Why use both the spread operator and destructuring assignment in a function parameter?

Question

I keep coming across this syntax and I'm having trouble understanding exactly what it's doing:

export class SomeClass extends SomeParent {
    constructor(...[configuration]) {
        // lines of code that reference only "configuration"
    }
}

Playing around in the Node REPL I can't find a difference between:

function foo(...[bar]) { console.log(bar); console.log(arguments) }

...and...

function foo(bar) { console.log(bar); console.log(arguments) }

...so what is it for?

…and it's not even spread syntax – Bergi Aug 16 '23 at 02:36 — Bergi, Aug 16 '23 at 02:36

score 0 · Answer 1 · answered Aug 16 '23 at 02:35

It seems pretty pointless indeed. You'll need to ask the author of the code what they intended with this, they should at least have left a comment.

However, there actually is a minuscule difference: rest parameters do not count for the function's arity. So (function(bar){}).length is 1, whereas (function(...[bar]){}).length is 0.

Why use both the spread operator and destructuring assignment in a function parameter?

1 Answers1