Why isPrototypeOf() returns false? Unicorn Meta Zoo #1: Why another podcast? ...
What do you call an IPA symbol that lacks a name (e.g. ɲ)?
What's parked in Mil Moscow helicopter plant?
Stretch a Tikz tree
Arriving in Atlanta after US Preclearance in Dublin. Will I go through TSA security in Atlanta to transfer to a connecting flight?
Married in secret, can marital status in passport be changed at a later date?
Why isPrototypeOf() returns false?
What helicopter has the most rotor blades?
RIP Packet Format
Coin Game with infinite paradox
Processing ADC conversion result: DMA vs Processor Registers
What *exactly* is electrical current, voltage, and resistance?
What is the numbering system used for the DSN dishes?
How did Elite on the NES work?
Was Objective-C really a hindrance to Apple software development?
Can gravitational waves pass through a black hole?
What is the purpose of the side handle on a hand ("eggbeater") drill?
Why do people think Winterfell crypts is the safest place for women, children & old people?
Is Bran literally the world's memory?
Mechanism of the formation of peracetic acid
Is it accepted to use working hours to read general interest books?
Why does the Cisco show run command not show the full version, while the show version command does?
Why I cannot instantiate a class whose constructor is private in a friend class?
TV series episode where humans nuke aliens before decrypting their message that states they come in peace
What happened to Viserion in Season 7?
Why isPrototypeOf() returns false?
Unicorn Meta Zoo #1: Why another podcast?
Announcing the arrival of Valued Associate #679: Cesar Manara
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
javascript
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
1
Not sure to get my head all clear here, butx === SubType.prototype
how do you expect it to be its own prototype?
– Kaiido
2 hours ago
Updated my question, sorry about that type
– Gautam
2 hours ago
tryconsole.log(x.isPrototypeOf(new SubType))
for example of how it's used.
– dandavis
2 hours 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
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
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
javascript
edited 2 hours ago
Gautam
asked 3 hours ago
GautamGautam
605413
605413
1
Not sure to get my head all clear here, butx === SubType.prototype
how do you expect it to be its own prototype?
– Kaiido
2 hours ago
Updated my question, sorry about that type
– Gautam
2 hours ago
tryconsole.log(x.isPrototypeOf(new SubType))
for example of how it's used.
– dandavis
2 hours ago
add a comment |
1
Not sure to get my head all clear here, butx === SubType.prototype
how do you expect it to be its own prototype?
– Kaiido
2 hours ago
Updated my question, sorry about that type
– Gautam
2 hours ago
tryconsole.log(x.isPrototypeOf(new SubType))
for example of how it's used.
– dandavis
2 hours ago
1
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
2 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
2 hours ago
Updated my question, sorry about that type
– Gautam
2 hours ago
Updated my question, sorry about that type
– Gautam
2 hours ago
try
console.log(x.isPrototypeOf(new SubType))
for example of how it's used.– dandavis
2 hours ago
try
console.log(x.isPrototypeOf(new SubType))
for example of how it's used.– dandavis
2 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
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
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
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
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
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
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
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
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 2 hours ago
KaiidoKaiido
46.4k468109
46.4k468109
add a comment |
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
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
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
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
New contributor
New contributor
answered 2 hours ago
David KlingeDavid Klinge
664
664
New contributor
New contributor
add a comment |
add a comment |
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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
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
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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
2 hours ago
Updated my question, sorry about that type
– Gautam
2 hours ago
try
console.log(x.isPrototypeOf(new SubType))
for example of how it's used.– dandavis
2 hours ago