It doesn't make sense because then `super` isn't really referring to the super class that the method is used on. I think it makes more sense for `super` to take meaning depending on where the method is found, not where it is defined.
What if there was also something like `Function.prototype.bind` like `Function.prototype.with`, so `someFunc.with(homeObject)` returns a new function who's [[HomeObject]] is the specified `homeObject`. It would be possible to do `someFunc.with(...).bind(...)` to configure both the home object and `this`. */#!/*JoePea On Thu, Jul 21, 2016 at 3:21 AM, Claude Pache <[email protected]> wrote: > There is probably no way to make the `super` semantics just work in all > use cases. But I think that the choice made in ES6 is the right one in > order to have it correctly working in the most naïve (and maybe the most > common?) use cases of method borrowing, i.e., when you don’t have knowledge > of — or even you wilfully ignore — its implementation details. Consider for > example: > > ```js > class MyArray extends Array { > forEach(callback, thisArg) { > // ... > super.forEach(callback, thisArg) > // ... > } > } > ``` > > and somewhere else: > > ```js > NodeList.prototype.forEach = MyArray.prototype.forEach // or: (new > MyArray).forEach > > HTMLCollection.prototype.forEach = MyArray.prototype.forEach > ``` > > Here, I expect that `super.forEach` will continue to point to > `Array.prototype.forEach`. This is because I’m thinking of > `MyArray.prototype.forEach` as a black box that should not change its > behaviour just because I’m moving it around, as it is generally the case > with methods of the builtin library. > > —Claude
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