number-limits.html}

Answer 4

In JavaScript, there is a number called Infinity.

Examples:

(Infinity>100)
=> true

// Also worth noting
Infinity - 1 == Infinity
=> true

Math.pow(2,1024) === Infinity
=> true

This may be sufficient for some questions regarding this topic.

Answer 5

Many earlier answers have shown 9007199254740992 === 9007199254740992 + 1 is true to verify that 9,007,199,254,740,991 is the maximum and safe integer.

But what if we keep doing accumulation:

input: 9007199254740992 + 1  output: 9007199254740992  // expected: 9007199254740993
input: 9007199254740992 + 2  output: 9007199254740994  // expected: 9007199254740994
input: 9007199254740992 + 3  output: 9007199254740996  // expected: 9007199254740995
input: 9007199254740992 + 4  output: 9007199254740996  // expected: 9007199254740996

We can see that among numbers greater than 9,007,199,254,740,992, only even numbers are representable.

It's an entry to explain how the double-precision 64-bit binary format works. Let's see how 9,007,199,254,740,992 be held (represented) by using this binary format.

Using a brief version to demonstrate it from 4,503,599,627,370,496:

  1 . 0000 ---- 0000  *  2^52            =>  1  0000 ---- 0000.  
     |-- 52 bits --|    |exponent part|        |-- 52 bits --|

On the left side of the arrow, we have bit value 1, and an adjacent radix point. By consuming the exponent part on the left, the radix point is moved 52 steps to the right. The radix point ends up at the end, and we get 4503599627370496 in pure binary.

Now let's keep incrementing the fraction part with 1 until all the bits are set to 1, which equals 9,007,199,254,740,991 in decimal.

  1 . 0000 ---- 0000  *  2^52  =>  1  0000 ---- 0000.  
                       (+1)
  1 . 0000 ---- 0001  *  2^52  =>  1  0000 ---- 0001.  
                       (+1)
  1 . 0000 ---- 0010  *  2^52  =>  1  0000 ---- 0010.  
                       (+1)
                        . 
                        .
                        .
  1 . 1111 ---- 1111  *  2^52  =>  1  1111 ---- 1111. 

Because the 64-bit double-precision format strictly allots 52 bits for the fraction part, no more bits are available if we add another 1, so what we can do is setting all bits back to 0, and manipulate the exponent part:

  ┏━━▶ This bit is implicit and persistent.
  ┃        
  1 . 1111 ---- 1111  *  2^52      =>  1  1111 ---- 1111. 
     |-- 52 bits --|                     |-- 52 bits --|

                          (+1)

  1 . 0000 ---- 0000  *  2^52 * 2  =>  1  0000 ---- 0000. * 2  
     |-- 52 bits --|                     |-- 52 bits --|
                                      (By consuming the 2^52, radix
                                       point has no way to go, but
                                       there is still one 2 left in
                                       exponent part)
  =>  1 . 0000 ---- 0000  *  2^53 
         |-- 52 bits --| 

Now we get the 9,007,199,254,740,992, and for the numbers greater than it, the format can only handle increments of 2 because every increment of 1 on the fraction part ends up being multiplied by the left 2 in the exponent part. That's why double-precision 64-bit binary format cannot hold odd numbers when the number is greater than 9,007,199,254,740,992:

                            (consume 2^52 to move radix point to the end)
  1 . 0000 ---- 0001  *  2^53  =>  1  0000 ---- 0001.  *  2
     |-- 52 bits --|                 |-- 52 bits --|

Following this pattern, when the number gets greater than 9,007,199,254,740,992 * 2 = 18,014,398,509,481,984 only 4 times the fraction can be held:

input: 18014398509481984 + 1  output: 18014398509481984  // expected: 18014398509481985
input: 18014398509481984 + 2  output: 18014398509481984  // expected: 18014398509481986
input: 18014398509481984 + 3  output: 18014398509481984  // expected: 18014398509481987
input: 18014398509481984 + 4  output: 18014398509481988  // expected: 18014398509481988

How about numbers between [ 2 251 799 813 685 248, 4 503 599 627 370 496 )?

 1 . 0000 ---- 0001  *  2^51  =>  1 0000 ---- 000.1
     |-- 52 bits --|                |-- 52 bits  --|

The value 0.1 in binary is exactly 2^-1 (=1/2) (=0.5) So when the number is less than 4,503,599,627,370,496 (2^52), there is one bit available to represent the 1/2 times of the integer:

input: 4503599627370495.5   output: 4503599627370495.5  
input: 4503599627370495.75  output: 4503599627370495.5  
            

Less than 2,251,799,813,685,248 (2^51)

input: 2251799813685246.75   output: 2251799813685246.8  // expected: 2251799813685246.75 
input: 2251799813685246.25   output: 2251799813685246.2  // expected: 2251799813685246.25 
input: 2251799813685246.5    output: 2251799813685246.5
/**
   Please note that if you try this yourself and, say, log 
   these numbers to the console, they will get rounded. JavaScript
   rounds if the number of digits exceed 17. The value 
   is internally held correctly:
*/
            
input: 2251799813685246.25.toString(2) 
output: "111111111111111111111111111111111111111111111111110.01"
input: 2251799813685246.75.toString(2) 
output: "111111111111111111111111111111111111111111111111110.11"
input: 2251799813685246.78.toString(2)   
output: "111111111111111111111111111111111111111111111111110.11"

And what is the available range of exponent part? 11 bits allotted for it by the format.

From Wikipedia (for more details, go there)

So to make the exponent part be 2^52, we exactly need to set e = 1075.

Answer 6

Jimmy's answer correctly represents the continuous JavaScript integer spectrum as -9007199254740992 to 9007199254740992 inclusive (sorry 9007199254740993, you might think you are 9007199254740993, but you are wrong! Demonstration below or in jsfiddle).

console.log(9007199254740993);

However, there is no answer that finds/proves this programatically (other than the one CoolAJ86 alluded to in his answer that would finish in 28.56 years ;), so here's a slightly more efficient way to do that (to be precise, it's more efficient by about 28.559999999968312 years :), along with a test fiddle:

/**
 * Checks if adding/subtracting one to/from a number yields the correct result.
 *
 * @param number The number to test
 * @return true if you can add/subtract 1, false otherwise.
 */
var canAddSubtractOneFromNumber = function(number) {
    var numMinusOne = number - 1;
    var numPlusOne = number + 1;
    
    return ((number - numMinusOne) === 1) && ((number - numPlusOne) === -1);
}

//Find the highest number
var highestNumber = 3; //Start with an integer 1 or higher

//Get a number higher than the valid integer range
while (canAddSubtractOneFromNumber(highestNumber)) {
    highestNumber *= 2;
}

//Find the lowest number you can't add/subtract 1 from
var numToSubtract = highestNumber / 4;
while (numToSubtract >= 1) {
    while (!canAddSubtractOneFromNumber(highestNumber - numToSubtract)) {
        highestNumber = highestNumber - numToSubtract;
    }
    
    numToSubtract /= 2;
}        

//And there was much rejoicing.  Yay.    
console.log('HighestNumber = ' + highestNumber);

Answer 7

To be safe

var MAX_INT = 4294967295;

Reasoning

I thought I'd be clever and find the value at which x + 1 === x with a more pragmatic approach.

My machine can only count 10 million per second or so... so I'll post back with the definitive answer in 28.56 years.

If you can't wait that long, I'm willing to bet that

• Most of your loops don't run for 28.56 years
• 9007199254740992 === Math.pow(2, 53) + 1 is proof enough
• You should stick to 4294967295 which is Math.pow(2,32) - 1 as to avoid expected issues with bit-shifting

Finding x + 1 === x:

(function () {
  "use strict";

  var x = 0
    , start = new Date().valueOf()
    ;

  while (x + 1 != x) {
    if (!(x % 10000000)) {
      console.log(x);
    }

    x += 1
  }

  console.log(x, new Date().valueOf() - start);
}());

Answer 8

The short answer is “it depends.”

If you’re using bitwise operators anywhere (or if you’re referring to the length of an Array), the ranges are:

Unsigned: 0…(-1>>>0)

Signed: (-(-1>>>1)-1)…(-1>>>1)

(It so happens that the bitwise operators and the maximum length of an array are restricted to 32-bit integers.)

If you’re not using bitwise operators or working with array lengths:

Signed: (-Math.pow(2,53))…(+Math.pow(2,53))

These limitations are imposed by the internal representation of the “Number” type, which generally corresponds to IEEE 754 double-precision floating-point representation. (Note that unlike typical signed integers, the magnitude of the negative limit is the same as the magnitude of the positive limit, due to characteristics of the internal representation, which actually includes a negative 0!)

Answer 9

ECMAScript 6:

Number.MAX_SAFE_INTEGER = Math.pow(2, 53)-1;
Number.MIN_SAFE_INTEGER = -Number.MAX_SAFE_INTEGER;

Answer 10

Other may have already given the generic answer, but I thought it would be a good idea to give a fast way of determining it :

for (var x = 2; x + 1 !== x; x *= 2);
console.log(x);

Which gives me 9007199254740992 within less than a millisecond in Chrome 30.

It will test powers of 2 to find which one, when 'added' 1, equals himself.

Answer 11

Anything you want to use for bitwise operations must be between 0x80000000 (-2147483648 or -2^31) and 0x7fffffff (2147483647 or 2^31 - 1).

The console will tell you that 0x80000000 equals +2147483648, but 0x80000000 & 0x80000000 equals -2147483648.

Answer 12

JavaScript has received a new data type in ECMAScript 2020: BigInt. It introduced numerical literals having an "n" suffix and allows for arbitrary precision:

var a = 123456789012345678901012345678901n;

Precision will still be lost, of course, when such big integer is (maybe unintentionally) coerced to a number data type.

And, obviously, there will always be precision limitations due to finite memory, and a cost in terms of time in order to allocate the necessary memory and to perform arithmetic on such large numbers.

For instance, the generation of a number with a hundred thousand decimal digits, will take a noticeable delay before completion:

console.log(BigInt("1".padEnd(100000,"0")) + 1n)

...but it works.

Answer 13

Try:

maxInt = -1 >>> 1

In Firefox 3.6 it's 2^31 - 1.

Answer 14

I did a simple test with a formula, X-(X+1)=-1, and the largest value of X I can get to work on Safari, Opera and Firefox (tested on OS X) is 9e15. Here is the code I used for testing:

javascript: alert(9e15-(9e15+1));

Answer 15

I write it like this:

var max_int = 0x20000000000000;
var min_int = -0x20000000000000;
(max_int + 1) === 0x20000000000000;  //true
(max_int - 1) < 0x20000000000000;    //true

Same for int32

var max_int32 =  0x80000000;
var min_int32 = -0x80000000;

Answer 16

Let's get to the sources

Description

The MAX_SAFE_INTEGER constant has a value of 9007199254740991 (9,007,199,254,740,991 or ~9 quadrillion). The reasoning behind that number is that JavaScript uses double-precision floating-point format numbers as specified in IEEE 754 and can only safely represent numbers between -(2^53 - 1) and 2^53 - 1.

Safe in this context refers to the ability to represent integers exactly and to correctly compare them. For example, Number.MAX_SAFE_INTEGER + 1 === Number.MAX_SAFE_INTEGER + 2 will evaluate to true, which is mathematically incorrect. See Number.isSafeInteger() for more information.

Because MAX_SAFE_INTEGER is a static property of Number, you always use it as Number.MAX_SAFE_INTEGER, rather than as a property of a Number object you created.

Browser compatibility

Answer 17

In JavaScript the representation of numbers is 2^53 - 1.

However, Bitwise operation are calculated on 32 bits ( 4 bytes ), meaning if you exceed 32bits shifts you will start loosing bits.

Answer 18

In the Google Chrome built-in javascript, you can go to approximately 2^1024 before the number is called infinity.

Answer 19

Scato wrotes:

anything you want to use for bitwise operations must be between 0x80000000 (-2147483648 or -2^31) and 0x7fffffff (2147483647 or 2^31 - 1).

the console will tell you that 0x80000000 equals +2147483648, but 0x80000000 & 0x80000000 equals -2147483648

Hex-Decimals are unsigned positive values, so 0x80000000 = 2147483648 - thats mathematically correct. If you want to make it a signed value you have to right shift: 0x80000000 >> 0 = -2147483648. You can write 1 << 31 instead, too.

Answer 20

Node.js and Google Chrome seem to both be using 1024 bit floating point values so:

Number.MAX_VALUE = 1.7976931348623157e+308

Answer 21

Firefox 3 doesn't seem to have a problem with huge numbers.

1e+200 * 1e+100 will calculate fine to 1e+300.

Safari seem to have no problem with it as well. (For the record, this is on a Mac if anyone else decides to test this.)

Unless I lost my brain at this time of day, this is way bigger than a 64-bit integer.