Are private fields of subclasses of built-in classes not accesible inside overriden methods?

Question

I have the following class that extends the built in data type Set. It adds a little bit of extra functionality by checking that every element that wants to be added to the set conforms to the type specified as the first parameter of the constructor. It does this silly "type checking" through the typeof operator. I save the type of the Set in a private field. If I try to override the .add() method so that it also does this little type checking, I get an error stating Uncaught TypeError: can't access private field or method: object is not the right class

class TypedSet extends Set {
    #type;
    constructor(type, elements) {
        if (elements && !elements.every(e => typeof e === type)) throw new TypeError(`Not all elements conform to type t->${type}`)
        super(elements)
        this.#type = type
    }
    add(e) {
        return typeof e === this.#type
            ? super.add(e)
            : new TypeError(`Type of element e does not conform to type t: ${this.#type}`)
    }
}

const s = new TypedSet('string', ['hello', 'world'])
s.add('!')

Why do I get that error if i try to access the private field -- the one I declared in the subclass -- in the overriden .add() method ? If I rename the .add() method to something like .typeAdd() referencing the private field doesn't throw an error

score 2 · Accepted Answer · answered Apr 14 '22 at 19:22

The problem is that super(elements) will call the add method to add the elements to the set¹. At that point, your subclass constructor did not yet create the field (this.#type = type).

As a workaround, you can use

class TypedSet extends Set {
    #type;
    constructor(type, elements) {
        super()
        this.#type = type
        for (const e of elements ?? []) {
            if (typeof e !== type) throw new TypeError(`Not all elements conform to type t->${type}`)
            super.add(e) // or this.add(e)
        }
    }
    add(e) {
        return typeof e === this.#type
            ? super.add(e)
            : new TypeError(`Type of element e does not conform to type t: ${this.#type}`)
    }
}

const s = new TypedSet('string', ['hello', 'world'])
s.add('!')

_{1: Calling an overrideable method from the constructor is usually considered an antipattern, for exactly the reason that your use case doesn't work. No idea why ECMAScript did specify this regardless. On the other hand, it's also not recommended - even if possible - to extend builtin classes, there are too many surprises such as this. Composition is usually a better approach.}

Thanks for the reply! Your explanation makes sense but what I don't get is why calling ```super(elements)``` in my original code would call the ```.add()``` method of the subclass -- where, as you explained, the private field doesn't exist yet --. Wouldn't calling the constructor of the superclass to initialize the instance also call only methods of the superclass ?? -- Also , I know it's probably not the best idea to extend built-in classes, I'm doing this mostly as an exercise but the error message about the private field caught me off guard -- — Ariel, Apr 14 '22 at 20:01
Well maybe the constructor *should* use `Set.prototype.add.call(this, …)`, but really it's just doing `this.add(…)`, and the instance (`this`) does already inherit from `TypedSet.prototype`. — Bergi, Apr 14 '22 at 22:29

Are private fields of subclasses of built-in classes not accesible inside overriden methods?

1 Answers1