FAQ: Arrays - Arrays and Functions


This community-built FAQ covers the “Arrays and Functions” exercise from the lesson “Arrays”.

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This exercise can be found in the following Codecademy content:

Web Development

Introduction To JavaScript

FAQs on the exercise Arrays and Functions

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split this topic #2

2 posts were split to a new topic: Where did arr and newArr come from and what do they do or mean?

split this topic #4

2 posts were split to a new topic: Why can .pop() method change the structure of array that was defined as const variable?


Hello everyone. I’m at Arrays and Function, exercise 10. I just want to confirm that what is happening in this exercise is that we are changing the value of the global scope by changing the string ‘mutated’ to ‘MUTATED’ and then .pop it off. Considering where the braces {} are I just want to confirm.

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Hey everyone, i don’t understand why it is used the ‘=>’ operator before the body of the funcion (to be an operator shouldnt ‘=’ be after the ‘>’ signal?).

const removeElement = newArr => {
newArr.pop() _______________ ^


That is arrow function syntax.

const foo = param => {
    // function body

Much the same as,

const foo = function (param) {


with a few exceptions.

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We are changing the elements within the array in global scope from within the function’s local/block scope. Even though the scopes are different, the function can do this because of pass-by-reference = if an array is passed into a function using an argument/parameter, any changes to the array’s elements within the function will also be made to the array outside the function. I also rationalise this by remembering that a child’s scope has access to its parent’s and ancestors’ scopes (but not vice versa) i.e. the scope hierarchy.

The first change the function makes is to reassign the value of the fourth element (index position 3) to 'MUTATED'.
The second change is to remove the last element using .pop()pop it off as you say - love it!! :joy::rofl: … which happens to be our friend 'MUTATED' (our old friend 'mutated') again. :wink:


This is because there are not two arrays, but one. The one seen inside the function is the global one referenced in the argument.

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