Why is array destructuring not working in this case?

Question

I am trying to swap members of arr. But I am getting [0, 0] as output. I am expecting [1, 0].

const arr = [0, 1];
[arr[1],arr[0]] = arr;
console.log(arr)

But while doing this, it outputs [1, 0] which I think I understand.

const arr = [0, 1];
[arr[1],arr[0]] = [arr[0], arr[1]];
console.log(arr)

The destructuring assignment is modifying the array. `arr[1]` becomes `0`, and then so does `arr[0]` — evolutionxbox, Apr 20 '22 at 15:31
What are you trying to do? The code you posted is weird and doesn't make a lot of sense. — Pointy, Apr 20 '22 at 15:33
@Utshav - I had the array indexes backward in my answer, which may have made it hard to understand. I've fixed them. — T.J. Crowder, Apr 21 '22 at 06:34

T.J. Crowder · Answer 1 · 2022-04-21T06:33:43.243

Your first example is both modifying the array arr points to and getting its values from that same array, as though you did this:

const arr = [0, 1];
// Here, `arr` is `[0, 1]`
arr[1] = arr[0];      // <=== Changes `arr[1]`...
// Now, `arr` is `[0, 0]`
arr[0] = arr[1];      // <=== ...so the new value of `arr[1]` is used here.
// Still `[0, 0]`

Your destructuring is doing the same thing: Using arr[1] after it's been changed by the destructuring assignment.

Your second example is modifying the array arr points to, but only after getting the old values from the array into a new array:

const arr = [0, 1];
const newArray = [arr[0], arr[1]];
// Here, `arr` is `[0, 1]`, `newArray` is `[0, 1]`
arr[1] = newArray[0]; // <=== Changes `arr[1]`...
// Here, `arr` is `[0, 0]`, `newArray` is still `[0, 1]`
arr[0] = newArray[1]; // <=== ...but that doesn't affect `newArray[1]`, so this works.
// Here, `arr` is `[1, 0]`, `newArray` is still `[0, 1]`

That works, because even though arr[1] was changed by the destructuring assignment, the new value for arr[0] isn't coming from arr[1], it's coming from newArray[1].

Why is array destructuring not working in this case?

1 Answers1