parseInt behaviour in Javascript [duplicate]

Question

This question already has an answer here:

• What is JavaScript's highest integer value that a number can go to without losing precision? 24 answers

I open terminal browser (Chrome for example).

I write this:

var y = "11000011010101011";

"11000011010101011"

parseInt(y)

11000011010101012

I expected 11000011010101011 but it returns me 11000011010101012.

Does anybody know why?

Answer 1

Only 9007199254740991 is the safe integer in javascript. This is case something like

9007199254740992 + 1 // 9007199254740992
9007199254740993 + 1 // 9007199254740992
9007199254740994 + 1 //9007199254740996

Please see the links for more details
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Number http://www.2ality.com/2012/04/number-encoding.html

Also see this one
how addition of Number works on max limit numbers?
This is the dublicate questions see its original

Answer 2

According to this page.

All number in Javascript are 64-bit floating point number, and integers are represented by the 53-bit mantisa. Because of that, you can't store a integer larger than 2^53 -1 and smaller than -2^53 +1 without losing precission (Javascript rounds your number in order to be able to store it).

Your number is larger than 2^53 -1. Even though a String can store it, in order to store it in a "Number" variable, it has to be rounded, losing the precision and returning you a slightly different number.

Answer 3

Every number in Javascript is represented as a double precision floating point. JavaScript can accurately represent integers only up to 9007199254740991 (2^53 - 1). Once you get over that limit, you will loose precision.