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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
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
1 hour ago
Updated my question, sorry about that type
– Gautam
1 hour ago
tryconsole.log(x.isPrototypeOf(new SubType))
for example of how it's used.
– dandavis
1 hour 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 1 hour ago
Gautam
asked 1 hour ago
GautamGautam
600413
600413
1
Not sure to get my head all clear here, butx === 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
tryconsole.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, butx === 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
tryconsole.log(x.isPrototypeOf(new SubType))
for example of how it's used.
– dandavis
1 hour 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
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
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
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
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 |
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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 1 hour 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 51 mins ago
David KlingeDavid Klinge
564
564
New contributor
New contributor
add a comment |
add a comment |
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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