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Home/ Questions/Q 8394941
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Editorial Team
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Editorial Team
Asked: June 9, 20262026-06-09T20:10:45+00:00 2026-06-09T20:10:45+00:00

As an example: $( function(){ //do stuff $( function(){ //do other stuff }); });

  • 0

As an example:

$( function(){
    //do stuff
    $( function(){
        //do other stuff
    });
});

Of course written out in code this seems to make no sense. But a plugin which works with HTML elements may use .ready() while itself being executed on an element in the main script’s .ready(). How does jQuery handle this exactly? It clearly works, but does it do anything special?

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  1. Editorial Team
    Editorial Team
    2026-06-09T20:10:46+00:00Added an answer on June 9, 2026 at 8:10 pm

    If .ready() is called after the DOM has been initialized, the new
    handler passed in will be executed immediately.

    This means that when the first function is executed, the inner one will be executed immediately

    Additionally, jQuery allows you to bind multiple functions to a single event and it will call them all (assuming that no one produces an error)

    You can test the behaviour by executing on already loaded page:

    jQuery(document).ready(function(){
            for(i=0;i<1000000;i++);
            console.log('2');
        });
    console.log('1');
    
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