What is the need to declare the constructor/super of an extended class?

Question

I have a class, example:

class Animal {
  constructor(raca) {
    this._raca = raca;
  }

  get raca() {
    return this._raca;
  }

  printRaca = function() {
    console.log(this._raca);
  }
}

And then I extend it this way:

class Cachorro extends Animal {
	
}

Notice that I did not call the builder, nor the super(). However, if I instantiate the class Cachorro it works the same way as class Animal, and I can take the this._raca even without having called the constructor(raca) { super(raca) } and use it normally, see:

class Animal {
  constructor(raca) {
    this._raca = raca;
  }

  get raca() {
    return this._raca;
  }

  printRaca = function() {
    console.log(this._raca);
  }
}

class Cachorro extends Animal {
	
}

var dog = new Cachorro('poodle');
console.log(dog.raca); //'poodle'
dog.printRaca(); // 'poodle'

What is the need then to call the builder and the super()? Is there any consequence in not calling them?

Answer 1

Every class has a default constructor created that you don’t even see, if you didn’t have it you couldn’t instantiate the class. In classes that are derived from other constructors this constructor makes the call to the constructor of the base class even without you seeing it there. You only need to do this if you need a builder to do something more specific. In this case it is necessary because you need to initialize the base constructor with an argument, the derived class has no way of guessing this. But the JS tries to guess and he can hit. In this case language so make your life easier, at the risk of doing something wrong, and creates a constructor for you that gets the same arguments you have to pass to the base constructor.

Most codes do much more than is visible. For some reason, especially Javascript programmers believe that the less code you see the less you’re doing and the faster it is, when it’s the opposite, "always" has something hidden running.

If you create your own builder and do not call the super() properly will not build the object in a way that can be used properly or will even generate an error. So you need to understand What good is a builder?. See that, contrary to the premise of the question, so does not work:

class Animal {
    constructor(raca) {
        this._raca = raca;
    }
    get raca() {
        return this._raca;
    }
}

class Cachorro extends Animal {}

var x = new Cachorro();
console.log(x.raca);

And so it works:

class Animal {
    constructor(raca) {
        this._raca = raca;
    }
    get raca() {
        return this._raca;
    }
}

class Cachorro extends Animal {
    constructor(raca) { super(raca) }
}

var x = new Cachorro("raça");
console.log(x.raca);

I put in the Github for future reference.

Note that it has error in the question code and does not work in every browser, so for all intents and purposes for the web it is an invalid code and should not be used.