I'm a bit confused in how functions operate in javascript. I understand that they're all objects but how does that change how I would use them as arguments?
For instance, if I'm trying to use a callback function where the 2nd argument is evaluated after 1000ms...
$(this).fadeIn(1000,function(){alert('done fading in');});
Why can't I achieve the same effect with:
$(this).fadeIn(1000,alert('done fading in'));
If I do, it evaluates both at the same time. That is, (this) element fades in and the alert pops up at the same time.
When I'm calling alert(arg), aren't I creating a new object which gets passed into fadeIn()?
How exactly does this work?
In this
$(this).fadeIn(1000,alert('done fading in'));
what does fadeIn() see as its second argument? It's the result of calling
alert('done fading in')
we are making the call to alert() before calling fadeIn().
In this case
$(this).fadeIn(1000,function(){alert('done fading in');});
we have an object
function(){alert('done fading in');}
which fadeIn() calls at the right time.
When you write:
$(this).fadeIn(1000,alert('done fading in'));
you call function called alert immadiately by putting function name and parentheses after this name. And to fadeIn the result of this call is passed - this is undefined, because alert returns always undefined.
When you write
$(this).fadeIn(1000,function(){alert('done fading in');});
you create a function object that and pass this function object to fadeIn. So after fadeIn is done it can call this function.
It is the same as:
// create function
var callback = function () { alert('done fading in'); };
// and pass this function to fadeIn
$(this).fadeIn(1000, callback);
but when you write:
var callback = alert('done fading in');
$(this).fadeIn(1000, callback);
then you will call alert immadiately and pass to fadeIn value which alert returns - undefined.
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