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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 4 hours ago

asked 4 hours ago

Gautam

610413

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
4 hours ago

Updated my question, sorry about that type

– Gautam
4 hours ago

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

– dandavis
4 hours 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 4 hours ago

asked 4 hours ago

Gautam

610413

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
4 hours ago

Updated my question, sorry about that type

– Gautam
4 hours ago

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

– dandavis
4 hours 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 4 hours ago

asked 4 hours ago

Gautam

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

javascript

edited 4 hours ago

asked 4 hours ago

Gautam

610413

edited 4 hours ago

asked 4 hours ago

Gautam

610413

edited 4 hours ago

asked 4 hours ago

Gautam

610413

asked 4 hours ago

Gautam

610413

asked 4 hours ago

Gautam

610413

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
4 hours ago

Updated my question, sorry about that type

– Gautam
4 hours ago

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

– dandavis
4 hours 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
4 hours ago

Updated my question, sorry about that type

– Gautam
4 hours ago

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

– dandavis
4 hours 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
4 hours ago

Updated my question, sorry about that type

– Gautam
4 hours ago

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

– dandavis
4 hours 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 4 hours 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 3 hours ago

David Klinge

864

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 4 hours 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 4 hours 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 4 hours 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 4 hours ago

Kaiido

46.4k468109

answered 4 hours ago

Kaiido

46.4k468109

answered 4 hours ago

Kaiido

46.4k468109

answered 4 hours 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 3 hours ago

David Klinge

864

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 3 hours ago

David Klinge

864

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 3 hours ago

David Klinge

864

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 3 hours ago

David Klinge

864

New contributor

answered 3 hours ago

David Klinge

864

New contributor

answered 3 hours ago

David Klinge

864

answered 3 hours ago

David Klinge

864

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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