Why isPrototypeOf() returns false? Announcing the arrival of Valued Associate #679: Cesar...

When speaking, how do you change your mind mid-sentence?

Why I cannot instantiate a class whose constructor is private in a friend class?

Has a Nobel Peace laureate ever been accused of war crimes?

How to keep bees out of canned beverages?

Is there a verb for listening stealthily?

Why did Europeans not widely domesticate foxes?

Is there an efficient way for synchronising audio events real-time with LEDs using an MCU?

Protagonist's race is hidden - should I reveal it?

Raising a bilingual kid. When should we introduce the majority language?

Putting Ant-Man on house arrest

How to compute a Jacobian using polar coordinates?

A journey... into the MIND

Marquee sign letters

Did war bonds have better investment alternatives during WWII?

Is it appropriate to mention a relatable company blog post when you're asked about the company?

Married in secret, can marital status in passport be changed at a later date?

Writing a T-SQL stored procedure to receive 4 numbers and insert them into a table

Why do people think Winterfell crypts is the safest place for women, children & old people?

Does using the Inspiration rules for character defects encourage My Guy Syndrome?

What's parked in Mil Moscow helicopter plant?

Why is arima in R one time step off?

Co-worker works way more than he should

When I export an AI 300x60 art board it saves with bigger dimensions

What to do with someone that cheated their way though university and a PhD program?

Why isPrototypeOf() returns false?

Announcing the arrival of Valued Associate #679: Cesar Manara

Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30 pm US/Eastern)

Data science time! April 2019 and salary with experience

The Ask Question Wizard is Live!Why is using “for…in” with array iteration a bad idea?Why does isNaN(“ ”) equal falseevent.preventDefault() vs. return falseWhat is JSONP, and why was it created?Why does Google prepend while(1); to their JSON responses?Why does ++[[]][+[]]+[+[]] return the string “10”?Is Safari on iOS 6 caching $.ajax results?How do I return the response from an asynchronous call?jQuery.inArray(), how to use it right?isPrototypeOf in Javascript

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box;
}

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 33 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
35 mins ago

Updated my question, sorry about that type

– Gautam
33 mins ago

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

– dandavis
27 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 33 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
35 mins ago

Updated my question, sorry about that type

– Gautam
33 mins ago

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

– dandavis
27 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 33 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 33 mins ago

asked 1 hour ago

Gautam

600413

edited 33 mins ago

asked 1 hour ago

Gautam

600413

edited 33 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
35 mins ago

Updated my question, sorry about that type

– Gautam
33 mins ago

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

– dandavis
27 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
35 mins ago

Updated my question, sorry about that type

– Gautam
33 mins ago

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

– dandavis
27 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
35 mins ago

Updated my question, sorry about that type

– Gautam
33 mins ago

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

– dandavis
27 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 27 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 13 mins ago

David Klinge

564

New contributor

add a comment |

Your Answer

StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});

}
});

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f55821319%2fwhy-isprototypeof-returns-false%23new-answer', 'question_page');
}
);

Post as a guest

Name

Required, but never shown

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 27 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 27 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 27 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 27 mins ago

Kaiido

46.4k468109

answered 27 mins ago

Kaiido

46.4k468109

answered 27 mins ago

Kaiido

46.4k468109

answered 27 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 13 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 13 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 13 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 13 mins ago

David Klinge

564

New contributor

answered 13 mins ago

David Klinge

564

New contributor

answered 13 mins ago

David Klinge

564

answered 13 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 |

draft saved

draft discarded

Thanks for contributing an answer to Stack Overflow!

Please be sure to answer the question. Provide details and share your research!

But avoid …

Asking for help, clarification, or responding to other answers.

Making statements based on opinion; back them up with references or personal experience.

To learn more, see our tips on writing great answers.

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Name

Required, but never shown

Name

Required, but never shown

This page is only for reference, If you need detailed information, please check here

搜尋此網誌

Ryfujk