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Why can't i set the 'prototype' of a function created using 'bind'?

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6

3

consider the code:

function foo(something) {

  this.a = something;

}



var obj1 = {};



var bar = foo.bind(obj1);

now the following statement doesn't execute:

bar.prototype.newprop = "new";//cannot execute this

as i understood,every function has a prototype object. then why can't we execute the above statement.
and bar is indeed a function as we can call it:

bar(2);

console.log(obj1.a); // 2

javascript prototype

share|improve this question

edited 1 hour ago

vikrant

asked 1 hour ago

vikrantvikrant

448613

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  What are you trying to do by adding a prototype property to that function?
  
  – Pointy
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Pointy just trying to understand prototypes and bind
  
  – vikrant
  1 hour ago
  

  
  
  
  

add a comment | 

6

3

consider the code:

function foo(something) {

  this.a = something;

}



var obj1 = {};



var bar = foo.bind(obj1);

now the following statement doesn't execute:

bar.prototype.newprop = "new";//cannot execute this

as i understood,every function has a prototype object. then why can't we execute the above statement.
and bar is indeed a function as we can call it:

bar(2);

console.log(obj1.a); // 2

javascript prototype

share|improve this question

edited 1 hour ago

vikrant

asked 1 hour ago

vikrantvikrant

448613

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  What are you trying to do by adding a prototype property to that function?
  
  – Pointy
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Pointy just trying to understand prototypes and bind
  
  – vikrant
  1 hour ago
  

  
  
  
  

add a comment | 

consider the code:

function foo(something) {

  this.a = something;

}



var obj1 = {};



var bar = foo.bind(obj1);

now the following statement doesn't execute:

bar.prototype.newprop = "new";//cannot execute this

as i understood,every function has a prototype object. then why can't we execute the above statement.
and bar is indeed a function as we can call it:

bar(2);

console.log(obj1.a); // 2

javascript prototype

share|improve this question

edited 1 hour ago

vikrant

asked 1 hour ago

vikrantvikrant

448613

function foo(something) {

  this.a = something;

}



var obj1 = {};



var bar = foo.bind(obj1);

bar.prototype.newprop = "new";//cannot execute this

bar(2);

console.log(obj1.a); // 2

share|improve this question

edited 1 hour ago

vikrant

asked 1 hour ago

vikrantvikrant

448613

share|improve this question

edited 1 hour ago

vikrant

asked 1 hour ago

vikrantvikrant

448613

asked 1 hour ago

vikrantvikrant

448613

asked 1 hour ago

vikrantvikrant

448613

4 Answers
4

active

oldest

votes

9

as i understood, every function has a prototype object

Well, there are exceptions to every rule :-) You found one: bound functions don't have a .prototype property because they don't need it. When you call a bound function with new, it calls the original function as a constructor, using the original's .prototype object as the prototype of the new instance.

In fact, since ES6 many functions don't have a .prototype property with an object, because they are not constructors - they cannot be called with new so they don't need it. Among those are

• arrow functions (() => {…})


• methods (method() { … } in object literals and classes)


• builtin non-constructor functions (like Math.sin)

share|improve this answer

answered 1 hour ago

BergiBergi

373k58561889

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Can you tell us, when to use bind method instead of call and apply.
  
  – Bear Nithi
  4 mins ago
  

  
  
  
  

add a comment | 

4

See the specification:

Function.prototype.bind ( thisArg , ...args)

[...]

NOTE 1 Function objects created using Function.prototype.bind are exotic objects. They also do not have a prototype property.

share|improve this answer

answered 1 hour ago

strstr

17.8k65579

add a comment | 

3

The returned function from .bind() has no prototype object. You can give it one:

bar.prototype = { newprop: "new" };

It is not true that "every function has a prototype object". Every function can have a prototype object, but the value of the "prototype" property can be anything, including null or undefined.

Additionally, there are "special" functions that may not behave like ordinary functions in all cases.

share|improve this answer

answered 1 hour ago

PointyPointy

318k44458521

add a comment | 

2

Adding properties to the prototype means you want to create an object by using the function as a constructor.

When you create an object like by calling new on a function, this value is the new object being created. So it makes no sense to bind this to another value

share|improve this answer

answered 1 hour ago

YoukouleleYYoukouleleY

2,5981826

add a comment | 

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9

as i understood, every function has a prototype object

Well, there are exceptions to every rule :-) You found one: bound functions don't have a .prototype property because they don't need it. When you call a bound function with new, it calls the original function as a constructor, using the original's .prototype object as the prototype of the new instance.

In fact, since ES6 many functions don't have a .prototype property with an object, because they are not constructors - they cannot be called with new so they don't need it. Among those are

• arrow functions (() => {…})


• methods (method() { … } in object literals and classes)


• builtin non-constructor functions (like Math.sin)

share|improve this answer

answered 1 hour ago

BergiBergi

373k58561889

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Can you tell us, when to use bind method instead of call and apply.
  
  – Bear Nithi
  4 mins ago
  

  
  
  
  

add a comment | 

9

as i understood, every function has a prototype object

Well, there are exceptions to every rule :-) You found one: bound functions don't have a .prototype property because they don't need it. When you call a bound function with new, it calls the original function as a constructor, using the original's .prototype object as the prototype of the new instance.

In fact, since ES6 many functions don't have a .prototype property with an object, because they are not constructors - they cannot be called with new so they don't need it. Among those are

• arrow functions (() => {…})


• methods (method() { … } in object literals and classes)


• builtin non-constructor functions (like Math.sin)

share|improve this answer

answered 1 hour ago

BergiBergi

373k58561889

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Can you tell us, when to use bind method instead of call and apply.
  
  – Bear Nithi
  4 mins ago
  

  
  
  
  

add a comment | 

as i understood, every function has a prototype object

Well, there are exceptions to every rule :-) You found one: bound functions don't have a .prototype property because they don't need it. When you call a bound function with new, it calls the original function as a constructor, using the original's .prototype object as the prototype of the new instance.

In fact, since ES6 many functions don't have a .prototype property with an object, because they are not constructors - they cannot be called with new so they don't need it. Among those are

• arrow functions (() => {…})


• methods (method() { … } in object literals and classes)


• builtin non-constructor functions (like Math.sin)

share|improve this answer

answered 1 hour ago

BergiBergi

373k58561889

share|improve this answer

answered 1 hour ago

BergiBergi

373k58561889

answered 1 hour ago

BergiBergi

373k58561889

answered 1 hour ago

BergiBergi

373k58561889

4

See the specification:

Function.prototype.bind ( thisArg , ...args)

[...]

NOTE 1 Function objects created using Function.prototype.bind are exotic objects. They also do not have a prototype property.

share|improve this answer

answered 1 hour ago

strstr

17.8k65579

add a comment | 

4

See the specification:

Function.prototype.bind ( thisArg , ...args)

[...]

NOTE 1 Function objects created using Function.prototype.bind are exotic objects. They also do not have a prototype property.

share|improve this answer

answered 1 hour ago

strstr

17.8k65579

add a comment | 

See the specification:

Function.prototype.bind ( thisArg , ...args)

[...]

NOTE 1 Function objects created using Function.prototype.bind are exotic objects. They also do not have a prototype property.

share|improve this answer

answered 1 hour ago

strstr

17.8k65579

share|improve this answer

answered 1 hour ago

strstr

17.8k65579

answered 1 hour ago

strstr

17.8k65579

answered 1 hour ago

strstr

17.8k65579

3

The returned function from .bind() has no prototype object. You can give it one:

bar.prototype = { newprop: "new" };

It is not true that "every function has a prototype object". Every function can have a prototype object, but the value of the "prototype" property can be anything, including null or undefined.

Additionally, there are "special" functions that may not behave like ordinary functions in all cases.

share|improve this answer

answered 1 hour ago

PointyPointy

318k44458521

add a comment | 

3

The returned function from .bind() has no prototype object. You can give it one:

bar.prototype = { newprop: "new" };

It is not true that "every function has a prototype object". Every function can have a prototype object, but the value of the "prototype" property can be anything, including null or undefined.

Additionally, there are "special" functions that may not behave like ordinary functions in all cases.

share|improve this answer

answered 1 hour ago

PointyPointy

318k44458521

add a comment | 

The returned function from .bind() has no prototype object. You can give it one:

bar.prototype = { newprop: "new" };

It is not true that "every function has a prototype object". Every function can have a prototype object, but the value of the "prototype" property can be anything, including null or undefined.

Additionally, there are "special" functions that may not behave like ordinary functions in all cases.

share|improve this answer

answered 1 hour ago

PointyPointy

318k44458521

bar.prototype = { newprop: "new" };

share|improve this answer

answered 1 hour ago

PointyPointy

318k44458521

answered 1 hour ago

PointyPointy

318k44458521

answered 1 hour ago

PointyPointy

318k44458521

2

Adding properties to the prototype means you want to create an object by using the function as a constructor.

When you create an object like by calling new on a function, this value is the new object being created. So it makes no sense to bind this to another value

share|improve this answer

answered 1 hour ago

YoukouleleYYoukouleleY

2,5981826

add a comment | 

2

Adding properties to the prototype means you want to create an object by using the function as a constructor.

When you create an object like by calling new on a function, this value is the new object being created. So it makes no sense to bind this to another value

share|improve this answer

answered 1 hour ago

YoukouleleYYoukouleleY

2,5981826

add a comment | 

Adding properties to the prototype means you want to create an object by using the function as a constructor.

When you create an object like by calling new on a function, this value is the new object being created. So it makes no sense to bind this to another value

share|improve this answer

answered 1 hour ago

YoukouleleYYoukouleleY

2,5981826

share|improve this answer

answered 1 hour ago

YoukouleleYYoukouleleY

2,5981826

answered 1 hour ago

YoukouleleYYoukouleleY

2,5981826

answered 1 hour ago

YoukouleleYYoukouleleY

2,5981826