JavaScript: Get value inside Number prototype function

Question

I try to use includes in prototype but it does not work but I do not get the expected results:

Number.prototype.inList = function (list) {
  return list.includes(this);
}

let value = 0;

let find1 = value.inList([0,1]);
let find2 = [0,1].includes(value);

console.log("find1", find1); // false
console.log("find2", find2); // true

I assume you understand the risks of overriding built in prototypes.? — Keith, Nov 14 '21 at 10:06
I don't override includes, it is a new function inList. I edit my question which could be confusing... — A.Baudouin, Nov 14 '21 at 10:10
Your overriding a built in Number. Have you ever wondered why there is no `Array.contains`?, A very popular lib called mooTools is to blame for that one, IOW: And the ES spec had to be changed and called it includes instead , all because a popular lib decided to alter Array.prototype with a function it decided to implement in its way. Not saying you don't do this, but just pointing out the risks of overriding a built-in's prototype, and is the reason most libs avoid doing it. For example underscore could have implemented it's lib via the prototype, let's say it's a good job they didn't. — Keith, Nov 15 '21 at 08:56

Majed Badawi · Answer 1 · 2021-11-14T10:10:51.917

3

Using Number#valueOf:

The valueOf() method returns the wrapped primitive value of a Number object.

Number.prototype.inList = function (list) {
  return list.includes(this.valueOf());
}

let value = 0;
console.log(value.inList([0,1]));

edited Nov 14 '21 at 10:10

answered Nov 14 '21 at 10:02

Majed Badawi

27,616
4
25
48

No, don't do that. Prevent it from being wrapped in the first place. – Bergi Nov 14 '21 at 10:37
@Bergi can you explain why it's wrong to do this? – Majed Badawi Nov 14 '21 at 10:55
1

It's inefficient and only a workaround instead of solving the root issue. Also for the edge case `const n = new Number(); n.inList([n])` I would expect `true`. – Bergi Nov 14 '21 at 11:03

trincot · Accepted Answer · 2021-11-14T10:29:57.803

This happens in non strict mode, where compatibility requires that this is always an object, not a primitive. So in your case this is an instance of the Number constructor, and of course that object does not occur in your list.

In strict mode your code works, as this will be the primitive value (number)

Number.prototype.inList = function (list) {
  "use strict";
  return list.includes(this);
}

let value = 0;

let find1 = value.inList([0,1]);
let find2 = [0,1].includes(value);

console.log("find1", find1); // true
console.log("find2", find2); // true

Here the strict mode is only effective on the function -- and this is enough for this issue. If your script has no "legacy dependencies", you should better set it globally.

JavaScript: Get value inside Number prototype function

2 Answers2