CODEWITHSUNDEEP
javascript
and-k
and-k

Reputation: 607

Constructor and prototype in javascript

What is the difference between this two codes and which one should I use?

function Test() {}
Test.method = function() {};

With Prototype:

function Test() {}
Test.prototype.method = function() {};

Upvotes: 4

Views: 99

Answers (6)

Raj Nathani
Raj Nathani

Reputation: 2845

Will NOT Work (for an instance)

function Test() {}
Test.method = function() {};

Will Work (for an instance)

function Test() {}
Test.prototype.method = function() {};

Why?

Assuming that you create an instance of Test:

t = new Test()

On calling a method:

t.method()

JavaScript will first look within Test, for example the 3 properties and methods will be recognized below:

function Test() {} {

    this.a = "hello";
    this.b = "world";
    this.c = function() { alert( this.a + this.b) }

}

and then the second, the look up will be done on prototype:

Test.prototype.method = function() { console.log('prototype defined method') }

In JavaScript the below does not register to the instance t but instead Test:

Test.method = function() { console.log('something unsual') };

therefore

Test.method()  // works
t.method()     // no works

Upvotes: 0

Wayne
Wayne

Reputation: 60414

The first example merely assigns a function to a new property of Test called method. There is nothing special or magic about it (by which I mean, the language does not do anything interesting with it). You might call this a static method, because it's shared across all instances, but it's really just a normal property. (Functions are objects and objects have properties.)

In the second case, a new method is added to Test's prototype, which makes it available to all instances of Test. This is special behavior. Every object in JavaScript has an internal link to another object, which is called its prototype. This link is established using the prototype of the constructor that created the object, so any instance created by Test will have the new property in its prototype chain.

For example:

function Test() {}
Test.method = function() {};

var a = new Test();
console.log(a.method); // => undefined

But:

function Test() {} 
Test.prototype.method = function() {};

var a = new Test();
console.log(a.method); // => function

Upvotes: 1

Josh Beam
Josh Beam

Reputation: 19772

If you define a method in the prototype, any instances will be able to use that method. If you define the method as simply a part of the original object, it allows you to simply use that in a manner that one could reasonably refer to as a static method.

For example:

Array.isArray( [] )  //true

[].isArray() //throws a TypeError

The second example throws a TypeError because isArray() is not in the Array.prototype (remember, instances only access methods defined in that instance, or in that instance's prototype or __proto__ chain).

In short:
* A method defined in the original object can be accessed by that object, but not instances of that object.
* A method defined in the original object's prototype can be accessed by any instance of the original object.

Upvotes: 0

user1636522
user1636522

Reputation:

1st case : static method.

function Test() {}
Test.method = function () { alert(1); };
var t = new Test;
Test.method(); // alerts "1"
t.method(); // TypeError: Object #<Test> has no method 'method'

2nd case : instance method.

function Test() {}
Test.prototype.method = function () { alert(1); };
var t1 = new Test;
var t2 = new Test;
t1.method(); // alerts "1"
t2.method(); // alerts "1"
Test.method(); // TypeError: Object function Test() {} has no method 'method'

Upvotes: 5

ppoliani
ppoliani

Reputation: 4906

The difference goes back to how the prototype chaining works. If you create instances of Test then they will be able to access the methods that are defined on the prototype.

The first code, simply adds a "static" method on the constructor; You won't be able to access that method doing something like this:

var testInst = new Test();
testInst.method() // Error Undefined is not a function

The only way to invoke that method is via the constructor reference i.e.

Test.method();

As i mentioned, the above is feasible if you add the method to the prototype of the constructor.

Upvotes: 0

Sterling Archer
Sterling Archer

Reputation: 22395

The prototype method is shared between all instances of Test.

Upvotes: 0

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