On 01.11.2011 21:30, Jake Verbaten wrote:
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 calls.
Haven't I just showed a solution for this problem in my previous letter?
This issue is well known for years and have (at least) three working and
elegant solutions -- i.e. w/o hardcoding names of classes.
Dmitry.
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