jQuery's return value is Array, but constructor is Object. how?

Question

for example,

$(document) // [ #document ] : document object in array
typeof $(document) // "object"
$(document).constructor // function Object() { [native code] } or function (a,b) { return some function; }

If value is array, it have to be Array constructor.

It's not an array like object. because array like object just has array property, not like [].

How can it be?

add: If you can, show simple example code, please. like

a = ... 
console.log(a) // [ ... ]
console.log(a.constructor) // function Object or something

I don't understand your reasoning for why the return from $(document) is "not an array like object" - if you look at the actual jQuery source you'll find it is an array-like object and not an array... — nnnnnn
– nnnnnn, Commented Jul 9, 2012 at 6:07
"Shown" where? The return from $(document) is an instance of jQuery.fn.init which has had 0 and length properties added. If it was an Array it would have array methods like .join() and it doesn't. Look at the source code. — nnnnnn
– nnnnnn, Commented Jul 9, 2012 at 6:20
Thanks your answer. Actually I confused about concept of return instance. — Ephemera
– Ephemera, Commented Jul 9, 2012 at 6:30

Robert · Accepted Answer · 2012-07-09 06:17:34Z

2

Take a look at the jQuery source. $(document) creates a cloned jQuery object of the document element and then makes and returns an array-like object.

the jQuery factory function $() returns a jQuery object that has many of the properties of an array (a length, the [] array access operator, etc.), but is not exactly the same as an array and lacks some of an array's built-in methods (such as .pop() and .reverse()).

edited Jul 9, 2012 at 6:17

answered Jul 9, 2012 at 6:07

Robert

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4 Comments

user235273 Over a year ago

Now I am confused. If it is returning an array of objects shouldn't its constructor give function Array() {//..}?

nnnnnn Over a year ago

It doesn't make or return an array, it returns an instance of jQuery.fn.init after adding 0 and length properties. That is, it returns an array-like object.

user235273 Over a year ago

Right, got it! The output is an object that have some properties of an array, but it is not an array.

nnnnnn Over a year ago

Note: in the case of $(document) (and indeed most other cases) the $() function returns before it gets to the last line that returns the result of jQuery.makeArray(). The array access operator [] isn't really "array access" because it applies to all JS objects.

elclanrs · Accepted Answer · 2012-07-09 05:57:22Z

0

An array, as in [], is also an object in JavaScript. In JS mostly everything is an object, such as functions, arrays, literal objects, regular expressions... To really know what you're dealing with is an array you can do this:

Object.prototype.toString.apply(myarray) === '[object Array]'

answered Jul 9, 2012 at 5:57

elclanrs

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