Why wrapping into a function is required in .change() of jQuery?

Why this works in jQuery :

$('#selCars').change(function(){
     alert( "I have changed!" );
}) 

but not this one :

$('#selCars').change(alert( "I have chan开发者_JAVA百科ged!" ) ); 

You pass a function reference to .change(). Your second example just has code there, not a function reference.

Your first example works because it passes a function reference which IS what is required.

A function reference is required because this is a callback that will be called at a later time. The .change() function which executes immediately needs to save the callback reference into it's own variable and then call it later when the change event actually occurs. To do that, it needs a function to call at that later time, not a raw piece of code.

And, the other answer is because, .change() was written to require a function reference. That's how the developers that spec'ed and wrote it designed it. If you want it to work, you have to follow their rules.

Because it's a callback, i.e. you're passing something that be called back later, so what you've to pass is a reference to a function, and that reference will be stored and called when the event will fire.

The change method doesn't store some code, it stores only a pointer to the function. Your function is called an event handler.

It's because .change() attaches an event handler to an element. The handler won't be called until the event has occurred.

Since in JavaScript, functions are just another datatype, you could also do this:

var handler = function(event) {
    alert("I have changed!");
}
$('#selCars').change(handler);

Note that handler is a function, Whereas alert() would just return undefined.