Isn't super-constructor null in this case?
From step 4 in
https://people.mozilla.org/~jorendorff/es6-draft.html#sec-getsuperconstructor
<https://people.mozilla.org/~jorendorff/es6-draft.html#sec-getsuperconstructor>
superConstructor is C.[[GetPrototypeOf]]()
which should be `null` after the class definition, if I'm not wrong. (But it
finally throws due to type error, so technically speaking, you can't even
reference the super constructor)
I can't think of a better way of extending null, but I wonder what's the use
case? null usually represents lacking of a value, and extending the lacking of
a value? My brain hurts.
> On May 7, 2015, at 4:25 PM, Axel Rauschmayer <a...@rauschma.de> wrote:
>
> Is this the best way to use `extends null`?
>
> ```js
> class C extends null {
> constructor() {
> let _this = Object.create(C.prototype);
> return _this;
> }
> }
> ```
>
> You can’t use `this`, because it can’t be initialized via a super-constructor
> call: the super-constructor is `Function.prototype` (which can’t be
> constructor-called).
>
> --
> Dr. Axel Rauschmayer
> a...@rauschma.de <mailto:a...@rauschma.de>
> rauschma.de
>
>
>
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