Why my JavaScript function won't end?

Question

So I have the following code, from which I expect the x function to return null after being called 3 times but it keeps returning the the same function:

const repeat = (n, tailFn) => {
  for (let i = 0; i < n; i++) {
    tailFn = () => tailFn;
  }
  return tailFn;
};

const x = repeat(2, x => null);

console.log(x());           // function tailFn() { return _tailFn }
console.log(x()());         // function tailFn() { return _tailFn }
console.log(x()()()()()()); // function tailFn() { return _tailFn }

What am I doing wrong? See it on CodePen.

Michał Perłakowski · Accepted Answer · 2016-12-22 16:49:13Z

6

Your function just assigns () => tailFn to tailFn three times and then returns it. Instead, you should return a function which returns repeat(n - 1, tailFn) if n is not 0, and tailFn otherwise.

const repeat = (n, tailFn) => {
  return n !== 0 ? () => repeat(n - 1, tailFn) : tailFn;
};

const x = repeat(2, x => null);

console.log(x());           // () => repeat(n - 1, tailFn)
console.log(x()());         // x => null
console.log(x()()());       // null

answered Dec 22, 2016 at 16:49

Michał Perłakowski

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Comments

Pablo Lozano · Accepted Answer · 2016-12-22 16:51:50Z

4

You have created a function that ALWAYS returns itself,

tailFn=()=>tailFn;

actually the loop is meaningless.Its behavior is similar to a recursive function without a base case.

answered Dec 22, 2016 at 16:51

Pablo Lozano

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2 Comments

Sam R. Over a year ago

Am I not creating a new function tailFn that is returning the old tailFn?

Sam R. Over a year ago

I think I got it. I just need to keep tailFn in another variable and return that from new tailFn;

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