How to limit function calls in JS?

Question

I need a function limitCalls (fn, maxCalls) that takes a function fn and returns a new function that can be called no more than the number of times specified in maxCalls. Test example:

 it('limitCalls', () => {
const makeIncrement = () => {
  let count = 0;

  return () => {
    count += 1;
    return count;
  };
};

const limitedIncrementA = limitCalls(makeIncrement(), 3);

expect(limitedIncrementA()).toBe(1);
expect(limitedIncrementA()).toBe(2);
expect(limitedIncrementA()).toBe(3);
expect(limitedIncrementA()).toBe(undefined);
expect(limitedIncrementA()).toBe(undefined);

const limitedIncrementB = limitCalls(makeIncrement(), 1);

expect(limitedIncrementB()).toBe(1);
expect(limitedIncrementB()).toBe(undefined);
expect(limitedIncrementB()).toBe(undefined);

});

I have:

var calls = 0;
export default function limitCalls(fn, maxCalls) {
  if (calls >= maxCalls) {
    return undefined;
  }
  calls += 1;
  return fn();
}

And error is limitedIncrementA is not a function. Help me please to realise it.

Guerric P · Accepted Answer · 2021-10-03 15:13:13Z

1

Instead of conditionally returning a function, always return a function that conditionally executes the fn callback:

function limitCalls(fn, maxCalls) {
  let count = 0;
  
  return function(...args) {
    return count++ < maxCalls ? fn(...args) : undefined;
  }
}

const limited = limitCalls(console.log, 3);

limited('one');
limited('two');
limited('three');
limited('four');

edited Oct 3, 2021 at 15:13

answered Oct 3, 2021 at 14:23

Guerric P

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Comments

Lucas Coelho · Accepted Answer · 2021-10-03 13:43:56Z

0

In this snippet, limitedIncrementA isn't indeed a function. See this:

/* You're calling makeIncrement,
   so you're passing its return to 'limitCalls'
 */
const limitedIncrementA = limitCalls(makeIncrement(), 3);

/* Here, considering that makeIncrement exists,
   you're passing a reference to this functions,
   which can be called inside 'limitCalls'
 */
const limitedIncrementB = limitCalls(makeIncrement, 3);

So, supposing that makeIncrement returns 1, 2, 3, ..., your current code is equivalent to:

limitCalls(1, 3);

answered Oct 3, 2021 at 13:43

Lucas Coelho

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2 Comments

ivan239 Over a year ago

I got it. But according to the task I can't change test code. Is there any way to make limitedIncrementA a function inside limitCalls?

Lucas Coelho Over a year ago

You can return what you called 'fn' then, without invoking (without parenthesis) but you can also check this code with a console.log(makeIncrement()) to see if this is actually returning a function

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