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8

2

I have the below constructors and SubType prototype pointing to an instance of SuperType. When I do x.isPrototypeOf(SubType.prototype) it returns false. I am confused as I have explicitly set x as a prototype for SubType. Can someone tell me why this is happening?

function SuperType()
 
function SubType()

x = new SuperType();

SubType.prototype = x;
SubType.prototype.constructor = SubType;

console.log(x.isPrototypeOf(SubType)) // returns false
console.log(SuperType.prototype.isPrototypeOf(SubType.prototype)) // returns true

javascript

share|improve this question

edited Apr 24 at 12:58

Boann

37.7k1291123

asked Apr 24 at 1:56

GautamGautam

610413

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  Not sure to get my head all clear here, but x === SubType.prototype how do you expect it to be its own prototype?
  
  – Kaiido
  Apr 24 at 2:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Updated my question, sorry about that type
  
  – Gautam
  Apr 24 at 2:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  try console.log(x.isPrototypeOf(new SubType)) for example of how it's used.
  
  – dandavis
  Apr 24 at 2:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Notice that you better should be using x = Object.create(SuperType.prototype)
  
  – Bergi
  Apr 24 at 7:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Possible duplicate of [[Prototype]] vs prototype: ..what is the difference?
  
  – Bergi
  Apr 24 at 7:13
  
  

  
  
  
  

add a comment | 

8

2

I have the below constructors and SubType prototype pointing to an instance of SuperType. When I do x.isPrototypeOf(SubType.prototype) it returns false. I am confused as I have explicitly set x as a prototype for SubType. Can someone tell me why this is happening?

function SuperType()
 
function SubType()

x = new SuperType();

SubType.prototype = x;
SubType.prototype.constructor = SubType;

console.log(x.isPrototypeOf(SubType)) // returns false
console.log(SuperType.prototype.isPrototypeOf(SubType.prototype)) // returns true

javascript

share|improve this question

edited Apr 24 at 12:58

Boann

37.7k1291123

asked Apr 24 at 1:56

GautamGautam

610413

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  Not sure to get my head all clear here, but x === SubType.prototype how do you expect it to be its own prototype?
  
  – Kaiido
  Apr 24 at 2:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Updated my question, sorry about that type
  
  – Gautam
  Apr 24 at 2:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  try console.log(x.isPrototypeOf(new SubType)) for example of how it's used.
  
  – dandavis
  Apr 24 at 2:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Notice that you better should be using x = Object.create(SuperType.prototype)
  
  – Bergi
  Apr 24 at 7:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Possible duplicate of [[Prototype]] vs prototype: ..what is the difference?
  
  – Bergi
  Apr 24 at 7:13
  
  

  
  
  
  

add a comment | 

I have the below constructors and SubType prototype pointing to an instance of SuperType. When I do x.isPrototypeOf(SubType.prototype) it returns false. I am confused as I have explicitly set x as a prototype for SubType. Can someone tell me why this is happening?

function SuperType()
 
function SubType()

x = new SuperType();

SubType.prototype = x;
SubType.prototype.constructor = SubType;

console.log(x.isPrototypeOf(SubType)) // returns false
console.log(SuperType.prototype.isPrototypeOf(SubType.prototype)) // returns true

javascript

share|improve this question

edited Apr 24 at 12:58

Boann

37.7k1291123

asked Apr 24 at 1:56

GautamGautam

610413

function SuperType()
 
function SubType()

x = new SuperType();

SubType.prototype = x;
SubType.prototype.constructor = SubType;

console.log(x.isPrototypeOf(SubType)) // returns false
console.log(SuperType.prototype.isPrototypeOf(SubType.prototype)) // returns true

function SuperType()
 
function SubType()

x = new SuperType();

SubType.prototype = x;
SubType.prototype.constructor = SubType;

console.log(x.isPrototypeOf(SubType)) // returns false
console.log(SuperType.prototype.isPrototypeOf(SubType.prototype)) // returns true

function SuperType()
 
function SubType()

x = new SuperType();

SubType.prototype = x;
SubType.prototype.constructor = SubType;

console.log(x.isPrototypeOf(SubType)) // returns false
console.log(SuperType.prototype.isPrototypeOf(SubType.prototype)) // returns true

share|improve this question

edited Apr 24 at 12:58

Boann

37.7k1291123

asked Apr 24 at 1:56

GautamGautam

610413

share|improve this question

edited Apr 24 at 12:58

Boann

37.7k1291123

asked Apr 24 at 1:56

GautamGautam

610413

edited Apr 24 at 12:58

Boann

37.7k1291123

edited Apr 24 at 12:58

Boann

37.7k1291123

asked Apr 24 at 1:56

GautamGautam

610413

asked Apr 24 at 1:56

GautamGautam

610413

3 Answers
3

active

oldest

votes

7

SubType is a function. What you probably want to check is if an instance of SubType would inherit from x:

function SuperType()
 
function SubType()

x = new SuperType();

SubType.prototype = x;
SubType.prototype.constructor = SubType;

const instance = new SubType();
console.log(x.isPrototypeOf(instance)) // returns true
console.log(SuperType.prototype.isPrototypeOf(SubType.prototype)) // returns true

share|improve this answer

answered Apr 24 at 2:29

KaiidoKaiido

46.6k469110

add a comment | 

3

It helps to add properties to the objects to see what's happening. I fixed a little of your code. You can run this in the console:

function SuperType(foo) this.foo = foo ;
function SubType(bar) this.bar = bar ;

var x = new SubType("bar");

SuperType.prototype = x;
SuperType.prototype.constructor = SubType;

Now, you asked x.isPrototypeOf(SuperType) and it returns false, because x is not a property of the class SuperType. But when you instantiate a SuperType, x is a property of that new object:

var y = new SuperType("foo");
console.log(x.isPrototypeOf(y)) // returns true

In your example that is true, SubType.prototype is a prototype of SuperType.prototype and returns true.

console.log(SubType.prototype.isPrototypeOf(SuperType.prototype)) // returns true

share|improve this answer

answered Apr 24 at 2:43

David KlingeDavid Klinge

3346

New contributor

David Klinge is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

0

I think that this is misunderstanding of prototype and __proto__ properties.

Lets modify example above a little

function SuperType()

function SubType()

x = new SuperType();

SubType.prototype = x;
console.log(x.isPrototypeOf(SubType)) // returns false.

// Now most interesting piece
SubType.__proto__ = x;
console.log(x.isPrototypeOf(SubType)) // Now true.

Demo

In place of SubType.__proto__ = x we may use Object.setPrototypeOf(SubType, x) which will yeilds the same result

The trik is __proto__ holds real object which is prototype. prototype is used only in constructor fucntions for defining prototype for constructed objects. (See here)

So if we again modify first example

function SuperType()

function SubType()

x = new SuperType();

SubType.prototype = x;

console.log(x.isPrototypeOf(SubType)) // returns false

var x2 = new SubType();
console.log(x.isPrototypeOf(x2)) // returns true

Demo

share|improve this answer

answered Apr 24 at 8:22

FyodorFyodor

1708

add a comment | 

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7

SubType is a function. What you probably want to check is if an instance of SubType would inherit from x:

function SuperType()
 
function SubType()

x = new SuperType();

SubType.prototype = x;
SubType.prototype.constructor = SubType;

const instance = new SubType();
console.log(x.isPrototypeOf(instance)) // returns true
console.log(SuperType.prototype.isPrototypeOf(SubType.prototype)) // returns true

share|improve this answer

answered Apr 24 at 2:29

KaiidoKaiido

46.6k469110

add a comment | 

7

SubType is a function. What you probably want to check is if an instance of SubType would inherit from x:

function SuperType()
 
function SubType()

x = new SuperType();

SubType.prototype = x;
SubType.prototype.constructor = SubType;

const instance = new SubType();
console.log(x.isPrototypeOf(instance)) // returns true
console.log(SuperType.prototype.isPrototypeOf(SubType.prototype)) // returns true

share|improve this answer

answered Apr 24 at 2:29

KaiidoKaiido

46.6k469110

add a comment | 

SubType is a function. What you probably want to check is if an instance of SubType would inherit from x:

function SuperType()
 
function SubType()

x = new SuperType();

SubType.prototype = x;
SubType.prototype.constructor = SubType;

const instance = new SubType();
console.log(x.isPrototypeOf(instance)) // returns true
console.log(SuperType.prototype.isPrototypeOf(SubType.prototype)) // returns true

share|improve this answer

answered Apr 24 at 2:29

KaiidoKaiido

46.6k469110

function SuperType()
 
function SubType()

x = new SuperType();

SubType.prototype = x;
SubType.prototype.constructor = SubType;

const instance = new SubType();
console.log(x.isPrototypeOf(instance)) // returns true
console.log(SuperType.prototype.isPrototypeOf(SubType.prototype)) // returns true

function SuperType()
 
function SubType()

x = new SuperType();

SubType.prototype = x;
SubType.prototype.constructor = SubType;

const instance = new SubType();
console.log(x.isPrototypeOf(instance)) // returns true
console.log(SuperType.prototype.isPrototypeOf(SubType.prototype)) // returns true

function SuperType()
 
function SubType()

x = new SuperType();

SubType.prototype = x;
SubType.prototype.constructor = SubType;

const instance = new SubType();
console.log(x.isPrototypeOf(instance)) // returns true
console.log(SuperType.prototype.isPrototypeOf(SubType.prototype)) // returns true

share|improve this answer

answered Apr 24 at 2:29

KaiidoKaiido

46.6k469110

answered Apr 24 at 2:29

KaiidoKaiido

46.6k469110

answered Apr 24 at 2:29

KaiidoKaiido

46.6k469110

3

It helps to add properties to the objects to see what's happening. I fixed a little of your code. You can run this in the console:

function SuperType(foo) this.foo = foo ;
function SubType(bar) this.bar = bar ;

var x = new SubType("bar");

SuperType.prototype = x;
SuperType.prototype.constructor = SubType;

Now, you asked x.isPrototypeOf(SuperType) and it returns false, because x is not a property of the class SuperType. But when you instantiate a SuperType, x is a property of that new object:

var y = new SuperType("foo");
console.log(x.isPrototypeOf(y)) // returns true

In your example that is true, SubType.prototype is a prototype of SuperType.prototype and returns true.

console.log(SubType.prototype.isPrototypeOf(SuperType.prototype)) // returns true

share|improve this answer

answered Apr 24 at 2:43

David KlingeDavid Klinge

3346

New contributor

David Klinge is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

3

It helps to add properties to the objects to see what's happening. I fixed a little of your code. You can run this in the console:

function SuperType(foo) this.foo = foo ;
function SubType(bar) this.bar = bar ;

var x = new SubType("bar");

SuperType.prototype = x;
SuperType.prototype.constructor = SubType;

Now, you asked x.isPrototypeOf(SuperType) and it returns false, because x is not a property of the class SuperType. But when you instantiate a SuperType, x is a property of that new object:

var y = new SuperType("foo");
console.log(x.isPrototypeOf(y)) // returns true

In your example that is true, SubType.prototype is a prototype of SuperType.prototype and returns true.

console.log(SubType.prototype.isPrototypeOf(SuperType.prototype)) // returns true

share|improve this answer

answered Apr 24 at 2:43

David KlingeDavid Klinge

3346

New contributor

David Klinge is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

It helps to add properties to the objects to see what's happening. I fixed a little of your code. You can run this in the console:

function SuperType(foo) this.foo = foo ;
function SubType(bar) this.bar = bar ;

var x = new SubType("bar");

SuperType.prototype = x;
SuperType.prototype.constructor = SubType;

Now, you asked x.isPrototypeOf(SuperType) and it returns false, because x is not a property of the class SuperType. But when you instantiate a SuperType, x is a property of that new object:

var y = new SuperType("foo");
console.log(x.isPrototypeOf(y)) // returns true

In your example that is true, SubType.prototype is a prototype of SuperType.prototype and returns true.

console.log(SubType.prototype.isPrototypeOf(SuperType.prototype)) // returns true

share|improve this answer

answered Apr 24 at 2:43

David KlingeDavid Klinge

3346

New contributor

David Klinge is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

function SuperType(foo) this.foo = foo ;
function SubType(bar) this.bar = bar ;

var x = new SubType("bar");

SuperType.prototype = x;
SuperType.prototype.constructor = SubType;

var y = new SuperType("foo");
console.log(x.isPrototypeOf(y)) // returns true

console.log(SubType.prototype.isPrototypeOf(SuperType.prototype)) // returns true

share|improve this answer

answered Apr 24 at 2:43

David KlingeDavid Klinge

3346

New contributor

David Klinge is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered Apr 24 at 2:43

David KlingeDavid Klinge

3346

New contributor

David Klinge is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered Apr 24 at 2:43

David KlingeDavid Klinge

3346

0

I think that this is misunderstanding of prototype and __proto__ properties.

Lets modify example above a little

function SuperType()

function SubType()

x = new SuperType();

SubType.prototype = x;
console.log(x.isPrototypeOf(SubType)) // returns false.

// Now most interesting piece
SubType.__proto__ = x;
console.log(x.isPrototypeOf(SubType)) // Now true.

Demo

In place of SubType.__proto__ = x we may use Object.setPrototypeOf(SubType, x) which will yeilds the same result

The trik is __proto__ holds real object which is prototype. prototype is used only in constructor fucntions for defining prototype for constructed objects. (See here)

So if we again modify first example

function SuperType()

function SubType()

x = new SuperType();

SubType.prototype = x;

console.log(x.isPrototypeOf(SubType)) // returns false

var x2 = new SubType();
console.log(x.isPrototypeOf(x2)) // returns true

Demo

share|improve this answer

answered Apr 24 at 8:22

FyodorFyodor

1708

add a comment | 

0

I think that this is misunderstanding of prototype and __proto__ properties.

Lets modify example above a little

function SuperType()

function SubType()

x = new SuperType();

SubType.prototype = x;
console.log(x.isPrototypeOf(SubType)) // returns false.

// Now most interesting piece
SubType.__proto__ = x;
console.log(x.isPrototypeOf(SubType)) // Now true.

Demo

In place of SubType.__proto__ = x we may use Object.setPrototypeOf(SubType, x) which will yeilds the same result

The trik is __proto__ holds real object which is prototype. prototype is used only in constructor fucntions for defining prototype for constructed objects. (See here)

So if we again modify first example

function SuperType()

function SubType()

x = new SuperType();

SubType.prototype = x;

console.log(x.isPrototypeOf(SubType)) // returns false

var x2 = new SubType();
console.log(x.isPrototypeOf(x2)) // returns true

Demo

share|improve this answer

answered Apr 24 at 8:22

FyodorFyodor

1708

add a comment | 

I think that this is misunderstanding of prototype and __proto__ properties.

Lets modify example above a little

function SuperType()

function SubType()

x = new SuperType();

SubType.prototype = x;
console.log(x.isPrototypeOf(SubType)) // returns false.

// Now most interesting piece
SubType.__proto__ = x;
console.log(x.isPrototypeOf(SubType)) // Now true.

Demo

In place of SubType.__proto__ = x we may use Object.setPrototypeOf(SubType, x) which will yeilds the same result

The trik is __proto__ holds real object which is prototype. prototype is used only in constructor fucntions for defining prototype for constructed objects. (See here)

So if we again modify first example

function SuperType()

function SubType()

x = new SuperType();

SubType.prototype = x;

console.log(x.isPrototypeOf(SubType)) // returns false

var x2 = new SubType();
console.log(x.isPrototypeOf(x2)) // returns true

Demo

share|improve this answer

answered Apr 24 at 8:22

FyodorFyodor

1708

function SuperType()

function SubType()

x = new SuperType();

SubType.prototype = x;
console.log(x.isPrototypeOf(SubType)) // returns false.

// Now most interesting piece
SubType.__proto__ = x;
console.log(x.isPrototypeOf(SubType)) // Now true.

function SuperType()

function SubType()

x = new SuperType();

SubType.prototype = x;

console.log(x.isPrototypeOf(SubType)) // returns false

var x2 = new SubType();
console.log(x.isPrototypeOf(x2)) // returns true

share|improve this answer

answered Apr 24 at 8:22

FyodorFyodor

1708

answered Apr 24 at 8:22

FyodorFyodor

1708

answered Apr 24 at 8:22

FyodorFyodor

1708