Javascript References Array

Question

I have a big multidimensional array that holds references to objects and I need to be able to move a reference from one spot in the array to another and then delete the original reference (or set it to undefined). The problem with this is it then deletes the recent reference.

var data = [[new Obj, new Obj], [new Obj, new Obj]];

function move(fromx, fromy, tox, toy) {
   data[tox][toy] = data[fromx][fromy];
   delete data[fromx][fromy];
}

Edit: I mean both are gone. data[tox][toy] === undefined; Both references are destroyed, not just data[fromx][fromy]

Randy the Dev · Accepted Answer · 2010-09-25 13:51:48Z

1

Yeah, that's just the delete operator doing what it is supposed to do, which is delete the object which is referred to at [fromx,fromy]. Try simply setting data[fromx][fromy] to null or an undefined variable such as allYourBaseBelongToUs (At least I'd hope it's undefined) or if you're boring undefined is also undefined.

Sources:

edited Sep 25, 2010 at 13:51

answered Sep 25, 2010 at 13:17

Randy the Dev

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2 Comments

T.J. Crowder Over a year ago

Or, you know...undefined. ;-)

Randy the Dev Over a year ago

I find using pop/code culture references as undefined variables is much more amusing though.

letronje · Accepted Answer · 2010-09-25 13:19:50Z

0

var x = [[{'a':'1'}, {'b':'1'}], [{'c':'1'}, {'d':'1'}]]
x[1][0] = x[0][0]
x[1][1] = x[0][1]
delete x[0][0]
delete x[0][1]

console.log(x)

prints [[undefined, undefined], [Object { a="1"}, Object { b="1"}]]

What's your expected output ?

answered Sep 25, 2010 at 13:19

letronje

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Comments

T.J. Crowder · Accepted Answer · 2010-09-25 13:28:32Z

delete doesn't delete the object, it deletes the property that was referencing the object. In your move function, since you've just assigned that object to another property, you still have the object. (Whatever was previously in that slot in the array is toast, though.)

Your example above mostly works, here's an example (using a custom object so we can easily print out the contents of the arrays):

function T(id) {
  this.id = id;
}
T.prototype.toString = function() {
  return this.id;
};

var data = [[new T("0x0"), new T("0x1")],
            [new T("1x0"), new T("1x1")]];

display("0: " + data[0].join(","));
display("1: " + data[1].join(","));

display("Moving 0x0 to 1x2");
move(0, 0, 1, 2);

display("0: " + data[0].join(","));
display("1: " + data[1].join(","));

function move(fromx, fromy, tox, toy) {
   data[tox][toy] = data[fromx][fromy];
   delete data[fromx][fromy];
}

function display(msg) {
  var p = document.createElement('p');
  p.innerHTML = msg;
  document.body.appendChild(p);
}

Live copy

I wouldn't recommend delete in this scenario, though, since it's very unlikely it's what you actually want. If you want to keep the "cell" but remove its contents, just assign undefined to it. delete actually removes it.

Somewhat off-topic, but: JavaScript doesn't have multi-dimensional arrays. What you have there is an array of arrays. It's a distinction that does actually have a difference, in that the arrays held by the outermost array can be of different lengths. (Also, JavaScript arrays are sparse — they're not really arrays at all, barring an implementation choosing to make them so — so the arrays could have the same length but gaps.)

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