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Prototype : difference between using constructor and just object itself

What is the reason to use the 'new' keyword at Derived.prototype = new BaseWhat techniques can be used to define a class in JavaScript, and what are their trade-offs?Difference between == and === in JavaScriptHow does JavaScript .prototype work?What's the difference between using “let” and “var”?What is the difference between call and apply?Understanding the difference between Object.create() and new SomeFunction()What is the explanation for these bizarre JavaScript behaviours mentioned in the 'Wat' talk for CodeMash 2012?Differences between lodash and underscoreWhat is the difference between Bower and npm?Trying to understand the difference between prototype and constructor in JavaScript

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0

I'm studying javascript prototype.

CASE 01 - works!

var A = function() 
A.prototype.hello = function() 
 console.log('hello');


var B = function() 
B.prototype = new A();

var C = function ()
C.prototype = new B();

var c = new C();
c.hello();

CASE 02 - doesn't work!

var A = 
 hello : function() 
 console.log('hello');
 

var B = 
B.prototype = A;
var C = 
C.prototype = B;
C.hello();

I think I didn't understand the concept of prototype and constructor or how the objects inherit each other. Can anybody explain why CASE 02 doesn't work?

javascript

share|improve this question

asked Mar 26 at 3:39

LemileLemile

2033 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I'd argue that it doesn't actually work
  
  – Bergi
  Mar 26 at 7:30
  

  
  
  
  

add a comment | 

0

I'm studying javascript prototype.

CASE 01 - works!

var A = function() 
A.prototype.hello = function() 
 console.log('hello');


var B = function() 
B.prototype = new A();

var C = function ()
C.prototype = new B();

var c = new C();
c.hello();

CASE 02 - doesn't work!

var A = 
 hello : function() 
 console.log('hello');
 

var B = 
B.prototype = A;
var C = 
C.prototype = B;
C.hello();

I think I didn't understand the concept of prototype and constructor or how the objects inherit each other. Can anybody explain why CASE 02 doesn't work?

javascript

share|improve this question

asked Mar 26 at 3:39

LemileLemile

2033 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I'd argue that it doesn't actually work
  
  – Bergi
  Mar 26 at 7:30
  

  
  
  
  

add a comment | 

I'm studying javascript prototype.

CASE 01 - works!

var A = function() 
A.prototype.hello = function() 
 console.log('hello');


var B = function() 
B.prototype = new A();

var C = function ()
C.prototype = new B();

var c = new C();
c.hello();

CASE 02 - doesn't work!

var A = 
 hello : function() 
 console.log('hello');
 

var B = 
B.prototype = A;
var C = 
C.prototype = B;
C.hello();

I think I didn't understand the concept of prototype and constructor or how the objects inherit each other. Can anybody explain why CASE 02 doesn't work?

javascript

share|improve this question

asked Mar 26 at 3:39

LemileLemile

2033 bronze badges

var A = function() 
A.prototype.hello = function() 
 console.log('hello');


var B = function() 
B.prototype = new A();

var C = function ()
C.prototype = new B();

var c = new C();
c.hello();

var A = 
 hello : function() 
 console.log('hello');
 

var B = 
B.prototype = A;
var C = 
C.prototype = B;
C.hello();

share|improve this question

asked Mar 26 at 3:39

LemileLemile

2033 bronze badges

share|improve this question

asked Mar 26 at 3:39

LemileLemile

2033 bronze badges

asked Mar 26 at 3:39

LemileLemile

2033 bronze badges

asked Mar 26 at 3:39

LemileLemile

2033 bronze badges

3 Answers
3

active

oldest

votes

0

B.prototype is just going to be a key on B object since B.prototype = A is going to create a key name prototype on B. Same happens with C , it will create a nested key of prototype

var A = 
 hello: function() 
 console.log('hello');
 

var B = 
B.prototype = A;
console.log(B)
var C = 
C.prototype = B;
console.log(C)
C.prototype.prototype.hello();

share|improve this answer

answered Mar 26 at 3:46

brkbrk

32.3k33 gold badges2525 silver badges4747 bronze badges

add a comment | 

0

I can not really explain, but for me it's just a question of syntax:

var A = 
 hello : function() 
 console.log('hello');
 


var X = Object.create(A);
X.hello();

share|improve this answer

answered Mar 26 at 3:54

Mister JojoMister Jojo

2,28322 gold badges22 silver badges1818 bronze badges

add a comment | 

0

I think the point to note is that prototype is a property of function AND not an object. Case 2 does NOT work mainly because of this reason.

share|improve this answer

answered Mar 26 at 3:58

82Tuskers82Tuskers

1,08766 silver badges1212 bronze badges

add a comment | 

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0

B.prototype is just going to be a key on B object since B.prototype = A is going to create a key name prototype on B. Same happens with C , it will create a nested key of prototype

var A = 
 hello: function() 
 console.log('hello');
 

var B = 
B.prototype = A;
console.log(B)
var C = 
C.prototype = B;
console.log(C)
C.prototype.prototype.hello();

share|improve this answer

answered Mar 26 at 3:46

brkbrk

32.3k33 gold badges2525 silver badges4747 bronze badges

add a comment | 

0

B.prototype is just going to be a key on B object since B.prototype = A is going to create a key name prototype on B. Same happens with C , it will create a nested key of prototype

var A = 
 hello: function() 
 console.log('hello');
 

var B = 
B.prototype = A;
console.log(B)
var C = 
C.prototype = B;
console.log(C)
C.prototype.prototype.hello();

share|improve this answer

answered Mar 26 at 3:46

brkbrk

32.3k33 gold badges2525 silver badges4747 bronze badges

add a comment | 

B.prototype is just going to be a key on B object since B.prototype = A is going to create a key name prototype on B. Same happens with C , it will create a nested key of prototype

var A = 
 hello: function() 
 console.log('hello');
 

var B = 
B.prototype = A;
console.log(B)
var C = 
C.prototype = B;
console.log(C)
C.prototype.prototype.hello();

share|improve this answer

answered Mar 26 at 3:46

brkbrk

32.3k33 gold badges2525 silver badges4747 bronze badges

var A = 
 hello: function() 
 console.log('hello');
 

var B = 
B.prototype = A;
console.log(B)
var C = 
C.prototype = B;
console.log(C)
C.prototype.prototype.hello();

var A = 
 hello: function() 
 console.log('hello');
 

var B = 
B.prototype = A;
console.log(B)
var C = 
C.prototype = B;
console.log(C)
C.prototype.prototype.hello();

var A = 
 hello: function() 
 console.log('hello');
 

var B = 
B.prototype = A;
console.log(B)
var C = 
C.prototype = B;
console.log(C)
C.prototype.prototype.hello();

share|improve this answer

answered Mar 26 at 3:46

brkbrk

32.3k33 gold badges2525 silver badges4747 bronze badges

answered Mar 26 at 3:46

brkbrk

32.3k33 gold badges2525 silver badges4747 bronze badges

answered Mar 26 at 3:46

brkbrk

32.3k33 gold badges2525 silver badges4747 bronze badges

0

I can not really explain, but for me it's just a question of syntax:

var A = 
 hello : function() 
 console.log('hello');
 


var X = Object.create(A);
X.hello();

share|improve this answer

answered Mar 26 at 3:54

Mister JojoMister Jojo

2,28322 gold badges22 silver badges1818 bronze badges

add a comment | 

0

I can not really explain, but for me it's just a question of syntax:

var A = 
 hello : function() 
 console.log('hello');
 


var X = Object.create(A);
X.hello();

share|improve this answer

answered Mar 26 at 3:54

Mister JojoMister Jojo

2,28322 gold badges22 silver badges1818 bronze badges

add a comment | 

I can not really explain, but for me it's just a question of syntax:

var A = 
 hello : function() 
 console.log('hello');
 


var X = Object.create(A);
X.hello();

share|improve this answer

answered Mar 26 at 3:54

Mister JojoMister Jojo

2,28322 gold badges22 silver badges1818 bronze badges

var A = 
 hello : function() 
 console.log('hello');
 


var X = Object.create(A);
X.hello();

var A = 
 hello : function() 
 console.log('hello');
 


var X = Object.create(A);
X.hello();

var A = 
 hello : function() 
 console.log('hello');
 


var X = Object.create(A);
X.hello();

share|improve this answer

answered Mar 26 at 3:54

Mister JojoMister Jojo

2,28322 gold badges22 silver badges1818 bronze badges

answered Mar 26 at 3:54

Mister JojoMister Jojo

2,28322 gold badges22 silver badges1818 bronze badges

answered Mar 26 at 3:54

Mister JojoMister Jojo

2,28322 gold badges22 silver badges1818 bronze badges

0

I think the point to note is that prototype is a property of function AND not an object. Case 2 does NOT work mainly because of this reason.

share|improve this answer

answered Mar 26 at 3:58

82Tuskers82Tuskers

1,08766 silver badges1212 bronze badges

add a comment | 

0

I think the point to note is that prototype is a property of function AND not an object. Case 2 does NOT work mainly because of this reason.

share|improve this answer

answered Mar 26 at 3:58

82Tuskers82Tuskers

1,08766 silver badges1212 bronze badges

add a comment | 

I think the point to note is that prototype is a property of function AND not an object. Case 2 does NOT work mainly because of this reason.

share|improve this answer

answered Mar 26 at 3:58

82Tuskers82Tuskers

1,08766 silver badges1212 bronze badges

share|improve this answer

answered Mar 26 at 3:58

82Tuskers82Tuskers

1,08766 silver badges1212 bronze badges

answered Mar 26 at 3:58

82Tuskers82Tuskers

1,08766 silver badges1212 bronze badges

answered Mar 26 at 3:58

82Tuskers82Tuskers

1,08766 silver badges1212 bronze badges