[Laszlo-dev] init vs oninit vs super.init

[Elliot Winard]

Hrm.  Actually, subclass init method gets called.  Superclass init 
method doesn't get called unless the subclass calls superclass.init().

The below example is a bit misleading because brusselssprouts calls 
super.init() before Debug.outputting.
-e


---=---===-------
Elliot Winard
Sr. Software Engineer
Webtop Team
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On Tue, Aug 7, 2007 at  2:45 PM, Benjamin Shine wrote:

> Bret and I were going through a code change and wanted a clarification 
> on what order things were expected to happen:
>
> <canvas>
> <class name="cabbage">
> 	<method name="init">
> 		// called FIRST
> 		Debug.write("cabbage.init (method name=init) on %w", this);
> 		this.setAttribute("deliciousness", 0.35);
> 	</method>
> 	<handler name="oninit">
> 		// called SECOND
> 		Debug.write("cabbage.oninit (handler name=oninit) on %w", this);
> 	</handler>	
> </class>
>
> <class name="brusselssprout" extends="cabbage">	
> 	<method name="init">
> 		super.init(); // calls method name="init" on cabbage
> 		Debug.write("brusselssprout.oninit method on %w,  deliciousness is 
> %f", this, this.deliciousness);
> 	</method>	
> 	<handler name="oninit">
> 		// called *after* cabbage's oninit handler
> 		Debug.write("brusselssprout.oninit handler on %w, deliciousness is 
> %f", this, this.deliciousness);
> 	</handler>	
> </class>
>
> <cabbage id="cabby" />
> <brusselssprout id="sprouty" />\
> </canvas>
>
> This produces the following output:
> cabbage.init (method name=init) on #cabby
> cabbage.oninit (handler name=oninit) on #cabby
> cabbage.init (method name=init) on #sprouty
> brusselssprout.oninit method on #sprouty,  deliciousness is 0.350000
> cabbage.oninit (handler name=oninit) on #sprouty
> brusselssprout.oninit handler on #sprouty, deliciousness is 0.350000
>
>
> ...which is just what I expect: superclass methods called before 
> sublcass methods, method name="init" called before oninit handler.
>
> Is this the guaranteed order of execution?
>
> -ben