[removed] Class.method vs. Class.prototype.method

Question

What is the difference between the following two declarations?

Class.method = function () { /* code */ }
Class.prototype.method = function () { /* code using this.values */ }

Is it okay to think of the first statement as a declaration of a static method, and the second statement as a declaration of an instance method?

Answer 1

For visual learners, when defining the function without .prototype

ExampleClass = function(){};
ExampleClass.method = function(customString){
             console.log((customString !== undefined)? 
                          customString : 
                          "called from func def.");}
ExampleClass.method(); // >> output: `called from func def.`  

var someInstance = new ExampleClass();
someInstance.method('Called from instance');
    // >> error! `someInstance.method is not a function`  

---

With same code, if .prototype is added,

ExampleClass.prototype.method = function(customString){
             console.log((customString !== undefined)? 
                          customString : 
                          "called from func def.");}
ExampleClass.method();  
      // > error! `ExampleClass.method is not a function.`  

var someInstance = new ExampleClass();
someInstance.method('Called from instance');
                 // > output: `Called from instance`

---

To make it clearer,

ExampleClass = function(){};
ExampleClass.directM = function(){}  //M for method
ExampleClass.prototype.protoM = function(){}

var instanceOfExample = new ExampleClass();

ExampleClass.directM();     ✓ works
instanceOfExample.directM();   x Error!

ExampleClass.protoM();     x Error!
instanceOfExample.protoM();  ✓ works

****Note for the example above, someInstance.method() won't be executed as,
ExampleClass.method() causes error & execution cannot continue.
But for the sake of illustration & easy understanding, I've kept this sequence.****

Results generated from chrome developer console & JS Bin
Click on the jsbin link above to step through the code.
Toggle commented section with ctrl+/

Answer 2

Yes, the first one is a static method also called class method, while the second one is an instance method.

Consider the following examples, to understand it in more detail.

In ES5

function Person(firstName, lastName) {
   this.firstName = firstName;
   this.lastName = lastName;
}

Person.isPerson = function(obj) {
   return obj.constructor === Person;
}

Person.prototype.sayHi = function() {
   return "Hi " + this.firstName;
}

In the above code, isPerson is a static method, while sayHi is an instance method of Person.

Below, is how to create an object from Person constructor.

var aminu = new Person("Aminu", "Abubakar");

Using the static method isPerson.

Person.isPerson(aminu); // will return true

Using the instance method sayHi.

aminu.sayHi(); // will return "Hi Aminu"

In ES6

class Person {
   constructor(firstName, lastName) {
      this.firstName = firstName;
      this.lastName = lastName;
   }

   static isPerson(obj) {
      return obj.constructor === Person;
   }

   sayHi() {
      return `Hi ${this.firstName}`;
   }
}

Look at how static keyword was used to declare the static method isPerson.

To create an object of Person class.

const aminu = new Person("Aminu", "Abubakar");

Using the static method isPerson.

Person.isPerson(aminu); // will return true

Using the instance method sayHi.

aminu.sayHi(); // will return "Hi Aminu"

NOTE: Both examples are essentially the same, JavaScript remains a classless language. The class introduced in ES6 is primarily a syntactical sugar over the existing prototype-based inheritance model.

Answer 3

Yes, the first function has no relationship with an object instance of that constructor function, you can consider it like a 'static method'.

In JavaScript functions are first-class objects, that means you can treat them just like any object, in this case, you are only adding a property to the function object.

The second function, as you are extending the constructor function prototype, it will be available to all the object instances created with the new keyword, and the context within that function (the this keyword) will refer to the actual object instance where you call it.

Consider this example:

// constructor function
function MyClass () {
  var privateVariable; // private member only available within the constructor fn

  this.privilegedMethod = function () { // it can access private members
    //..
  };
}

// A 'static method', it's just like a normal function 
// it has no relation with any 'MyClass' object instance
MyClass.staticMethod = function () {};

MyClass.prototype.publicMethod = function () {
  // the 'this' keyword refers to the object instance
  // you can access only 'privileged' and 'public' members
};

var myObj = new MyClass(); // new object instance

myObj.publicMethod();
MyClass.staticMethod();

Answer 4

When you create more than one instance of MyClass , you will still only have only one instance of publicMethod in memory but in case of privilegedMethod you will end up creating lots of instances and staticMethod has no relationship with an object instance.

That's why prototypes save memory.

Also, if you change the parent object's properties, is the child's corresponding property hasn't been changed, it'll be updated.