simple swap program not working javascript

Question

I've made the following swap function:

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

It is supposed to swap 2 numbers. I have the follwing code:

var x=5;
var y=10;
swap(x,y);

The problem is that when I output the vaues of these variables after swap I still get 5 for x and 10 for y. Any ideas?

The **values** of `x` and `y` are passed into `swap`, not the variables themselves (or a reference to them). — T.J. Crowder, May 15 '16 at 13:44

score 1 · Answer 1 · answered May 15 '16 at 13:44

1

Since the parameters are passed by value you cannot write a function that replaces the following:

var a, b;
var temp = a;
a = b;
b = temp;

You can also use a one-liner:

b = [a, a = b][0];

answered May 15 '16 at 13:44

Idos

Or just: `[a, b] = [b, a]` (as of late 2014 (Safari) to mid March 2016 (Chromium)) – Darryl Noakes Aug 11 '22 at 20:15

1 Answers1