Javascript: hiding prototype methods in for loop?

Question

So lets say I've added some prototype methods to the Array class:

Array.prototype.containsKey = function(obj) {
    for(var key in this)
        if (key == obj) return true;
    return false;
}

Array.prototype.containsValue = function(obj) {
    for(var key in this)
        if (this[key] == obj) return true;
    return false;
}

then I create an associative array and attempt to loop through it's keys:

var arr = new Array();
arr['One'] = 1;
arr['Two'] = 2;
arr['Three'] = 3;

for(var key in arr)
   alert(key);

this returns five items:

  -One
  -Two
  -Three
  -containsKey
  -containsValue

but I want (expect?) only three. Am I approaching this wrong? is there a way to "hide" the prototype methods? or should I be doing something differently?

Answer 1

You can achieve desired outcome from the other end by making the prototype methods not enumerable:

Object.defineProperty(Array.prototype, "containsKey", {
  enumerable: false,
  value: function(obj) {
      for(var key in this)
        if (key == obj) return true;
      return false;
    }
});

This usually works better if you have control over method definitions, and in particular if you have no control over how your code will be called by other people, which is a common assumption in library code development.

Answer 2

You can use JavaScript's hasOwnProperty method to achieve this in the loop, like this:

for(var key in arr) {
    if (arr.hasOwnProperty(key)) {
        ...
    }
}

Reference: This YUI blog article.

Answer 3

Javascript doesn't support associative arrays the way you think they do. http://ajaxian.com/archives/javascript-associative-arrays-considered-harmful

for (var i in .. gets all of the properties of an object (an array is just another object) which is why you're seeing the other objects you've prototyped to it.

As the article suggests you should use an object:

var assoc = {'One' : 1, 'Two' : 2};
assoc['Three'] = 3;

for(var key in assoc)
   alert(key+' => '+assoc[key]);