Re: [Laszlo-user] Mixins revisited

[P T Withington]

On 2009-11-20, at 11:06, Rami Ojares / AMG Oy wrote:

> > It works like this:
> > <class name="a" extends="b" with="c, d, e">
> > To find a property (attribute, method, etc.) in an instance, look in
> > this order:  <instance>, a, c, d, e, b, <b's inheritance list>
> 
> So let's say I am now at class:a level.
> Can I call super.foo() to call c.foo()
> And further can I call super.super.foo() to call d's foo method and so forth?
> (I know this is not possible in java)

When you call super.foo in any class or mixin, it looks up the (automatically 
computed) "inheritance chain".

<canvas>
  <class name="etc">
    <method name="foo">return "&lt;b's inheritance list&gt;"</method>
  </class>
  <class name="b" extends="etc">
    <method name="foo">return "b -> " + super.foo()</method>
  </class>
  <class name="c">
    <method name="foo">return "c -> " + super.foo()</method>
  </class>
  <class name="d">
    <method name="foo">return "d -> " + super.foo()</method>
  </class>
  <class name="e">
    <method name="foo">return "e -> " + super.foo()</method>
  </class>
  <class name="a" extends="b" with="c, d, e">
    <method name="foo">return "a -> " + super.foo()</method>
  </class>
  <a oninit="foo()">
    <method name="foo">Debug.info("&lt;instance&gt; -> " + super.foo())</method>
  </a>
</canvas>

Will display:

INFO @mixample.lzx#21: <instance> -> a -> c -> d -> e -> b -> <b's inheritance 
list>

Try playing with that example.  Shuffle the mixin's around, remove a foo method 
or two.  You'll get the idea. 

You can't say `super.super`, you can't get around the built-in inheritance 
chain.  This is part of the power (and danger) of mixins.  With a 
single-inheritance language, it should not be too surprising or confusing.  It 
gets more complicated if your language is a multiple-inheritance language (more 
than you probably every want to know about the issue here: 
http://bit.ly/6iixlA).