If I were Tucker, I would say, "You should never override init,
because if you forget to call super.init(), things are deeply hosed."
(Tucker's offline for the day so we'll have to wait till tomorrow to
have him in this discussion.)
On Aug 7, 2007, at 12:30 PM, Bret Simister wrote:
This is about a choice of design patterns. When extending a class
that needs to have new functionality during its initialization,
but the subclass also needs to execute the init functionality of
its super class, which of the methods below is the preferred
design pattern?
<method name="init" >
super.init()
...
</method>
or
<handler name="oninit" >
...
</handler>
and rely on the fact that when using the handler, the super.init()
of the class is effectively being called because it has not been
overridden.
Even if the functionality is identical in both cases, I prefer the
explicit design pattern of <method name=" ..." > over the implicit
design pattern of <handler name="oninit" >
my two cents,
Bret
On Aug 7, 2007, at 12:08 PM, Elliot Winard wrote:
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
---=---===-------
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
