swap() function for variables's values

Question

I am not able to achieve the desired result for this swap function below, where I want the values printed as 3,2

function swap(x,y){
 var t = x;
 x = y;
 y = t;
}

console.log(swap(2,3));

Any clue will be appreciated !

It does swap your variables, but that's about it, it doesn't return anything, and the variables are local to the function ? — adeneo, Oct 20 '16 at 21:56

score 8 · Answer 1 · edited Oct 21 '16 at 00:47

8

Your function is swapping the values internally, but the function does not return a value.

The following method returns an array of values in reverse order of what was supplied.

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

console.log(swap(2, 3));

However, you could easily do something like the following snippet, because - based on your supplied code - there seems to be no need to actually swap the values of the arguments.

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

console.log(swap(2, 3));

edited Oct 21 '16 at 00:47

answered Oct 20 '16 at 22:04

Sathvik Chinnu

565
1
7
23

Mind explaining why this works? – Andrew Li Oct 21 '16 at 00:37
First snippet or second snippet? @AndrewLi – Sathvik Chinnu Oct 21 '16 at 14:31

score 1 · Answer 2 · answered Oct 20 '16 at 22:10

1

If you don't actually need the values to swap:

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

If you do need the values to swap, but you don't want to declare another variable:

function swap (x, y)
{
  x = x + y;
  y = x - y;
  x = x - y;
  return [x, y];
}

answered Oct 20 '16 at 22:10

Dominic Aquilina

617
3
13

1

The second algorithm will not swap strings. Besides, it comes from an era when RAM memory was an issue. It shouldn't be used for modern JS apps. – Alex Sicoe Feb 21 '19 at 10:39
1

y = [x, x = y][0]; one line solution – Vi8L Sep 03 '19 at 10:34

swap() function for variables's values

2 Answers2

Linked