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I thought to swap the elements of a tuple in place using destructuring assignment as follows:

var a = [1,2];
[a[1], a[0]] = a;

However, this yields [1, 1].

Babel piles this as

a[1] = a[0];
a[0] = a[1];

I would have thought this should be piled as

let tmp0 = a[0];
let tmp1 = a[1];
a[0] = tmp1;
a[1] = tmp0;

Traceur behaves identically to babel. So I guess this is the specified behavior?

I want to swap the two elements in place. So is the only way...

let tmp = a[0];
a[0] = a[1];
a[1] = tmp;

But I thought the above was what destructuring assignment was supposed to let me avoid having to do.

I'm perfectly capable of reversing the order of the two elements of the array, so that is not my question. I could do something as simple as a.push(a.shift()), which meets the criteria of the swapping being in-place.

I'm most interested here in why destructuring doesn't work the way it seems that it ought to.

I thought to swap the elements of a tuple in place using destructuring assignment as follows:

var a = [1,2];
[a[1], a[0]] = a;

However, this yields [1, 1].

Babel piles this as

a[1] = a[0];
a[0] = a[1];

I would have thought this should be piled as

let tmp0 = a[0];
let tmp1 = a[1];
a[0] = tmp1;
a[1] = tmp0;

Traceur behaves identically to babel. So I guess this is the specified behavior?

I want to swap the two elements in place. So is the only way...

let tmp = a[0];
a[0] = a[1];
a[1] = tmp;

But I thought the above was what destructuring assignment was supposed to let me avoid having to do.

I'm perfectly capable of reversing the order of the two elements of the array, so that is not my question. I could do something as simple as a.push(a.shift()), which meets the criteria of the swapping being in-place.

I'm most interested here in why destructuring doesn't work the way it seems that it ought to.

• javascript
• swap
• ecmascript-6
• destructuring

Share Improve this question Follow edited Aug 7, 2015 at 18:31 asked Aug 7, 2015 at 17:28 user663031user663031 4

• let a = 'A'; let b = 'B'; [b, a] = [a, b]; console.log('A:', a, ', B:', b); – Misha Tavkhelidze Commented Apr 16, 2017 at 6:42
• I know how to swap two values. The question is how to swap the first and second element of an array. – user663031 Commented Apr 16, 2017 at 7:16
• const a = [ 1, 2 ]; const b = a.reverse(); console.log(b);. Than modifies a unfortunately. – Misha Tavkhelidze Commented Apr 16, 2017 at 10:35
• I'm not sure why it's so hard to understand that my entire question is how to do this with array destructuring assignment. – user663031 Commented Apr 16, 2017 at 12:29

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5 Answers 5

Sorted by: Reset to default 8

I'd just use a function to do this -

const swap = ([a,b]) => [b,a];

console.log(swap([1,2])); // [2,1]

Of you can swap in place -

const a = [1,2];

[a[1], a[0]] = [a[0], a[1]];

console.log(a); // [2,1]

I would have thought this should be piled as

let tmp0 = a[0];
let tmp1 = a[1];
a[0] = tmp1;
a[1] = tmp0;

No, the values don't get retrieved before they are assigned. Or neither is a copy of a made from which the values are retrieved. There only is a single reference to the array; your code rather desugars to

{
    const tmp = a;
    a[1] = tmp[0];
    a[0] = tmp[1]; // here, tmp[1] has the "wrong" value because a===tmp
}

or in fact, it desugars to an iterator that iterates the array while the left-hand-side references in the […] are assigned to.

How about this:

var a = [1,2];
a = a.reverse();

As an alternative, you can use slice:

var a = [1,2];
[a[1], a[0]] = a.slice();
a; // [2,1]

This works in Firefox which supports destructuring:

(function () {
  "use strict";

  var a = [0, 1];
  [a[1], a[0]] = [a[0], a[1]];
  console.log(a);//[1, 0]
}());

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