super() on class that extends

[Allen Wirfs-Brock]

> On Apr 10, 2015, at 8:06 PM, Axel Rauschmayer <axel at rauschma.de> wrote:
> 
> ...
> If you do not call `super()`, you only get into trouble if you access `this` in some manner. Two examples:
> 
> ...
> 
> Therefore, there are two ways to avoid typing super-constructor calls.
> 
> First, you can avoid accessing `this` by explicitly returning an object from the derived class constructor. However, this is not what you want, because the object created via `new B()` does not inherit `A`’s methods.
> 
> ```js
> // Base class
> class A {
>     constructor() {}
> }
> // Derived class
> class B extends A {
>     constructor() {
>         // No super-constructor call here!
>         
>         return {}; // must be an object
>     }
> }
> ```
> 
> Second, you can let JavaScript create default constructors for you:
> 
> ```js
> // Base class
> class A {
> }
> // Derived class
> class B extends A {
> }
> ```
> 
> This code is equivalent to:
> 
> ```js
> // Base class
> class A {
>     constructor() {}
> }
> // Derived class
> class B extends A {
>     constructor(...args) {
>         super(...args);
>     }
> }
> ```

or third, about the super class to make sure that you correctly initialize the instance to work with inherited methods:

‘’’js
class B extends A {
   constructor(…args) {
         let newObj = Reflect.construct(Object, args, this.target);
         newObj.prop = something;
         return newObj;
   }
} 
```

Allen