javascript, jQuery: how to save values instead of the references

Question

hi,

I'm using the following lines to store the location of an object.

var lightboxTop = $('#lightbox').css('top');
var lightboxLeft = $('#lightbox').css('left');

I'm successively moving this object in my element, and I want to restore it previous position with the stored variables.

But, I'm afraid javascript is saving the values by reference so I lose the initial positions. Am I correct ? How can I solve this ?

thanks

Answer 1

In this case it's not storing any reference, but the actual value.

Answer 2

No, those values returned are not stored by reference. If you change the top and left style of the element, it will not affect your stored values.

Primitive types in javascript are not passed by reference.

    var a = "a";

    var b = a;

    a = "c";

    alert(b);  // alerts "a"
    alert(a); // alerts "c"

or

    var a = 1;

    var b = a;

    a = 3;

    alert(b);  // alerts "1"
    alert(a); // alerts "3"

Objects are passed by reference:

    var a = {one:"one"};

    var b = a;

    a.one = "three";

    alert(b.one);  // alerts "three"
    alert(a.one); // alerts "three"

Answer 3

Javascript has no support for 'by reference'.

var a = 1; // a is 1
var b = a; // b is set to the value of a, that is, 1
a = 2; // a is set to 2, b is still 1

The only way to pass 'references' is to share the object that the variable is a property of

var props = {};
props.a = 1;
var newprops = props; // props === newprops = true, both variables point to the same reference
newprops.a // is 1
props.a = 3;
newprops.a // is 3

And what happens if we replace one of the variables pointing to the reference to the object?

props = {}; // props === newprops = true, props is set to a NEW object, newprops still points to the old one
props.a = 2; // is 2
newprops.a; // is still 3

Answer 4

You could try jQuery.clone() method, like so:

var l = $('#lightbox');
var start = l.clone().hide();

then move l around, remove it afterwards and redisplay start.