From a non-Ruby guy, here is the scoop:
Ruby has a keyword "super" that is available within all methods. It
allows you to call the parent function of the same name.
In php I write:
public function mymethod($arg) {
parent::mymethod($arg); // equivalent to Ruby super($arg)
}
So this JS implementation basically has $super refer to the parent
method of the same name so you can call $super()
"$super" is used instead of "super" because "super" is a reserved word
in JS.
The sugar is that extended classes automagically pass a reference of the
parent function into child function only for methods that have a $super
argument (detected when the child function is converted to a string with
toString()).
- Ken Snyder
PS. great interjection of "confuzzled"!
Matt Foster wrote:
> Greetings,
>
> I had a question about the $super argument that I wasn't
> able to find in documentation, searching etc. What exactly is the
> $super argument. Is it a reference to the super class's prototype
> function? Or is it an instantiated object of the super class that is
> passed to the subclass's method?
>
> I am completely confuzzled on this one, my typical method of just
> reading the source has left me worse off. Can anyone explain this
> concept or perhaps point me to a resource which properly defines it?
>
> Thanks in advance,
> Matt
>
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