How can I ignore certain returned values from array destructuring?

Question

Can I avoid declaring a useless variable when array destructuring when I am only interested in array values beyond index 0?

In the following, I want to avoid declaring a, I am only interested in index 1 and beyond.

// How can I avoid declaring "a"?
const [a, b, ...rest] = [1, 2, 3, 4, 5];

console.log(a, b, rest);

Dichromatic

Related: Destructuring array get second value?, Object destructuring solution for long arrays? – Elfreda

Does this answer your question? Destructuring array get second value? – Liddy

Answer 1

Can I avoid declaring a useless variable when array destructuring when I am only interested in array values beyond index 0?

Yes, if you leave the first index of your assignment empty, nothing will be assigned. This behavior is explained here.

// The first value in array will not be assigned
const [, b, ...rest] = [1, 2, 3, 4, 5];

console.log(b, rest);

You can use as many commas as you like wherever you like, except after a rest element:

const [, , three] = [1, 2, 3, 4, 5];
console.log(three);

const [, two, , four] = [1, 2, 3, 4, 5];
console.log(two, four);

The following produces an error:

const [, ...rest,] = [1, 2, 3, 4, 5];
console.log(rest);

Dichromatic

Is it possible to ignore all items at the front and keep only a fixed bunch at the end? Something like const [..., fourth_to_last, , penultimate, last] = my_list; – Acorn

@BallpointBen, that's an interesting question. Unfortunately, rest elements must be the last element. A simple way to get the last items instead would be to reverse the array and vars: const [last, penultimate, , fourth_to_last] = my_list.reverse(); – Dichromatic

Answer 2

Ignoring some returned values

You can use ',' ignore return values that you're not interested in:

const [, b, ...rest] = [1, 2, 3, 4, 5];

console.log(b);
console.log(rest);

Caseworm

This doesn't really add much that isn't covered by the existing answer. – Gill