Making "super" work outside a literal?

Axel Rauschmayer axel at
Wed Jun 22 16:02:48 PDT 2011

> From: Brendan Eich <brendan at>
> Date: June 23, 2011 0:21:17 GMT+02:00
>> Is there no per-call cost whatsoever to adding static super?
> No, it's static -- an internal property of the function object set once on creation and (in my view) never changed thereafter unless unobservably (Allen's point about optimizing Object.defineMethod when "handing off" an otherwise-useless function expression).

How does a function get access to its properties? Doesn’t the ability to access "thisFunction" incur any costs? I would expect static super to work as follows:

var Sub = Super <| {
    foo: function me(x) {, x);

Does the named function expression cost nothing here?

Dr. Axel Rauschmayer

axel at


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