Class inheritance, super

Classes can extend one another. There’s a nice syntax, technically based on the prototypal inheritance.

To inherit from another class, we should specify "extends" and the parent class before the brackets {..}.

Here Rabbit inherits from Animal:

class Animal {

  constructor(name) {
    this.speed = 0;
    this.name = name;
  }

  run(speed) {
    this.speed += speed;
    alert(`${this.name} runs with speed ${this.speed}.`);
  }

  stop() {
    this.speed = 0;
    alert(`${this.name} stopped.`);
  }

}

// Inherit from Animal
class Rabbit extends Animal {
  hide() {
    alert(`${this.name} hides!`);
  }
}

let rabbit = new Rabbit("White Rabbit");

rabbit.run(5); // White Rabbit runs with speed 5.
rabbit.hide(); // White Rabbit hides!

The extends keyword actually adds a [[Prototype]] reference from Rabbit.prototype to Animal.prototype, just as you expect it to be, and as we’ve seen before.

So now rabbit has access both to its own methods and to methods of Animal.

Any expression is allowed after extends

Class syntax allows to specify not just a class, but any expression after extends.

For instance, a function call that generates the parent class:

function f(phrase) {
  return class {
    sayHi() { alert(phrase) }
  }
}

class User extends f("Hello") {}

new User().sayHi(); // Hello

Here class User inherits from the result of f("Hello").

That may be useful for advanced programming patterns when we use functions to generate classes depending on many conditions and can inherit from them.

Overriding a method

Now let’s move forward and override a method. As of now, Rabbit inherits the stop method that sets this.speed = 0 from Animal.

If we specify our own stop in Rabbit, then it will be used instead:

class Rabbit extends Animal {
  stop() {
    // ...this will be used for rabbit.stop()
  }
}

…But usually we don’t want to totally replace a parent method, but rather to build on top of it, tweak or extend its functionality. We do something in our method, but call the parent method before/after it or in the process.

Classes provide "super" keyword for that.

  • super.method(...) to call a parent method.
  • super(...) to call a parent constructor (inside our constructor only).

For instance, let our rabbit autohide when stopped:

class Animal {

  constructor(name) {
    this.speed = 0;
    this.name = name;
  }

  run(speed) {
    this.speed += speed;
    alert(`${this.name} runs with speed ${this.speed}.`);
  }

  stop() {
    this.speed = 0;
    alert(`${this.name} stopped.`);
  }

}

class Rabbit extends Animal {
  hide() {
    alert(`${this.name} hides!`);
  }

  stop() {
    super.stop(); // call parent stop
    hide(); // and then hide
  }
}

let rabbit = new Rabbit("White Rabbit");

rabbit.run(5); // White Rabbit runs with speed 5.
rabbit.stop(); // White Rabbit stopped. White rabbit hides!

Now Rabbit has the stop method that calls the parent super.stop() in the process.

Arrow functions have no super

As was mentioned in the chapter Arrow functions revisited, arrow functions do not have super.

If accessed, it’s taken from the outer function. For instance:

class Rabbit extends Animal {
  stop() {
    setTimeout(() => super.stop(), 1000); // call parent stop after 1sec
  }
}

The super in the arrow function is the same as in stop(), so it works as intended. If we specified a “regular” function here, there would be an error:

// Unexpected super
setTimeout(function() { super.stop() }, 1000);

Overriding constructor

With constructors, things are is a little bit tricky.

Till now, Rabbit had no its own constructor.

According to the specification, if a class extends another class and has no constructor, then the following constructor is generated:

class Rabbit extends Animal {
  // generated for extending classes without own constructors
  constructor(...args) {
    super(...args);
  }
}

As we can see, it basically calls the parent constructor passing it all the arguments. That happens if we don’t write a constructor of our own.

Now let’s add a custom constructor to Rabbit. It will specify the earLength in addition to name:

class Animal {
  constructor(name) {
    this.speed = 0;
    this.name = name;
  }
  // ...
}

class Rabbit extends Animal {

  constructor(name, earLength) {
    this.speed = 0;
    this.name = name;
    this.earLength = earLength;
  }

  // ...
}

// Doesn't work!
let rabbit = new Rabbit("White Rabbit", 10); // Error: this is not defined.

Whoops! We’ve got an error. Now we can’t create rabbits. What went wrong?

The short answer is: constructors in inheriting classes must call super(...), and (!) do it before using this.

…But why? What’s going on here? Indeed, the requirement seems strange.

Of course, there’s an explanation. Let’s get into details, so you’d really understand what’s going on.

In JavaScript, there’s a distinction between a “constructor function of an inheriting class” and all others. In an inheriting class, the corresponding constructor function is labelled with a special internal property [[ConstructorKind]]:"derived".

The difference is:

  • When a normal constructor runs, it creates an empty object as this and continues with it.
  • But when a derived constructor runs, it doesn’t do it. It expects the parent constructor to do this job.

So if we’re making a constructor of our own, then we must call super, because otherwise the object with this reference to it won’t be created. And we’ll get an error.

For Rabbit to work, we need to call super() before using this, like here:

class Animal {

  constructor(name) {
    this.speed = 0;
    this.name = name;
  }

  // ...
}

class Rabbit extends Animal {

  constructor(name, earLength) {
    super(name);
    this.earLength = earLength;
  }

  // ...
}

// now fine
let rabbit = new Rabbit("White Rabbit", 10);
alert(rabbit.name); // White Rabbit
alert(rabbit.earLength); // 10

Super: internals, [[HomeObject]]

Let’s get a little deeper under the hood of super. We’ll see some interesting things by the way.

First to say, from all that we’ve learned till now, it’s impossible for super to work.

Yeah, indeed, let’s ask ourselves, how it could technically work? When an object method runs, it gets the current object as this. If we call super.method() then, how to retrieve that method? In other words, we need to take the method from the parent prototype of the current object. How, technically, we (or a JavaScript engine) can do it?

Maybe we can get it [[Prototype]] of this, as this.__proto__.method? Unfortunately, that won’t work.

Let’s try to do it. Without classes, using plain objects for sheer simplicity.

Here, rabbit.eat() should call animal.eat() method of the parent object:

let animal = {
  name: "Animal",
  eat() {
    alert(this.name + " eats.");
  }
};

let rabbit = {
  __proto__: animal,
  name: "Rabbit",
  eat() {
    this.__proto__.eat.call(this); // (*)
  }
};

rabbit.eat(); // Rabbit eats.

At the line (*) we take eat from the prototype (animal) and call it in the context of the current object. Please note that .call(this) is important here, because a simple this.__proto__.eat() would execute parent eat in the context of the prototype, not the current object.

And here it works.

Now let’s add one more object to the chain. We’ll see how things break:

let animal = {
  name: "Animal",
  eat() {
    alert(this.name + " eats.");
  }
};

let rabbit = {
  __proto__: animal,
  eat() {
    // ...bounce around rabbit-style and call parent (animal) method
    this.__proto__.eat.call(this); // (*)
  }
};

let longEar = {
  __proto__: rabbit,
  eat() {
    // ...do something with long ears and call parent (rabbit) method
    this.__proto__.eat.call(this); // (**)
  }
};

longEar.eat(); // Error: Maximum call stack size exceeded

The code doesn’t work any more! We can see the error trying to call longEar.eat().

It may be not that obvious, but if we trace longEar.eat() call, then we can see why. In both lines (*) and (**) the value of this is the current object (longEar). That’s essential: all object methods get the current object as this, not a prototype or something.

So, in both lines (*) and (**) the value of this.__proto__ is exactly the same: rabbit. They both call rabbit.eat without going up the chain.

In other words:

  1. Inside longEar.eat(), we pass the call up to rabbit.eat giving it the same this=longEar.

    // inside longEar.eat() we have this = longEar
    this.__proto__.eat.call(this) // (**)
    // becomes
    longEar.__proto__.eat.call(this)
    // or
    rabbit.eat.call(this);
  2. Inside rabbit.eat, we want to pass the call even higher in the chain, but this=longEar, so this.__proto__.eat is rabbit.eat!

    // inside rabbit.eat() we also have this = longEar
    this.__proto__.eat.call(this) // (*)
    // becomes
    longEar.__proto__.eat.call(this)
    // or (again)
    rabbit.eat.call(this);
  3. …So rabbit.eat calls itself in the endless loop, because it can’t ascend any further.

There problem is unsolvable, because this must always be the calling object itself, no matter which parent method is called. So its prototype will always be the immediate parent of the object. We can’t go up the chain.

[[HomeObject]]

To provide the solution, JavaScript adds one more special internal property for functions: [[HomeObject]].

When a function is specified as a class or object method, its [[HomeObject]] property becomes that object.

This actually violates the idea of “unbound” functions, because methods remember their objects. And [[HomeObject]] can’t be changed, so this bound is forever. So that’s a very important change in the language.

But this change is safe. [[HomeObject]] is used only for calling parent methods in super, to resolve the prototype. So it doesn’t break compatibility.

Let’s see how it works for super – again, using plain objects:

let animal = {
  name: "Animal",
  eat() {         // [[HomeObject]] == animal
    alert(this.name + " eats.");
  }
};

let rabbit = {
  __proto__: animal,
  name: "Rabbit",
  eat() {         // [[HomeObject]] == rabbit
    super.eat();
  }
};

let longEar = {
  __proto__: rabbit,
  name: "Long Ear",
  eat() {         // [[HomeObject]] == longEar
    super.eat();
  }
};

longEar.eat();  // Long Ear eats.

Every method remembers its object in the internal [[HomeObject]] property. Then super uses it to resolve the parent prototype.

[[HomeObject]] is defined for methods defined both in classes and in plain objects. But for objects, methods must be specified exactly the given way: as method(), not as "method: function()".

In the example below a non-method syntax is used for comparison. [[HomeObject]] property is not set and the inheritance doesn’t work:

let animal = {
  eat: function() { // should be the short syntax: eat() {...}
    // ...
  }
};

let rabbit = {
  __proto__: animal,
  eat: function() {
    super.eat();
  }
};

rabbit.eat();  // Error calling super (because there's no [[HomeObject]])

Static methods and inheritance

The class syntax supports inheritance for static properties too.

For instance:

class Animal {

  constructor(name, speed) {
    this.speed = speed;
    this.name = name;
  }

  run(speed = 0) {
    this.speed += speed;
    alert(`${this.name} runs with speed ${this.speed}.`);
  }

  static compare(animalA, animalB) {
    return animalA.speed - animalB.speed;
  }

}

// Inherit from Animal
class Rabbit extends Animal {
  hide() {
    alert(`${this.name} hides!`);
  }
}

let rabbits = [
  new Rabbit("White Rabbit", 10),
  new Rabbit("Black Rabbit", 5)
];

rabbits.sort(Rabbit.compare);

rabbits[0].run(); // Black Rabbit runs with speed 5.

Now we can call Rabbit.compare assuming that the inherited Animal.compare will be called.

How does it work? Again, using prototypes. As you might have already guessed, extends also gives Rabbit the [[Prototype]] reference to Animal.

So, Rabbit function now inherits from Animal function. And Animal function normally has [[Prototype]] referencing Function.prototype, because it doesn’t extend anything.

Here, let’s check that:

class Animal {}
class Rabbit extends Animal {}

// for static propertites and methods
alert(Rabbit.__proto__ == Animal); // true

// and the next step is Function.prototype
alert(Animal.__proto__ == Function.prototype); // true

// that's in addition to the "normal" prototype chain for object methods
alert(Rabbit.prototype.__proto__ === Animal.prototype);

This way Rabbit has access to all static methods of Animal.

Please note that built-in classes don’t have such static [[Prototype]] reference. For instance, Object has Object.defineProperty, Object.keys and so on, but Array, Date etc do not inherit them.

Here’s the picture structure for Date and Object:

Note, there’s no link between Date and Object. Both Object and Date exist independently. Date.prototype inherits from Object.prototype, but that’s all.

Such difference exists for historical reasons: there was no thought about class syntax and inheriting static methods at the dawn of JavaScript language.

Natives are extendable

Built-in classes like Array, Map and others are extendable also.

For instance, here PowerArray inherits from the native Array:

// add one more method to it (can do more)
class PowerArray extends Array {
  isEmpty() {
    return this.length == 0;
  }
}

let arr = new PowerArray(1, 2, 5, 10, 50);
alert(arr.isEmpty()); // false

let filteredArr = arr.filter(item => item >= 10);
alert(filteredArr); // 10, 50
alert(filteredArr.isEmpty()); // false

Please note one very interesting thing. Built-in methods like filter, map and others – return new objects of exactly the inherited type. They rely on the constructor property to do so.

In the example above,

arr.constructor === PowerArray

So when arr.filter() is called, it internally creates the new array of results exactly as new PowerArray. And we can keep using its methods further down the chain.

Even more, we can customize that behavior. The static getter Symbol.species, if exists, returns the constructor to use in such cases.

For example, here due to Symbol.species built-in methods like map, filter will return “normal” arrays:

class PowerArray extends Array {
  isEmpty() {
    return this.length == 0;
  }

  // built-in methods will use this as the constructor
  static get [Symbol.species]() {
    return Array;
  }
}

let arr = new PowerArray(1, 2, 5, 10, 50);
alert(arr.isEmpty()); // false

// filter creates new array using arr.constructor[Symbol.species] as constructor
let filteredArr = arr.filter(item => item >= 10);

// filteredArr is not PowerArray, but Array
alert(filteredArr.isEmpty()); // Error: filteredArr.isEmpty is not a function

We can use it in more advanced keys to strip extended functionality from resulting values if not needed. Or, maybe, to extend it even further.

Tasks

importance: 5

Here’s the code with Rabbit extending Animal.

Unfortunately, Rabbit objects can’t be created. What’s wrong? Fix it.

class Animal {

  constructor(name) {
    this.name = name;
  }

}

class Rabbit extends Animal {
  constructor(name) {
    this.name = name;
    this.created = Date.now();
  }
}

let rabbit = new Rabbit("White Rabbit"); // Error: this is not defined
alert(rabbit.name);

That’s because the child constructor must call super().

Here’s the corrected code:

class Animal {

  constructor(name) {
    this.name = name;
  }

}

class Rabbit extends Animal {
  constructor(name) {
    super(name);
    this.created = Date.now();
  }
}

let rabbit = new Rabbit("White Rabbit"); // ok now
alert(rabbit.name); // White Rabbit
importance: 5

We’ve got a Clock class. As of now, it prints the time every second.

Create a new class ExtendedClock that inherits from Clock and adds the parameter precision – the number of ms between “ticks”. Should be 1000 (1 second) by default.

  • Your code should be in the file extended-clock.js
  • Don’t modify the original clock.js. Extend it.

Open a sandbox for the task.

importance: 5

As we know, all objects normally inherit from Object.prototype and get access to “generic” object methods.

Like demonstrated here:

class Rabbit {
  constructor(name) {
    this.name = name;
  }
}

let rabbit = new Rabbit("Rab");

// hasOwnProperty method is from Object.prototype
// rabbit.__proto__ === Object.prototype
alert( rabbit.hasOwnProperty('name') ); // true

So, is it correct to say that "class Rabbit extends Object" does exactly the same as "class Rabbit", or not?

Will it work?

class Rabbit extends Object {
  constructor(name) {
    this.name = name;
  }
}

let rabbit = new Rabbit("Rab");

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

If it won’t please fix the code.

Open a sandbox for the task.

The answer has two parts.

The first, an easy one is that the inheriting class needs to call super() in the constructor. 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 important difference in "class Rabbit extends Object" versus 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 itself (for static methods).

In our case, for 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 we can access 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

And if we don’t use extends, then class Rabbit does not get the second reference.

Please compare with it:

class Rabbit {}

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

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

For the simple class Rabbit, the Rabbit function has the same prototype

class Rabbit {}

// instead of (2) that's correct for Rabbit (just like any function):
alert( Rabbit.__proto__ === Function.prototype );

By the way, Function.prototype 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:

class Rabbit class Rabbit extends Object
needs to call super() in constructor
Rabbit.__proto__ === Function.prototype Rabbit.__proto__ === Object

Open the solution in a sandbox.

Tutorial map

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