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Why isPrototypeOf() returns false?

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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

edited 39 mins ago

asked 1 hour ago

Gautam

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
41 mins ago

Updated my question, sorry about that type

– Gautam
38 mins ago

try console.log(x.isPrototypeOf(new SubType)) for example of how it's used.

– dandavis
32 mins ago

add a comment |

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

edited 39 mins ago

asked 1 hour ago

Gautam

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
41 mins ago

Updated my question, sorry about that type

– Gautam
38 mins ago

try console.log(x.isPrototypeOf(new SubType)) for example of how it's used.

– dandavis
32 mins ago

add a comment |

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

edited 39 mins ago

asked 1 hour ago

Gautam

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

javascript

edited 39 mins ago

asked 1 hour ago

Gautam

600413

edited 39 mins ago

asked 1 hour ago

Gautam

600413

edited 39 mins ago

asked 1 hour ago

Gautam

600413

asked 1 hour ago

Gautam

600413

asked 1 hour ago

Gautam

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
41 mins ago

Updated my question, sorry about that type

– Gautam
38 mins ago

try console.log(x.isPrototypeOf(new SubType)) for example of how it's used.

– dandavis
32 mins ago

add a comment |

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
41 mins ago

Updated my question, sorry about that type

– Gautam
38 mins ago

try console.log(x.isPrototypeOf(new SubType)) for example of how it's used.

– dandavis
32 mins ago

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

– Kaiido
41 mins ago

Updated my question, sorry about that type

– Gautam
38 mins ago

try console.log(x.isPrototypeOf(new SubType)) for example of how it's used.

– dandavis
32 mins ago

add a comment |

2 Answers
2

active

oldest

votes

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

answered 32 mins ago

Kaiido

46.4k468109

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

answered 19 mins ago

David Klinge

564

New contributor

add a comment |

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

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

answered 32 mins ago

Kaiido

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

answered 32 mins ago

Kaiido

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

answered 32 mins ago

Kaiido

46.4k468109

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

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

answered 32 mins ago

Kaiido

46.4k468109

answered 32 mins ago

Kaiido

46.4k468109

answered 32 mins ago

Kaiido

46.4k468109

answered 32 mins ago

Kaiido

46.4k468109

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

answered 19 mins ago

David Klinge

564

New contributor

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

answered 19 mins ago

David Klinge

564

New contributor

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

answered 19 mins ago

David Klinge

564

New contributor

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

answered 19 mins ago

David Klinge

564

New contributor

answered 19 mins ago

David Klinge

564

New contributor

answered 19 mins ago

David Klinge

564

answered 19 mins ago

David 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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