closure example using setTimeout not recognizing function call

Question

I am trying to create a small application that closes over the delay function and defines a new function using a callback and a wait time. I then wanted to use this newly created inner setTimeout function to take a single parameter that would be run on the callback after the wait time.

function addTwo(num) {
  return num + 2
}

const delay = (callback, wait) => {
  return setTimeout((value) => callback(value), wait)
}
var delayAddByTwo = delay(addTwo, 100)
console.log(delayAddByTwo(6))
// expected result 8 
// actual result --- Uncaught TypeError: delayAddByTwo is not a function

As far as I can tell after the delayAddByTwo = delay(addTwo, 100) the only parameter left to create is the value passed to callback inside the 'inner' setTimeOut function. What am I missing about closure in this example?

setTimeout will return its id

Nick Parsons
– Nick Parsons

2020-01-30 02:54:56 +00:00
Commented Jan 30, 2020 at 2:54 — Nick Parsons
– Nick Parsons, Commented Jan 30, 2020 at 2:54

Mohamed Farouk · Accepted Answer · 2020-01-30 03:06:01Z

2

You need to replace var delayAddByTwo = delay(addTwo, 100) by var delayAddByTwo = (num) => { delay(() => {addTwo(num)}, 100);}

function addTwo(num) {
  console.log(num + 2)
  return num + 2
}

const delay = (callback, wait) => {
  setTimeout(callback, wait)
}
var delayAddByTwo = (num) => {
  delay(() => {addTwo(num)}, 100);
}
console.log(delayAddByTwo)
delayAddByTwo(6)
// expected result 8 
// actual result --- Uncaught TypeError: delayAddByTwo is not a function

edited Jan 30, 2020 at 3:06

answered Jan 30, 2020 at 2:54

Mohamed Farouk

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1 Comment

Nick Parsons Over a year ago

You can use setTimeout(callback, wait) instead of setTimeout((value) => callback(value), wait) - value isn't being used

Nick Parsons · Accepted Answer · 2020-01-30 03:22:11Z

0

You need to make delay return another function which accepts n. This n represents the number which will be passed into addTwo (ie the callback). setTimeout by default will return its timeoutID, so, instead, you can return a Promise, which resolves to the result of calling addTwo with n. In order to get your result from the Promise, you can await it within an async function.

See example below:

const addTwo = num => num + 2;
const delay = (callback, wait) => n => new Promise(res => setTimeout(res, wait, callback(n)));

(async () => {
  const delayAddByTwo = delay(addTwo, 1000);
  const res = await delayAddByTwo(6);
  console.log(res);
})();

Above I have used the third argument of setTimeout, which will be used as the first argument of the callback and passed to res.

edited Jan 30, 2020 at 3:22

answered Jan 30, 2020 at 3:16

Nick Parsons

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Comments

ReallyBoringPerson · Accepted Answer · 2020-01-30 02:55:10Z

0

Change

var delayAddByTwo = delay(addTwo, 100)

To

const delayAddByTwo = delay => (addTwo, 100)

answered Jan 30, 2020 at 2:55

ReallyBoringPerson

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