CODEWITHSUNDEEP
javascriptprototype
Julio Rodriguez
Julio Rodriguez

Reputation: 515

Number.isInteger(this) doesn't work in Number.prototype method

I needed to count decimals and came with this solution (please run the working snippet):

Number.prototype.decimalCounter = function() {
  if (!Number.isInteger(this)) {
    return this.toString().split(".")[1].length;
  } else return "not integer";
}

var x = 3.445;
console.log(x.decimalCounter())
console.log((3).decimalCounter())

And this works well if the number is a float. However, if the number is an integer, it throws an error. I don't know why, because in the first if statement I declared that only integers will fire that block of code, and if you remove the decimals of x variable, it should enter the else clause and print out "not an integer". But it won't work. Can you help me figure out where it's failing?

Upvotes: 1

Views: 180

Answers (1)

CertainPerformance
CertainPerformance

Reputation: 370989

In sloppy mode, this for a primitive method like decimalCounter will be the primitive wrapped in an object, so the Number.isInteger test fails, because you're not passing it a primitive, but an object.

console.log(
  Number.isInteger(new Number(5))
);

Enable strict mode, and it'll work as desired, because in strict mode, the primitive won't be wrapped when the method is called:

'use strict';

Number.prototype.decimalCounter = function() {
  if (Number.isInteger(this)) {
    return "not decimal"
  }
  return this.toString().split(".")[1].length;
}

console.log((3).decimalCounter())
console.log((3.45678).decimalCounter())

Upvotes: 4

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