GeezerVison

First, let’s see why the latter code doesn’t work.

The reason becomes obvious if we try to run it. An inheriting class constructor must call super(). Otherwise "this" won’t be “defined”.

So here’s the fix:

class Rabbit extends Object {
  constructor(name) {
    super(); // need to call the parent constructor when inheriting
    this.name = name;
  }
}

let rabbit = new Rabbit("Rab");

alert( rabbit.hasOwnProperty('name') ); // true

But that’s not all yet.

Even after the fix, there’s still an important difference between "class Rabbit extends Object" and class Rabbit.

As we know, the “extends” syntax sets up two prototypes:

1. Between "prototype" of the constructor functions (for methods).
2. Between the constructor functions themselves (for static methods).

In the case of class Rabbit extends Object it means:

class Rabbit extends Object {}

alert( Rabbit.prototype.__proto__ === Object.prototype ); // (1) true
alert( Rabbit.__proto__ === Object ); // (2) true

So Rabbit now provides access to the static methods of Object via Rabbit, like this:

class Rabbit extends Object {}

// normally we call Object.getOwnPropertyNames
alert ( Rabbit.getOwnPropertyNames({a: 1, b: 2})); // a,b

But if we don’t have extends Object, then Rabbit.__proto__ is not set to Object.

Here’s the demo:

class Rabbit {}

alert( Rabbit.prototype.__proto__ === Object.prototype ); // (1) true
alert( Rabbit.__proto__ === Object ); // (2) false (!)
alert( Rabbit.__proto__ === Function.prototype ); // as any function by default

// error, no such function in Rabbit
alert ( Rabbit.getOwnPropertyNames({a: 1, b: 2})); // Error

So Rabbit doesn’t provide access to static methods of Object in that case.

By the way, Function.prototype also has “generic” function methods, like call, bind etc. They are ultimately available in both cases, because for the built-in Object constructor, Object.__proto__ === Function.prototype.

Here’s the picture:

So, to put it short, there are two differences: