[javascript] JavaScript: Class.method vs. Class.prototype.method

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?

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.

A. Static Method:

      Class.method = function () { /* code */ }

1. method() here is a function property added to an another function (here Class).
2. You can directly access the method() by the class / function name. Class.method();
3. No need for creating any object/instance (new Class()) for accessing the method(). So you could call it as a static method.

B. Prototype Method (Shared across all the instances):

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

1. method() here is a function property added to an another function protype (here Class.prototype).
2. You can either directly access by class name or by an object/instance (new Class()).
3. Added advantage - this way of method() definition will create only one copy of method() in the memory and will be shared across all the object's/instance's created from the Class

C. Class Method (Each instance has its own copy):

   function Class () {
      this.method = function () { /* do something with the private members */};
   }

1. method() here is a method defined inside an another function (here Class).
2. You can't directly access the method() by the class / function name. Class.method();
3. You need to create an object/instance (new Class()) for the method() access.
4. This way of method() definition will create a unique copy of the method() for each and every objects created using the constructor function (new Class()).
5. Added advantage - Bcos of the method() scope it has the full right to access the local members(also called private members) declared inside the constructor function (here Class)

Example:

    function Class() {
        var str = "Constructor method"; // private variable
        this.method = function () { console.log(str); };
    }
    Class.prototype.method = function() { console.log("Prototype method"); };
    Class.method = function() { console.log("Static method"); };

    new Class().method();     // Constructor method
    // Bcos Constructor method() has more priority over the Prototype method()

    // Bcos of the existence of the Constructor method(), the Prototype method 
    // will not be looked up. But you call it by explicity, if you want.
    // Using instance
    new Class().constructor.prototype.method(); // Prototype method

    // Using class name
    Class.prototype.method(); // Prototype method

    // Access the static method by class name
    Class.method();           // Static method

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.

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.
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