> Le 11 mai 2015 à 05:26, Glen Huang <[email protected]> a écrit : > > No one care to comment? Hope I didn't mistake how super works. > >> On May 8, 2015, at 11:05 AM, Glen Huang <[email protected]> wrote: >> >> TLDR: If `this` is used without `super` (which should be statically >> analyzable), let it refer to Object.create(new.target.prototype). Otherwise, >> let super creates what it refers to. >> >> I know the reason to force `super` in derived class's constructor is to make >> sure `this` refers to the exotic object the base class might allocate. >> >> But I bet in real world, extending base classes who only create ordinary >> objects is more common than extending those create exotic objects. And >> forget to call super is going to be frequent mistake. In es5, when you >> extend a class like this >> >> ```js >> function Foo() { >> Bar.call(this) >> } >> Foo.prototype = Object.create(Bar.prototype); >> ``` >> >> `this` refers to Object.create(new.target.prototype). So I wonder if we can >> keep this behavior, making things less surprising when people transit to es >> 2015?
Yes, in the course of refactoring your classes, you are indeed forced to transform your old clunky `Bar.call(this)` invocations into new shiny `super()` invocations, and it is IMHO a good thing. I don't see personally the point to support deprecated patterns in new syntactic forms. But anyway, if refactoring your old classes is too difficult (for it might need more complicated changes that the aforementioned simple substitution), you are not obliged to: the old ES3/5 "classes" will continue to work, and will even be usable as superclass (in the `extends` clause) of new ES2015-classes. —Claude _______________________________________________ es-discuss mailing list [email protected] https://mail.mozilla.org/listinfo/es-discuss
