If have two javaScript functions in a page which are required to be called when document load is plete. Is is possible that any function can executed first or it will be the first function which is executed first always?

So if using jQuery if you have following code:

$(document).ready(function(){ function1(); });

$(document).ready(function(){ function2(); });

Can it happen that function2 is executed first or will function1 always be executed first ?

If have two javaScript functions in a page which are required to be called when document load is plete. Is is possible that any function can executed first or it will be the first function which is executed first always?

So if using jQuery if you have following code:

$(document).ready(function(){ function1(); });

$(document).ready(function(){ function2(); });

Can it happen that function2 is executed first or will function1 always be executed first ?

• javascript
• jquery
• onreadystatechange

• 1 Your script will be read and executed line by line – Mr. Alien Commented Oct 9, 2012 at 9:50
• 2 This got nothing to do with pure JavaScript, it's 100% dependent on how jQuery implement the $(document).ready() method. – Shadow Wizzard Commented Oct 9, 2012 at 9:52
• 1 @Mr.Alien - That's not the question. He basically wants to know the rules of event queuing (if any). – Álvaro González Commented Oct 9, 2012 at 9:53

3 Answers 3

jQuery ready uses the Deferred object system :

ready: function( fn ) {
    // Add the callback
    jQuery.ready.promise().done( fn );

    return this;
},

(from the source code)

And the documentation states that

Callbacks are executed in the order they were added

So yes, your callbacks will be executed in order of addition.

If you want these functions to be executed in order, why don't you just write:

$(document).ready(function(){ 
    function1();
    function2();
});

Better make use of callback function to be sure about the execution order of the functions