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Why isPrototypeOf() returns false?

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6

1

I have above 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 it's 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 42 mins ago

Gautam

asked 1 hour ago

GautamGautam

600413

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Not sure to get my head all clear here, but x === SubType.prototype how do you expect it to be its own prototype?
  
  – Kaiido
  44 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Updated my question, sorry about that type
  
  – Gautam
  42 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  try console.log(x.isPrototypeOf(new SubType)) for example of how it's used.
  
  – dandavis
  35 mins ago
  

  
  
  
  

add a comment | 

6

1

I have above 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 it's 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 42 mins ago

Gautam

asked 1 hour ago

GautamGautam

600413

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Not sure to get my head all clear here, but x === SubType.prototype how do you expect it to be its own prototype?
  
  – Kaiido
  44 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Updated my question, sorry about that type
  
  – Gautam
  42 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  try console.log(x.isPrototypeOf(new SubType)) for example of how it's used.
  
  – dandavis
  35 mins ago
  

  
  
  
  

add a comment | 

I have above 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 it's 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 42 mins ago

Gautam

asked 1 hour ago

GautamGautam

600413

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 42 mins ago

Gautam

asked 1 hour ago

GautamGautam

600413

share|improve this question

edited 42 mins ago

Gautam

asked 1 hour ago

GautamGautam

600413

asked 1 hour ago

GautamGautam

600413

asked 1 hour ago

GautamGautam

600413

2 Answers
2

active

oldest

votes

5

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 36 mins ago

KaiidoKaiido

46.4k468109

add a comment | 

2

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 22 mins ago

David KlingeDavid Klinge

564

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 | 

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5

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 36 mins ago

KaiidoKaiido

46.4k468109

add a comment | 

5

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 36 mins ago

KaiidoKaiido

46.4k468109

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 36 mins ago

KaiidoKaiido

46.4k468109

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 36 mins ago

KaiidoKaiido

46.4k468109

answered 36 mins ago

KaiidoKaiido

46.4k468109

answered 36 mins ago

KaiidoKaiido

46.4k468109

2

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 22 mins ago

David KlingeDavid Klinge

564

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 | 

2

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 22 mins ago

David KlingeDavid Klinge

564

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 22 mins ago

David KlingeDavid Klinge

564

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 22 mins ago

David KlingeDavid Klinge

564

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 22 mins ago

David KlingeDavid Klinge

564

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 22 mins ago

David KlingeDavid Klinge

564