December 12, 2021

F.prototype

Remember, new objects can be created with a constructor function, like new F().

If F.prototype is an object, then the new operator uses it to set [[Prototype]] for the new object.

JavaScript had prototypal inheritance from the beginning. It was one of the core features of the language.

But in the old times, there was no direct access to it. The only thing that worked reliably was a "prototype" property of the constructor function, described in this chapter. So there are many scripts that still use it.

Please note that F.prototype here means a regular property named "prototype" on F. It sounds something similar to the term “prototype”, but here we really mean a regular property with this name.

Here’s the example:

let animal = {
  eats: true
};

function Rabbit(name) {
  this.name = name;
}

Rabbit.prototype = animal;

let rabbit = new Rabbit("White Rabbit"); //  rabbit.__proto__ == animal

alert( rabbit.eats ); // true

Setting Rabbit.prototype = animal literally states the following: “When a new Rabbit is created, assign its [[Prototype]] to animal”.

That’s the resulting picture:

On the picture, "prototype" is a horizontal arrow, meaning a regular property, and [[Prototype]] is vertical, meaning the inheritance of rabbit from animal.

F.prototype property is only used when new F is called, it assigns [[Prototype]] of the new object.

If, after the creation, F.prototype property changes (F.prototype = <another object>), then new objects created by new F will have another object as [[Prototype]], but already existing objects keep the old one.

Default F.prototype, constructor property

Every function has the "prototype" property even if we don’t supply it.

The default "prototype" is an object with the only property constructor that points back to the function itself.

Like this:

          function Rabbit() {}

/* default prototype
Rabbit.prototype = { constructor: Rabbit };
*/
        

We can check it:

function Rabbit() {}
// by default:
// Rabbit.prototype = { constructor: Rabbit }

alert( Rabbit.prototype.constructor == Rabbit ); // true

Naturally, if we do nothing, the constructor property is available to all rabbits through [[Prototype]]:

function Rabbit() {}
// by default:
// Rabbit.prototype = { constructor: Rabbit }

let rabbit = new Rabbit(); // inherits from {constructor: Rabbit}

alert(rabbit.constructor == Rabbit); // true (from prototype)

We can use constructor property to create a new object using the same constructor as the existing one.

Like here:

function Rabbit(name) {
  this.name = name;
  alert(name);
}

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

let rabbit2 = new rabbit.constructor("Black Rabbit");

That’s handy when we have an object, don’t know which constructor was used for it (e.g. it comes from a 3rd party library), and we need to create another one of the same kind.

But probably the most important thing about "constructor" is that…

…JavaScript itself does not ensure the right "constructor" value.

Yes, it exists in the default "prototype" for functions, but that’s all. What happens with it later – is totally on us.

In particular, if we replace the default prototype as a whole, then there will be no "constructor" in it.

For instance:

function Rabbit() {}
Rabbit.prototype = {
  jumps: true
};

let rabbit = new Rabbit();
alert(rabbit.constructor === Rabbit); // false

So, to keep the right "constructor" we can choose to add/remove properties to the default "prototype" instead of overwriting it as a whole:

          function Rabbit() {}

// Not overwrite Rabbit.prototype totally
// just add to it
Rabbit.prototype.jumps = true
// the default Rabbit.prototype.constructor is preserved
        

Or, alternatively, recreate the constructor property manually:

          Rabbit.prototype = {
  jumps: true,
  constructor: Rabbit
};

// now constructor is also correct, because we added it
        

Summary

In this chapter we briefly described the way of setting a [[Prototype]] for objects created via a constructor function. Later we’ll see more advanced programming patterns that rely on it.

Everything is quite simple, just a few notes to make things clear:

The F.prototype property (don’t mistake it for [[Prototype]]) sets [[Prototype]] of new objects when new F() is called.
The value of F.prototype should be either an object or null: other values won’t work.
The "prototype" property only has such a special effect when set on a constructor function, and invoked with new.

On regular objects the prototype is nothing special:

          let user = {
  name: "John",
  prototype: "Bla-bla" // no magic at all
};
        

By default all functions have F.prototype = { constructor: F }, so we can get the constructor of an object by accessing its "constructor" property.

Tasks

Changing "prototype"

In the code below we create new Rabbit, and then try to modify its prototype.

In the start, we have this code:

function Rabbit() {}
Rabbit.prototype = {
  eats: true
};

let rabbit = new Rabbit();

alert( rabbit.eats ); // true

We added one more string (emphasized). What will alert show now?

          function Rabbit() {}
Rabbit.prototype = {
  eats: true
};

let rabbit = new Rabbit();

Rabbit.prototype = {};

alert( rabbit.eats ); // ?
        

…And if the code is like this (replaced one line)?

          function Rabbit() {}
Rabbit.prototype = {
  eats: true
};

let rabbit = new Rabbit();

Rabbit.prototype.eats = false;

alert( rabbit.eats ); // ?
        

And like this (replaced one line)?

          function Rabbit() {}
Rabbit.prototype = {
  eats: true
};

let rabbit = new Rabbit();

delete rabbit.eats;

alert( rabbit.eats ); // ?
        

The last variant:

          function Rabbit() {}
Rabbit.prototype = {
  eats: true
};

let rabbit = new Rabbit();

delete Rabbit.prototype.eats;

alert( rabbit.eats ); // ?
        

Answers:

true.

The assignment to Rabbit.prototype sets up [[Prototype]] for new objects, but it does not affect the existing ones.
false.

Objects are assigned by reference. The object from Rabbit.prototype is not duplicated, it’s still a single object referenced both by Rabbit.prototype and by the [[Prototype]] of rabbit.

So when we change its content through one reference, it is visible through the other one.
true.

All delete operations are applied directly to the object. Here delete rabbit.eats tries to remove eats property from rabbit, but it doesn’t have it. So the operation won’t have any effect.
undefined.

The property eats is deleted from the prototype, it doesn’t exist any more.

Create an object with the same constructor

Imagine, we have an arbitrary object obj, created by a constructor function – we don’t know which one, but we’d like to create a new object using it.

Can we do it like that?

let obj2 = new obj.constructor();

Give an example of a constructor function for obj which lets such code work right. And an example that makes it work wrong.

We can use such approach if we are sure that "constructor" property has the correct value.

For instance, if we don’t touch the default "prototype", then this code works for sure:

function User(name) {
  this.name = name;
}

let user = new User('John');
let user2 = new user.constructor('Pete');

alert( user2.name ); // Pete (worked!)

It worked, because User.prototype.constructor == User.

…But if someone, so to speak, overwrites User.prototype and forgets to recreate constructor to reference User, then it would fail.

For instance:

function User(name) {
  this.name = name;
}
User.prototype = {}; // (*)

let user = new User('John');
let user2 = new user.constructor('Pete');

alert( user2.name ); // undefined

Why user2.name is undefined?

Here’s how new user.constructor('Pete') works:

First, it looks for constructor in user. Nothing.
Then it follows the prototype chain. The prototype of user is User.prototype, and it also has no constructor (because we “forgot” to set it right!).
Going further up the chain, User.prototype is a plain object, its prototype is the built-in Object.prototype.
Finally, for the built-in Object.prototype, there’s a built-in Object.prototype.constructor == Object. So it is used.

Finally, at the end, we have let user2 = new Object('Pete').

Probably, that’s not what we want. We’d like to create new User, not new Object. That’s the outcome of the missing constructor.

(Just in case you’re curious, the new Object(...) call converts its argument to an object. That’s a theoretical thing, in practice no one calls new Object with a value, and generally we don’t use new Object to make objects at all).

Tutorial map