Javascript array declaration: new Array(), new Array(3), ['a', 'b', 'c'] create arrays that behave differently

The Solution to Javascript array declaration: new Array(), new Array(3), ['a', 'b', 'c'] create arrays that behave differently is


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.

~ Answered on 2011-11-23 17:21:22


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