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Not able to use 'super' with function defined on protoype object in JavaScript class

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0

I have class A which is parent of class B

class A 
 constructor(a)
 this.a=a;
 

 par()
 console.log("para");
 


class B extends A 
 constructor(a) 
 super(a)
 this.a = "child";
 

 par() 
 super.par();
 console.log("child");
 

When I use this code, it works fine.

But when I explicitly define the par function on B using this code:

B.prototype.par = function() 
 super.par();

I get the error

Uncaught SyntaxError: 'super' keyword unexpected here

Whether we create a function in class definition or in prototype object of function('class'), it should be the same thing.
What am I doing wrong here?

javascript class inheritance super

share|improve this question

edited Mar 26 at 7:37

TiiJ7

2,75733 gold badges1818 silver badges2929 bronze badges

asked Mar 26 at 6:33

Suraj Prakash JoshiSuraj Prakash Joshi

8811 silver badge77 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  No, you can't use super in a function. Just call A.prototype.par.
  
  – Bergi
  Mar 26 at 7:27
  

  
  
  
  

add a comment | 

0

I have class A which is parent of class B

class A 
 constructor(a)
 this.a=a;
 

 par()
 console.log("para");
 


class B extends A 
 constructor(a) 
 super(a)
 this.a = "child";
 

 par() 
 super.par();
 console.log("child");
 

When I use this code, it works fine.

But when I explicitly define the par function on B using this code:

B.prototype.par = function() 
 super.par();

I get the error

Uncaught SyntaxError: 'super' keyword unexpected here

Whether we create a function in class definition or in prototype object of function('class'), it should be the same thing.
What am I doing wrong here?

javascript class inheritance super

share|improve this question

edited Mar 26 at 7:37

TiiJ7

2,75733 gold badges1818 silver badges2929 bronze badges

asked Mar 26 at 6:33

Suraj Prakash JoshiSuraj Prakash Joshi

8811 silver badge77 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  No, you can't use super in a function. Just call A.prototype.par.
  
  – Bergi
  Mar 26 at 7:27
  

  
  
  
  

add a comment | 

I have class A which is parent of class B

class A 
 constructor(a)
 this.a=a;
 

 par()
 console.log("para");
 


class B extends A 
 constructor(a) 
 super(a)
 this.a = "child";
 

 par() 
 super.par();
 console.log("child");
 

When I use this code, it works fine.

But when I explicitly define the par function on B using this code:

B.prototype.par = function() 
 super.par();

I get the error

Uncaught SyntaxError: 'super' keyword unexpected here

Whether we create a function in class definition or in prototype object of function('class'), it should be the same thing.
What am I doing wrong here?

javascript class inheritance super

share|improve this question

edited Mar 26 at 7:37

TiiJ7

2,75733 gold badges1818 silver badges2929 bronze badges

asked Mar 26 at 6:33

Suraj Prakash JoshiSuraj Prakash Joshi

8811 silver badge77 bronze badges

class A 
 constructor(a)
 this.a=a;
 

 par()
 console.log("para");
 


class B extends A 
 constructor(a) 
 super(a)
 this.a = "child";
 

 par() 
 super.par();
 console.log("child");
 

B.prototype.par = function() 
 super.par();

share|improve this question

edited Mar 26 at 7:37

TiiJ7

2,75733 gold badges1818 silver badges2929 bronze badges

asked Mar 26 at 6:33

Suraj Prakash JoshiSuraj Prakash Joshi

8811 silver badge77 bronze badges

share|improve this question

edited Mar 26 at 7:37

TiiJ7

2,75733 gold badges1818 silver badges2929 bronze badges

asked Mar 26 at 6:33

Suraj Prakash JoshiSuraj Prakash Joshi

8811 silver badge77 bronze badges

edited Mar 26 at 7:37

TiiJ7

2,75733 gold badges1818 silver badges2929 bronze badges

edited Mar 26 at 7:37

TiiJ7

2,75733 gold badges1818 silver badges2929 bronze badges

asked Mar 26 at 6:33

Suraj Prakash JoshiSuraj Prakash Joshi

8811 silver badge77 bronze badges

asked Mar 26 at 6:33

Suraj Prakash JoshiSuraj Prakash Joshi

8811 silver badge77 bronze badges

1 Answer
1

active

oldest

votes

1

'super' is simply a syntactic sugar introduced in ES2015 along with class syntax.
It can only be used within functions of 'class' (constructor and methods) that extends another class.

class A 
 constructor()
 par() console.log('para') 


class B extends A 
 constructor()
 super()
 

Is equivalent to:

function A()
A.prototype.par = function()console.log('para')

var B = (function(parent)
 var _super = parent;
 function B()
 _super.call(this); // calls parent's constructor
 

 B.prototype = Object.create(_super.prototype); // Inherits parent's methods.

 B.prototype.par = function() // override parent's par.
 _super.prototype.par.call(this); // child still has access to parent's par method thanks to closure :)
 console.log('child');
 

 return B;
)(A);

var b = new B();
b.par()

You cannot do:

function()
 super // super is not defined...

share|improve this answer

edited Mar 26 at 8:58

answered Mar 26 at 6:47

dannyjeedannyjee

33822 silver badges77 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  i have doubt in this line 'B.prototype = new super();'..what is it trying to do? because 'b.__proto_' will be equal to B.prototype
  
  – Suraj Prakash Joshi
  Mar 26 at 7:57
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Bergi You're right. It should be Object.create(A.prototype). I corrected the answer.
  
  – dannyjee
  Mar 26 at 8:58
  

  
  
  
  

add a comment | 

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1

'super' is simply a syntactic sugar introduced in ES2015 along with class syntax.
It can only be used within functions of 'class' (constructor and methods) that extends another class.

class A 
 constructor()
 par() console.log('para') 


class B extends A 
 constructor()
 super()
 

Is equivalent to:

function A()
A.prototype.par = function()console.log('para')

var B = (function(parent)
 var _super = parent;
 function B()
 _super.call(this); // calls parent's constructor
 

 B.prototype = Object.create(_super.prototype); // Inherits parent's methods.

 B.prototype.par = function() // override parent's par.
 _super.prototype.par.call(this); // child still has access to parent's par method thanks to closure :)
 console.log('child');
 

 return B;
)(A);

var b = new B();
b.par()

You cannot do:

function()
 super // super is not defined...

share|improve this answer

edited Mar 26 at 8:58

answered Mar 26 at 6:47

dannyjeedannyjee

33822 silver badges77 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  i have doubt in this line 'B.prototype = new super();'..what is it trying to do? because 'b.__proto_' will be equal to B.prototype
  
  – Suraj Prakash Joshi
  Mar 26 at 7:57
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Bergi You're right. It should be Object.create(A.prototype). I corrected the answer.
  
  – dannyjee
  Mar 26 at 8:58
  

  
  
  
  

add a comment | 

1

'super' is simply a syntactic sugar introduced in ES2015 along with class syntax.
It can only be used within functions of 'class' (constructor and methods) that extends another class.

class A 
 constructor()
 par() console.log('para') 


class B extends A 
 constructor()
 super()
 

Is equivalent to:

function A()
A.prototype.par = function()console.log('para')

var B = (function(parent)
 var _super = parent;
 function B()
 _super.call(this); // calls parent's constructor
 

 B.prototype = Object.create(_super.prototype); // Inherits parent's methods.

 B.prototype.par = function() // override parent's par.
 _super.prototype.par.call(this); // child still has access to parent's par method thanks to closure :)
 console.log('child');
 

 return B;
)(A);

var b = new B();
b.par()

You cannot do:

function()
 super // super is not defined...

share|improve this answer

edited Mar 26 at 8:58

answered Mar 26 at 6:47

dannyjeedannyjee

33822 silver badges77 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  i have doubt in this line 'B.prototype = new super();'..what is it trying to do? because 'b.__proto_' will be equal to B.prototype
  
  – Suraj Prakash Joshi
  Mar 26 at 7:57
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Bergi You're right. It should be Object.create(A.prototype). I corrected the answer.
  
  – dannyjee
  Mar 26 at 8:58
  

  
  
  
  

add a comment | 

'super' is simply a syntactic sugar introduced in ES2015 along with class syntax.
It can only be used within functions of 'class' (constructor and methods) that extends another class.

class A 
 constructor()
 par() console.log('para') 


class B extends A 
 constructor()
 super()
 

Is equivalent to:

function A()
A.prototype.par = function()console.log('para')

var B = (function(parent)
 var _super = parent;
 function B()
 _super.call(this); // calls parent's constructor
 

 B.prototype = Object.create(_super.prototype); // Inherits parent's methods.

 B.prototype.par = function() // override parent's par.
 _super.prototype.par.call(this); // child still has access to parent's par method thanks to closure :)
 console.log('child');
 

 return B;
)(A);

var b = new B();
b.par()

You cannot do:

function()
 super // super is not defined...

share|improve this answer

edited Mar 26 at 8:58

answered Mar 26 at 6:47

dannyjeedannyjee

33822 silver badges77 bronze badges

class A 
 constructor()
 par() console.log('para') 


class B extends A 
 constructor()
 super()
 

function A()
A.prototype.par = function()console.log('para')

var B = (function(parent)
 var _super = parent;
 function B()
 _super.call(this); // calls parent's constructor
 

 B.prototype = Object.create(_super.prototype); // Inherits parent's methods.

 B.prototype.par = function() // override parent's par.
 _super.prototype.par.call(this); // child still has access to parent's par method thanks to closure :)
 console.log('child');
 

 return B;
)(A);

var b = new B();
b.par()

function()
 super // super is not defined...

share|improve this answer

edited Mar 26 at 8:58

answered Mar 26 at 6:47

dannyjeedannyjee

33822 silver badges77 bronze badges

answered Mar 26 at 6:47

dannyjeedannyjee

33822 silver badges77 bronze badges

answered Mar 26 at 6:47

dannyjeedannyjee

33822 silver badges77 bronze badges