Type Error on line 4: callback is not a function JavaScript callback

Question

I created a forEach function that takes an array and a callback function, and runs the callback on each element of the array. The forEach function does not return anything. There is also a separate helper function map that is used.

function forEach(array, callback) {
  for (let i = 0; i < array.length; i++) {
    callback(array[i]);
  }
}

function map(array, callback) {
  let newArr = []
  forEach(array, newArr.push(callback));
  return newArr;
}

console.log(typeof forEach); // should log: 'function'
forEach(['a', 'b', 'c'], i => console.log(i)); // should log: 'a', 'b', 'c'
console.log(typeof map); // should log: 'function'
console.log(map([3, 4, 5], n => n - 2)); // should log: [1, 2, 3]

I get the following error: Type Error on line 4: callback is not a function I'm quite close to resolving the issue but can't seem to figure out how to line 9 with the forEach function call. All test cases above pass, except the very last one.

blankart · Accepted Answer · 2020-07-03 12:27:17Z

2

Change your map function to this:

function map(array, callback) {
   let newArr = []
   forEach(array, (value) => newArr.push(callback(value)));
   return newArr;
}

answered Jul 3, 2020 at 12:27

blankart

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2 Comments

Codestudio Over a year ago

Can you explain how the "(value) =>" works? Trying to understand why it was running now and how rewriting it this way finally made it work. Thanks.

blankart Over a year ago

You may refer here to learn more about arrow functions.

Dragos Andrei · Accepted Answer · 2020-07-03 12:26:57Z

1

Problem resides in forEach(array, newArr.push(callback));. Try wrapping the second parameter, newArr.push(callback), in an arrow function expression. Resulting line will be forEach(array, () => newArr.push(callback));.

Here's a working example. Check out the console log.

Good luck!

answered Jul 3, 2020 at 12:26

Dragos Andrei

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2 Comments

Karan Over a year ago

Your code will not work. It will not return desired value as per question.

Dragos Andrei Over a year ago

Taking a second look, you're right. I have now edited it in order to work as intended. Thanks for the notice!

Karan · Accepted Answer · 2020-07-03 12:50:14Z

You just need to write return array.map(callback); and it will work.

As Array.prototype.map() map accepts callback as parameter and you are already having callback function as parameter. All you need to do is return array.map(callback);.

Similarly you can do with Array.prototype.forEach(). But forEach doesn't return anything so you can not write return array.forEach(callback);. Instead of that you need to use it as array.forEach(callback);.

function forEach(array, callback) {
  array.forEach(callback);
}

function map(array, callback) {
  return array.map(callback);
}

console.log(typeof forEach); // should log: 'function'
forEach(['a', 'b', 'c'], i => console.log(i)); // should log: 'a', 'b', 'c'
console.log(typeof map); // should log: 'function'
console.log(map([3, 4, 5], n => n - 2)); // should log: [1, 2, 3]

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