How to expose methods of a property object as if they are objects own methods? JavaScript

Question

I have a really big urge to do something like this:

class A {
    constructor(){}
    age() { return 15; }
}

class B {
    constructor() {
        this.a = new A();
    // DO SOMETHING THAT THIS IS POSSIBLE:
    }
}

B b = new B();
b.age();

How to expose the method as if it was object b's own method? I need to do this for every method of a property, regardless of number and name.

NOTE: I can't use inheritance.

@Oen44 Is there a way to combine the prototypes somehow? To expose the method `age` on the object `b`? — Shocky2, Dec 26 '18 at 21:21
@StudioKonKon Then it's pointless to create class B. It will work as Class A. Any methods that class B is using won't work. — Oen44, Dec 26 '18 at 21:27
https://stackoverflow.com/questions/8312459/iterate-through-object-properties Maybe this can help? — SoMa, Dec 26 '18 at 22:10
It should be more than a note to the reason why you cannot use inheritance, please expand on the reasons why you believe you cannot use and if possible, please give a sample of your code where you want to use it — Icepickle, Dec 27 '18 at 08:48

ic3b3rg · Answer 1 · 2018-12-26T21:31:17.557

3

extends provides inheritance functionality:

class A {
  constructor(){}
  age() { return 15; }
}

class B extends A {
  constructor() {
    super()
  }
}

const b = new B();

console.log(b.age());

I'd be interested to learn why you can't use normal inheritance. Here's a way to manually inherit from a base class:

class A {
  constructor(){}
  age() { return 15; }
}

class B {
  constructor() {
    Object.getOwnPropertyNames(A.prototype).forEach(prop => {
      if (prop !== 'constructor') {
        this[prop] = A.prototype[prop];
      }
    });
  }
}

const b = new B();

console.log(b.age());

edited Dec 26 '18 at 21:31

answered Dec 26 '18 at 21:23

ic3b3rg

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1

`NOTE: I can't use inheritance.` – Oen44 Dec 26 '18 at 21:25
You didn't even read the entire question... No inheritance. I can't use it in this situation because of other reasons. – Shocky2 Dec 26 '18 at 21:25
1

What are the reasons? – ic3b3rg Dec 26 '18 at 21:26
1

I updated my answer with a different method. I's be curious to hear why you can't use regular inheritance – ic3b3rg Dec 26 '18 at 21:35

score 3 · Answer 2 · answered Dec 26 '18 at 21:40

I would try implementing B as a Proxy.

class A {
    constructor(){}
    age() { return 15; }
}

const b = new Proxy(new A(), {
    get: function(object, property) {
        if (typeof object[property] === 'function') {
            return object.property();
        }
    }
    set: function(object, property) {}
}

You can read into it more at MDN.

How to expose methods of a property object as if they are objects own methods? JavaScript

2 Answers2