Javascript swap array elements

Question

Is there any simpler way to swap two elements in an array?

var a = list[x], b = list[y];
list[y] = a;
list[x] = b;

Answer 1

This seems ok....

var b = list[y];
list[y] = list[x];
list[x] = b;

Howerver using

var b = list[y];

means a b variable is going to be to be present for the rest of the scope. This can potentially lead to a memory leak. Unlikely, but still better to avoid.

Maybe a good idea to put this into Array.prototype.swap

Array.prototype.swap = function (x,y) {
  var b = this[x];
  this[x] = this[y];
  this[y] = b;
  return this;
}

which can be called like:

list.swap( x, y )

This is a clean approach to both avoiding memory leaks and DRY.

Answer 2

Well, you don't need to buffer both values - only one:

var tmp = list[x];
list[x] = list[y];
list[y] = tmp;

Answer 3

There is one interesting way of swapping:

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

(ES6 way)

Answer 4

Swap the first and last element in an array without temporary variable or ES6 swap method [a, b] = [b, a]

[a.pop(), ...a.slice(1), a.shift()]

Answer 5

Array.prototype.swap = function(a, b) {
  var temp = this[a];
  this[a] = this[b];
  this[b] = temp;
};

Usage:

var myArray = [0,1,2,3,4...];
myArray.swap(4,1);

Answer 6

You can swap any number of objects or literals, even of different types, using a simple identity function like this:

var swap = function (x){return x};
b = swap(a, a=b);
c = swap(a, a=b, b=c);

For your problem:

var swap = function (x){return x};
list[y]  = swap(list[x], list[x]=list[y]);

This works in JavaScript because it accepts additional arguments even if they are not declared or used. The assignments a=b etc, happen after a is passed into the function.