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?

Answer 1

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

Answer 2

Change

var delayAddByTwo = delay(addTwo, 100)

To

const delayAddByTwo = delay => (addTwo, 100)

Answer 3

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 .