How to disable pointers usage in JS?

Question

Why is the result {"a":"b","c":1}?

var foo = {"a":"b","c":0};
var bar = foo;
bar.c++;
alert(JSON.stringify(foo));

How to disable this behavior?

Answer 1

Both foo and bar variables reference the same object. It doesn't matter which reference you use to modify that object.

You cannot disable that behaviour, this is how JavaScript and many other major languages work. All you can do is to clone the object explicitly.

var foo = {"a":"b","c":0};
var bar = {"a":foo.a, "c": foo.c};
bar.c++;

Answer 2

First, Javascript doesn't pass pointers, it passes references, slightly different. Secondly, there's no way to modify Javascript's default behavior, unfortunately fortunately.

What you might want to do is create a constructor and use that to create two similar, but separate instances of an object.

function Foo(a, b) {
    this.a = a;
    this.b = b;
}

var bar1 = new Foo(0, 0);
var bar2 = new Foo(0, 0);
bar2.b++;

console.log(bar1);
console.log(bar2);

>> {a:0, b:0};
>> {a:0, b:1};

Answer 3

What you're doing is making a second reference to an object but what it seems you want is a copy of that object instead.

If you want that functionality then you really want a copy function that copies all of the properties, one by one, into a new object:

// Take all the properties of 'obj' and copy them to a new object,
// then return that object
function copy(obj) {
  var a = {};
  for (var x in obj) a[x] = obj[x];
  return a;
}

var foo = {"a":"b","c":0};
var bar = copy(foo);
bar.c++;
alert(JSON.stringify(foo));

and you'll get {"a":"b","c":0}

Answer 4

You can't disable the way javascript works.

If you change a reference object, it effects all the object references...