First of all, I think I fully understand the implications of using
Class.create to generate classes in Prototype together with how wrapping
works and the use of the $super argument in methods that want to call the
similarly -named method on the parent class's prototype.
Also, it's clear that the constructor of any class (in Prototype) just
delegates the task of creating instance methods to the initialize prototype
method by passing on any arguments it is getting at instantiation time.
So, if a class wants to use the instance- member- making capability of the
parent class, the parent's initialize method is called through $super
inside its initialize .
A very clear example of this is how Ajax.request uses Ajax.Base to make
sure all the necessary arguments are set to a default, (and some are then
normalized) if they are not taken in through the options hash.
Ajax.Request = Class.create(Ajax.Base, {
_complete: false,
initialize: function($super, url, options) {
$super(options);
this.transport = Ajax.getTransport();
this.request(url);
},
... cut ...
});
This is how far things are clear for me.
------------------------------------------------------------------------------------------------------
Now, Element.Layout inherits methods from Hash and in its initialize
method the parent Hash's initialize is called WITHOUT passing any
arguments to it!
initialize: function($super, element, preCompute) {
* $super();*
this.element = $(element);
......cut.....
}
At the same time, Hash.initialize is waiting for an object as an only
argument to store it on the instance, isn’t it?
*var Hash = Class.create(Enumerable, (function() {*
* function initialize(object) {*
* this._object = Object.isHash(object) ? object.toObject() :
Object.clone(object);*
* }*
*..cut..*
*});*
My question is: *How will the this._object ever be created on the
Element.Layout instance?*
*
*
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