I am new to JavaScript OOP. Can you please explain the difference between the following blocks of code? I tested and both blocks work. What's the best practice and why?
First block:
function Car(name){
this.Name = name;
}
Car.prototype.Drive = function(){
console.log("My name is " + this.Name + " and I'm driving.");
}
SuperCar.prototype = new Car();
SuperCar.prototype.constructor = SuperCar;
function SuperCar(name){
Car.call(this, name);
}
SuperCar.prototype.Fly = function(){
console.log("My name is " + this.Name + " and I'm flying!");
}
var myCar = new Car("Car");
myCar.Drive();
var mySuperCar = new SuperCar("SuperCar");
mySuperCar.Drive();
mySuperCar.Fly();
Second block:
function Car(name){
this.Name = name;
this.Drive = function(){
console.log("My name is " + this.Name + " and I'm driving.");
}
}
SuperCar.prototype = new Car();
function SuperCar(name){
Car.call(this, name);
this.Fly = function(){
console.log("My name is " + this.Name + " and I'm flying!");
}
}
var myCar = new Car("Car");
myCar.Drive();
var mySuperCar = new SuperCar("SuperCar");
mySuperCar.Drive();
mySuperCar.Fly();
Why did the author add the Drive
and Fly
methods using prototype
, and did not declare them as a this.Drive
method inside the Car
class and as this.Fly
in the SuperCar
class?
Why does SuperCar.prototype.constructor
need to be set back to SuperCar
? Is the constructor
property overridden when prototype
is set? I commented out this line and nothing changed.
Why call Car.call(this, name);
in the SuperCar
constructor? Won't properties and methods of Car
be 'inherited' when I do
var myCar = new Car("Car");
Prototype methods and property created for function abc
Creating new instances for function abc
http://astutejs.blogspot.in/2015/10/javascript-prototype-is-easy.html
Norbert, you should note that your first example is pretty much what Douglas Crockford calls pseudoclassical inheritance. Something things to note about this:
Lastly, I'd like to mention that I have several examples of TDD JavaScript Inheritance code that works here: TDD JavaScript Inheritance Code and Essay I'd love to get your feedback as I'm hoping to improve it and keep it open source. The goal is to help classical programmers get up to speed with JavaScript quickly and also supplement the study both Crockford and Zakas books.
There are several questions here:
The first only creates one
Drive
function, the second creates two of them: one onmyCar
and another one onmySuperCar
.Here is code that would give different results when either the first or second block was executed:
It is better practice to define methods on the prototype, because:
Object.create(Car.prototype)
);The
constructor
property is not overridden whenprototype
is set. But the constructor ofnew Car()
isCar
, so if you setnew Car()
toSuperCar.prototype
, then obviouslySuperCar.prototype.constructor
isCar
.As long as you don't reassign to
prototype
, there is an invariance:Constructor.prototype.constructor === Constructor
. For example, this is true forCar
:Car.prototype.constructor === Car
, but it is equally true forArray
,Object
,String
, ...etc.But if you reassign a different object to
prototype
, that invariance is broken. Usually this is not a problem (as you have noticed), but it is better to reinstate it, because it answers the question "Which constructor uses this prototype object when creating new instances?" Some code may do such inspection and depend on it. See "Why is it necessary to set the prototype constructor?" for such cases.If you don't do
Car.call(this, name);
then yourSuperCar
instance will not have a name property. You could of course decide to just dothis.name = name;
instead, which just copies the code that is in theCar
constructor, but in more complex situations it would be bad practice to have such code duplication.It would not be helpful to call
new Car(name)
within theSuperCar
constructor, as that will create another object, while you really need to extend thethis
object. By not usingnew
(usingcall
instead) you actually tell theCar
function to not run as a constructor (i.e. to not create a new object), but to use the object you pass to it instead.Times have changed
In modern versions of JavaScript you can use
super(name)
instead ofCar.call(this, name)
:Today, you would also use the
class
syntax and write the first code block from the question as follows:Note how you don't even have to mention the
prototype
property to achieve the goal. Theclass ... extends
syntax also takes care of setting theprototype.constructor
property as the first block in your question did.The two blocks differ in a way that in the first example
Drive()
will only exist once while at the second approachDrive()
will exist per instance (Every time you donew Car()
the functiondrive()
will be created again). Or different said the first uses the prototype to store the function and the second the constructor. The lookup for functions is constructor and then prototype. So for your lookup ofDrive()
it finds it regardless if it is in the constructor or in the prototype. Using the prototype is more efficient because usually you need a function only once per type.The
new
call in javascript automatically sets the constructor in the prototype. If you are overwriting the prototype so you have to set the constructor manually.Inheritance in javascript has nothing like
super
. So if you have a subclass the only chance to call the super constructor is by its name.I am not 100% sure, but I believe the difference is that the second example simply duplicates the contents of the Car class into the SuperCar object, while the first links the SuperCar prototype to the Car class, so that run-time changes to the Car class affect the SuperCar class as well.
To add to Norbert Hartl's answer, SuperCar.prototype.constructor isn't needed, but some people use it as a convenient way of getting the constructing function of an object (SuperCar objects in this case).
Just from the first example, Car.call(this, name) is in the SuperCar constructor function because when you do this:
This is what JavaScript does:
Notice how JavaScript didn't call Car for you. Prototypes being as they are, any property or method that you don't set yourself for SuperCar will be looked up in Car. Sometimes this is good, e.g. SuperCar doesn't have a Drive method, but it can share Car's one, so all SuperCars will use the same Drive method. Other times you don't want sharing, like each SuperCar having it's own Name. So how does one go about setting each SuperCar's name to it's own thing? You could set this.Name inside the SuperCar constructor function:
This works, but wait a second. Didn't we do exactly the same thing in the Car constructor? Don't want to repeat ourselves. Since Car sets the name already, let's just call it.
Whoops, you never want to change the special
this
object reference. Remember the 4 steps? Hang onto that object that JavaScript gave you, because it's the only way to keep the precious internal prototype link between your SuperCar object and Car. So how do we set Name, without repeating ourselves and without throwing away our fresh SuperCar object JavaScript spent so much special effort to prepare for us?Two things. One: the meaning of
this
is flexible. Two: Car is a function. It's possible to call Car, not with a pristine, fresh instantiated object, but instead with, say, a SuperCar object. That gives us the final solution, which is part of the first example in your question:As a function, Car is allowed to be invoked with the function's call method, which changes the meaning of
this
within Car to the SuperCar instance we're building up. Presto! Now each SuperCar gets it's own Name property.To wrap up,
Car.call(this, name)
in the SuperCar constructor gives each new SuperCar object it's own unique Name property, but without duplicating the code that's already in Car.Prototypes aren't scary once you understand them, but they're not much like the classic class/inheritence OOP model at all. I wrote an article about the prototypes concept in JavaScript. It's written for a game engine that uses JavaScript, but it's the same JavaScript engine used by Firefox, so it should all be relevant. Hope this helps.