How can I call a superclass method like it was defined on the derived class?

Question

I have these classes:

class Control {
  get code() {
    return 3;
  }
  getCodeChain() {
    var result = [this.code];
    if (super.getCodeChain) {
      result = result.concat(super.getCodeChain());
    }
    return result;
  }
}

class SubControl extends Control {
  get code() {
    return 2;
  }
}

class AnotherControl extends SubControl {
  get code() {
    return 1;
  }
}

console.log((new AnotherControl()).getCodeChain()); // prints 1

When I call getCodeChain on AnotherControl instance, it goes all the way up to Control context, so the recursion ignores AnotherControl and SubControl contexts.

I need to get the CodeChain, but I don't want/can implement the getCodeChain() method in all subclasses. The results I expect is [1,2,3].

How can I call a superclass method like it was defined on the derived class?

Answer 1

You can follow the prototype chain using Object.getPrototypeOf:

class Control {
    get code() { return 3; }
    getCodeChain() {
        const result = [];
        for (let proto = Object.getPrototypeOf(this); Object.hasOwn(proto, "code"); proto = Object.getPrototypeOf(proto)) {
            result.push(proto.code);
        }
        return result;
    }
}

class SubControl extends Control {
    get code() { return 2; }
}

class AnotherControl extends SubControl {
    get code() { return 1; }
}

console.log((new AnotherControl()).getCodeChain()); // [1, 2, 3]