JavaScript: how do I improve this repeat function using tail call optimization

Asked Oct 08 '21 at 19:31

Active Oct 08 '21 at 19:31

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Here is a repeat function that recursively calls itself

  const repeat = x => f => {
    if (x > 0) {
      f()
      return repeat (x - 1) (f)
    }
  }

The problem with it is it can cause stack overflow. I was wondering if I can rewrite it using tail call recursion so it can be stack-safe.

asked Oct 08 '21 at 19:31

Joji

JavaScript doesn’t have proper tail calls in its spec yet and the only major engine that does tail call elimination is Safari’s (IIRC), so no, you can’t. Have to use a loop. – Ry- Oct 08 '21 at 19:58
2

To elaborate this a bit, iirc, tail calls are [in the spec since ES6](https://tc39.es/ecma262/#sec-tail-position-calls), but there are [issues](https://stackoverflow.com/questions/42788139/es6-tail-recursion-optimisation-stack-overflow), which led to [barely any major browser implementing them](https://kangax.github.io/compat-table/es6/). – ASDFGerte Oct 08 '21 at 20:20
Your function already does use tail recursion. – Bergi Oct 08 '21 at 23:40

0 Answers0