48

Consider this example Javascript code:

a = new Array();
a['a1']='foo';
a['a2']='bar';

b = new Array(2);
b['b1']='foo';
b['b2']='bar';

c=['c1','c2','c3'];

console.log(a);
console.log(b);
console.log(c);

Results in the Firebug console are as follows:

For a (the '[]' had to be expanded by clicking on the '+' button):

[]      
a1  "foo"   
a2  "bar"

For b:

[undefined, undefined]

For c:

["c1", "c2", "c3"]

My questions are:

  1. Am I using the array['key']='value' syntax correctly?
  2. Why isn't array b working as expected?
  3. Why are arrays a and c displayed differently in the console? It also seems that jQuery is unable to iterate through the array a with it's .each() method.
  4. Could you reccomend any good tutorials on Javascript array behaviour?

NOTE: Google Chrome's Firebug displays only [] for array 'a', without the option to expand it.

EDIT: Alright, it seems that arrays in Javascript have only numerical keys, so adding a string as a key name makes an object out of an array. But why doesn't jQuery's .each work with it?

$.each(a, function ()
    {
    alert ('derp');
    })

This code, appended to the script, produces no alerts.

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6
  • 1
    You're confusing JavaScript's array objects with associative arrays available in languages like PHP. JavaScript arrays use numeric indices only. Using the syntax a['a1'] = val adds a property a1 to the object a, not an array value. Commented Nov 23, 2011 at 17:24
  • @P.Brian.Mackey: Probably because this sort of question has been asked plenty of times on SO. Commented Nov 23, 2011 at 20:23
  • 1
    @exizt: Firebug and similar tools make assumptions about what you want displayed. They're not necessarily a prefect representation of the language, though as answered below, it is usually best to only rely on numeric properties in JavaScript Arrays. Commented Nov 23, 2011 at 20:25
  • @RightSaidFred - So are you suggesting the question is a repeat, but the answer is new? Commented Nov 23, 2011 at 20:25
  • 1
    What exactly would count as "research" in this case? Could you show me what search queries I should've used to find the right answer? Commented Nov 23, 2011 at 22:15

2 Answers 2

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60

Arrays have numerical indexes. So,

a = new Array();
a['a1']='foo';
a['a2']='bar';

and

b = new Array(2);
b['b1']='foo';
b['b2']='bar';

are not adding elements to the array, but adding .a1 and .a2 properties to the a object (arrays are objects too). As further evidence, if you did this:

a = new Array();
a['a1']='foo';
a['a2']='bar';
console.log(a.length);   // outputs zero because there are no items in the array

Your third option:

c=['c1','c2','c3'];

is assigning the variable c an array with three elements. Those three elements can be accessed as: c[0], c[1] and c[2]. In other words, c[0] === 'c1' and c.length === 3.

Javascript does not use its array functionality for what other languages call associative arrays where you can use any type of key in the array. You can implement most of the functionality of an associative array by just using an object in javascript where each item is just a property like this.

a = {};
a['a1']='foo';
a['a2']='bar';

It is generally a mistake to use an array for this purpose as it just confuses people reading your code and leads to false assumptions about how the code works.

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5 Comments

And regarding the difference between a and b, I think it comes from the declaration b = new Array(2);. Here you're saying "my numerically-indexed array has 2 items". And since you're puting nothing into b[0] and b[1], that explains the 2 undefined you get in Firebug.
I see, thanks! But why doesn't jQuery's .each() work with the 'a' array? (See edit)
It is interesting to note how firebug treats a as an object that contains an array element. But, it treats b as an array with no visible properties (though the properties do exist). Where the only difference is the parameter passed to the Array constructor. I wonder why this is?
@exizt - jQuery's .each() method has to try to figure out if the item you pass it is an object or an array in order to figure out how to iterate it. If it's an array, it uses the array elements in the .each() iteration. In your case, you gave it an array so therefore it treats it as an array, but the array has no elements in the actual array so, there's nothing to iterate. If you change to a = {}; as the initialization, then it would work with jQuery's .each().
@jfriend clarified my own doubt in my answer :P
4

Arrays in JS have two types of properties:

Regular elements and associative properties (which are nothing but objects)

When you define a = new Array(), you are defining an empty array. Note that there are no associative objects yet

When you define b = new Array(2), you are defining an array with two undefined locations.

In both your examples of 'a' and 'b', you are adding associative properties i.e. objects to these arrays.

console.log (a) or console.log(b) prints the array elements i.e. [] and [undefined, undefined] respectively. But since a1/a2 and b1/b2 are associative objects inside their arrays, they can be logged only by console.log(a.a1, a.a2) kind of syntax

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