Number of bits in Javascript numbers

Question

I work in Javascript with integer numbers only (mainly adding numbers and shifting them). I wonder how big they can be without loosing any bits.

For example, how big X can be such that 1 << X will represent 2^X ?

Answer 1

All numbers in JavaScript are actually IEEE-754 compliant floating-point doubles. These have a 53-bit mantissa which should mean that any integer value with a magnitude of approximately 9 quadrillion or less -- more specifically, 9,007,199,254,740,991 -- will be represented accurately.

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NOTICE: in 2018 main browsers and NodeJS are working also with the new Javascript's primitive-type, BigInt , solving the problems with integer value magnitude.

Answer 2

All answers are partially wrong - Maybe due the new ES6/ES7 specs - , read why:

First of all, in JavaScript, the representation of the number is 2^53 - 1 that is true for @Luke answer, we can prove that by running Number.MAX_SAFE_INTEGER that will show a big number, then we do log2 to confirm that the number of bits is the same :

Number.MAX_SAFE_INTEGER
// 9007199254740991

Math.log2(9007199254740991)
// 53

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

Welcome to Javascript!

Answer 3

All numbers in JavaScript are 64-bit (double-precision) floating point numbers.

Here 'sa description of the format and what values can and can't be represented with it.