The `super` keyword doesn't work as it should?

Bergi a.d.bergi at
Mon Jul 18 21:46:22 UTC 2016

/#!/JoePea wrote:

> Why can't `super` simply be a shortcut
> for "look up the prototype of the object that the method is called on, then
> find the `.constructor` property and call it on `this`"? That seems to be
> simple.

Simple, yes, and broken in the case of multi-level inheritance:
const x = Object.assign(Object.create({
     method() {
}), {
     method() {
         Object.getPrototypeOf(this).method(); // super.method()
x.method(); // works as expected

const y = Object.create(x);
y.method(); // infinite loop/stack overflow
A `super` query must not depend on `this` (only), it must statically 
resolve the object on which the called method is defined.

In constructors, using the prototype of the currenctly called 
constructor for `super()` works well, but you'd need to use 
`Object.setPrototype` as there is currently no declarative way other 
than `class`es to define functions with custom prototypes.

In methods, there would need to be a way to populate the [[HomeObject]] 
other than declaring the method as part of a class/object literal.

Kind regards,

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