raynos2 at gmail.com
Tue Nov 1 10:30:02 PDT 2011
> Why isn't the |super| lookup-point |this.getPrototypeOf()|
Assume |super| is |this.getPrototypeOf()|
Let F be a "class", let f be an instance of the class.
inside f you have access to a method defined on F.
If you call a method defined on F from f and that method calls super, you
would be invoking getPrototypeOf(this) which is just F (since this is f,
and the prototype of f is F). So you would then end up calling the same
method again and recurse endlessly.
Basically because you apply methods with a value of this then if super were
tied to this you would get infinite super recursion if you chained super
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