Hey Leonardo, I think that if you know which elements you need it doesn't matter if the array will change size in the future. Destructuring will ignore any element which you do not use in the assignment e.g let array = [ 'cat' , 'dog' , 'bird' ] let [cat, dog] = array console .log(cat) // 'cat' console .log(dog) // 'dog' 'bird' is being ignored here. So even if the array changes later it doesn't really matter - you got elements you wanted to assign to variables. If you wanna use the rest of the array but you are not sure how many elements will be there you can use the rest pattern like this: let array = [ 'cat' , 'dog' , 'bird' ] let [cat, ...remainingAnimals] = array console .log(cat) // 'cat' console .log(remainingAnimals) // [ 'dog', 'bird'] As you can see the first element of the array will be assigned to the cat variable and the rest will be assigned as an array to remainingAnimals variable. Also, if for some reason there won't be any element to assign like here: let array = [ 'cat' , 'dog' ] let [cat, dog, bird] = array // we try to assign 3 variables from 2 element array console .log(cat) // 'cat' console .log(dog) // 'dog' console .log(bird) // undefined The variable that can't get a matching element will just get an undefined value. Hope this helps, let me know if you have any other questions.

