> Le 7 mai 2015 à 11:49, Glen Huang <curvedm...@gmail.com> a écrit :
>
> 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)
No, as a special case of the `extends` semantics, `C.[[GetPrototypeOf]]()` will
be %FunctionPrototype%; see step 6.e.ii of:
http://people.mozilla.org/~jorendorff/es6-draft.html#sec-runtime-semantics-classdefinitionevaluation
The reason is presumably that, since the constructor is a function, it should
always have the methods for functions (`.bind`, `.call`, etc.) on its prototype
chain.
The true meaning of `C extends null` is the following: The instances of `C`
won’t have %ObjectPrototype% in their prototype chain. (For the use cases,
don’t ask me.)
—Claude
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