Class inheritance

Class inheritance is a way for one class to extend another class.

So we can create new functionality on top of the existing.

The "extends" keyword

Let's say with have class 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} stands still.`);
  }
}

let animal = new Animal("My animal");

Here's how we can represent animal object and Animal class graphically:

...And we would like to create another class Rabbit.

As rabbits are animals, Rabbit class should be based on Animal, have access to animal methods, so that rabbits can do what "generic" animals can do.

The syntax to extend another class is: class Child extends Parent.

Let's create class Rabbit that inherits 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!

Object of Rabbit class have access to both Rabbit methods, such as rabbit.hide(), and also to Animal methods, such as rabbit.run().

Internally, extends keyword works using the good old prototype mechanics. It sets Rabbit.prototype.[[Prototype]] to Animal.prototype. So, if a method is not found in Rabbit.prototype, JavaScript takes it from Animal.prototype.

For instance, to find rabbit.run method, the engine checks (bottom-up on the picture):

The rabbit object (has no run).
Its prototype, that is Rabbit.prototype (has hide, but not run).
Its prototype, that is (due to extends) Animal.prototype, that finally has the run method.

As we can recall from the chapter info:native-prototypes, JavaScript itself uses prototypal inheritance for build-in objects. E.g. Date.prototype.[[Prototype]] is Object.prototype. That's why dates have access to generic object methods.

````smart header="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. By default, all methods that are not specified in `class Rabbit` are taken directly "as is" from `class Animal`.

But if we specify our own method in `Rabbit`, such as `stop()` then it will be used instead:

```js
class Rabbit extends Animal {
  stop() {
    // ...now this will be used for rabbit.stop()
    // instead of stop() from class Animal
  }
}
```

Usually we don't want to totally replace a parent method, but rather to build on top of it to 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:

```js run
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} stands still.`);
  }

}

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

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

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

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

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

````smart header="Arrow functions have no `super`"
As was mentioned in the chapter <info:arrow-functions>, arrow functions do not have `super`.

If accessed, it's taken from the outer function. For instance:
```js
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:

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

Overriding constructor

With constructors it gets a little bit tricky.

Till now, Rabbit did not have its own constructor.

According to the specification, if a class extends another class and has no constructor, then the following "empty" 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'll 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 labeled with a special internal property [[ConstructorKind]]:"derived".

The difference is:

When a normal constructor runs, it creates an empty object and assigns it to this.
But when a derived constructor runs, it doesn't do this. 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 for this won't be created. And we'll get an error.

For Rabbit constructor to work, it needs 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]]

If you're reading the tutorial for the first time - this section may be skipped.

It's about the internal mechanisms behind inheritance and `super`.

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

First to say, from all that we've learned till now, it's impossible for super to work at all!

Yeah, indeed, let's ask ourselves, how it should technically work? When an object method runs, it gets the current object as this. If we call super.method() then, the engine needs to get the method from the prototype of the current object. But how?

The task may seem simple, but it isn't. The engine knows the current object this, so it could get the parent method as this.__proto__.method. Unfortunately, such a "naive" solution won't work.

Let's demonstrate the problem. Without classes, using plain objects for the sake of simplicity.

You may skip this part and go below to the [[HomeObject]] subsection if you don't want to know the details. That won't harm. Or read on if you're interested in understanding things in-depth.

In the example below, rabbit.__proto__ = animal. Now let's try: in rabbit.eat() we'll call animal.eat(), using this.__proto__:

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

let rabbit = {
  __proto__: animal,
  name: "Rabbit",
  eat() {
*!*
    // that's how super.eat() could presumably work
    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 in the code above it actually works as intended: we have the correct alert.

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 anymore! 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 the endless loop.

Here's the picture of what happens:

Inside longEar.eat(), the line (**) calls rabbit.eat providing it with this=longEar.

// inside longEar.eat() we have this = longEar
this.__proto__.eat.call(this) // (**)
// becomes
longEar.__proto__.eat.call(this)
// that is
rabbit.eat.call(this);

Then in the line (*) of rabbit.eat, we'd like to pass the call even higher in the chain, but this=longEar, so this.__proto__.eat is again 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);

...So rabbit.eat calls itself in the endless loop, because it can't ascend any further.

The problem can't be solved by using this alone.

`[[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.

Then super uses it to resolve the parent prototype and its methods.

Let's see how it works, first with plain objects:

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

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

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

*!*
// works correctly
longEar.eat();  // Long Ear eats.
*/!*

It works as intended, due to [[HomeObject]] mechanics. A method, such as longEar.eat, knows its [[HomeObject]] and takes the parent method from its prototype. Without any use of this.

Methods are not "free"

As we've known before, generally functions are "free", not bound to objects in JavaScript. So they can be copied between objects and called with another this.

The very existance of [[HomeObject]] violates that principle, because methods remember their objects. [[HomeObject]] can't be changed, so this bond is forever.

The only place in the language where [[HomeObject]] is used -- is super. So, if a method does not use super, then we can still consider it free and copy between objects. But with super things may go wrong.

Here's the demo of a wrong super result after copying:

let animal = {
  sayHi() {
    console.log(`I'm an animal`);
  }
};

// rabbit inherits from animal
let rabbit = {
  __proto__: animal,
  sayHi() {
    super.sayHi();
  }
};

let plant = {
  sayHi() {
    console.log("I'm a plant");
  }
};

// tree inherits from plant
let tree = {
  __proto__: plant,
*!*
  sayHi: rabbit.sayHi // (*)
*/!*
};

*!*
tree.sayHi();  // I'm an animal (?!?)
*/!*

A call to tree.sayHi() shows "I'm an animal". Definitevely wrong.

The reason is simple:

In the line (*), the method tree.sayHi was copied from rabbit. Maybe we just wanted to avoid code duplication?
Its [[HomeObject]] is rabbit, as it was created in rabbit. There's no way to change [[HomeObject]].
The code of tree.sayHi() has super.sayHi() inside. It goes up from rabbit and takes the method from animal.

Here's the diagram of what happens:

Methods, not function properties

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

The difference may be non-essential for us, but it's important for JavaScript.

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() { // intentially writing like this instead of eat() {...
    // ...
  }
};

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

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

Summary

To extend a class: class Child extends Parent:
- That means Child.prototype.__proto__ will be Parent.prototype, so methods are inherited.
When overriding a constructor:
- We must call parent constructor as super() in Child constructor before using this.
When overriding another method:
- We can use super.method() in a Child method to call Parent method.
Internals:
- Methods remember their class/object in the internal [[HomeObject]] property. That's how super resolves parent methods.
- So it's not safe to copy a method with super from one object to another.

Also:

Arrow functions don't have own this or super, so they transparently fit into the surrounding context.

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Class inheritance

The "extends" keyword

Overriding constructor

Super: internals, [[HomeObject]]

`[[HomeObject]]`

Methods are not "free"

Methods, not function properties

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Class inheritance

The "extends" keyword

Overriding constructor

Super: internals, [[HomeObject]]

[[HomeObject]]

Methods are not "free"

Methods, not function properties

Summary

`[[HomeObject]]`