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 1 hour ago

asked 2 hours 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
1 hour ago

Updated my question, sorry about that type

– Gautam
1 hour ago

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

– dandavis
1 hour 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 1 hour ago

asked 2 hours 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
1 hour ago

Updated my question, sorry about that type

– Gautam
1 hour ago

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

– dandavis
1 hour 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 1 hour ago

asked 2 hours 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 1 hour ago

asked 2 hours ago

Gautam

600413

edited 1 hour ago

asked 2 hours ago

Gautam

600413

edited 1 hour ago

asked 2 hours ago

Gautam

600413

asked 2 hours ago

Gautam

600413

asked 2 hours 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
1 hour ago

Updated my question, sorry about that type

– Gautam
1 hour ago

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

– dandavis
1 hour 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
1 hour ago

Updated my question, sorry about that type

– Gautam
1 hour ago

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

– dandavis
1 hour 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
1 hour ago

Updated my question, sorry about that type

– Gautam
1 hour ago

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

– dandavis
1 hour 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 1 hour 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 1 hour 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 1 hour 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 1 hour 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 1 hour 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 1 hour ago

Kaiido

46.4k468109

answered 1 hour ago

Kaiido

46.4k468109

answered 1 hour ago

Kaiido

46.4k468109

answered 1 hour 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 1 hour 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 1 hour 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 1 hour 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 1 hour ago

David Klinge

564

New contributor

answered 1 hour ago

David Klinge

564

New contributor

answered 1 hour ago

David Klinge

564

answered 1 hour 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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