CODEWITHSUNDEEP
javascriptarraysforeachprototype
Addis
Addis

Reputation: 2530

Extending Array.prototype crashes

I was trying to solve the following problem which I got on a blog but the program crashes. What could be the reason? and is there any means of solving it? I have read warnings not to extend builtin objects, if that's the case, what could be the reason associated with this specific example.

const a = [1, 2, 3, 4, 5];

//this is what I tried
Array.prototype.multiply = function() {
  for (m of this) this.push(m * m);
}

a.multiply(); //this should not be changed
console.log(a); // [1, 2, 3, 4, 5, 1, 4, 9, 16, 25] (expected output)

Upvotes: 2

Views: 82

Answers (2)

Code Maniac
Code Maniac

Reputation: 37755

When you push the value to same array during loop, you end up in infinite loop, create a temp array push value to it, in the end add it to this

const a = [1, 2, 3, 4, 5];

//this is the what I tried
Array.prototype.multiply = function() {
  let newArr = []
  for (const m of this) {
    newArr.push(m * m)
  }
  this.push(...newArr)
}

a.multiply();
console.log(a);

That being said you should not override the prototype simply use a function and pass the parameters

const a = [1, 2, 3, 4, 5];

function multiply(arr) {
  return [...arr, ...arr.map(a => a * a)]
}

console.log(multiply(a));

Upvotes: 3

Nick
Nick

Reputation: 147206

Pushing a new value into an array in the middle of a for ... of loop creates an infinite loop as the loop includes the new values. You can use forEach instead as that ignores the new values added:

const a = [1, 2, 3, 4, 5];

Array.prototype.multiply = function() {
    this.forEach(v => this.push(v*v));
}

a.multiply(); //this should not be changed
console.log(a); // [1, 2, 3, 4, 5, 1, 4, 9, 16, 25] (expected output)

Upvotes: 1

Related Questions