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Ilya Kantor 2017-01-18 09:53:11 +01:00
parent 6c9c2219ba
commit a7c00e1c76
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The answer has two parts.
The first, an easy one is that the inheriting class needs to call `super()` in the constructor. Otherwise `"this"` won't be "defined".
So here's the fix:
```js run
class Rabbit extends Object {
constructor(name) {
*!*
super(); // need to call the parent constructor when inheriting
*/!*
this.name = name;
}
}
let rabbit = new Rabbit("Rab");
alert( rabbit.hasOwnProperty('name') ); // true
```
But that's not all yet.
Even after the fix, there's still important difference in `"class Rabbit extends Object"` versus `class Rabbit`.
As we know, the "extends" syntax sets up two prototypes:
1. Between `"prototype"` of the constructor functions (for methods).
2. Between the constructor functions itself (for static methods).
In our case, for `class Rabbit extends Object` it means:
```js run
class Rabbit extends Object {}
alert( Rabbit.prototype.__proto__ === Object.prototype ); // (1) true
alert( Rabbit.__proto__ === Object ); // (2) true
```
So we can access static methods of `Object` via `Rabbit`, like this:
```js run
class Rabbit extends Object {}
*!*
// normally we call Object.getOwnPropertyNames
alert ( Rabbit.getOwnPropertyNames({a: 1, b: 2})); // a,b
*/!*
```
And if we don't use `extends`, then `class Rabbit` does not get the second reference.
Please compare with it:
```js run
class Rabbit {}
alert( Rabbit.prototype.__proto__ === Object.prototype ); // (1) true
alert( Rabbit.__proto__ === Object ); // (2) false (!)
*!*
// error, no such function in Rabbit
alert ( Rabbit.getOwnPropertyNames({a: 1, b: 2})); // Error
*/!*
```
For the simple `class Rabbit`, the `Rabbit` function has the same prototype
```js run
class Rabbit {}
// instead of (2) that's correct for Rabbit (just like any function):
alert( Rabbit.__proto__ === Function.prototype );
```
By the way, `Function.prototype` has "generic" function methods, like `call`, `bind` etc. They are ultimately available in both cases, because for the built-in `Object` constructor, `Object.__proto__ === Function.prototype`.
Here's the picture:
![](rabbit-extends-object.png)
So, to put it short, there are two differences:
| class Rabbit | class Rabbit extends Object |
|--------------|------------------------------|
| -- | needs to call `super()` in constructor |
| `Rabbit.__proto__ === Function.prototype` | `Rabbit.__proto__ === Object` |

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importance: 5
---
# Class extends Object?
As we know, all objects normally inherit from `Object.prototype` and get access to "generic" object methods.
Like demonstrated here:
```js run
class Rabbit {
constructor(name) {
this.name = name;
}
}
let rabbit = new Rabbit("Rab");
*!*
// hasOwnProperty method is from Object.prototype
// rabbit.__proto__ === Object.prototype
alert( rabbit.hasOwnProperty('name') ); // true
*/!*
```
So, is it correct to say that `"class Rabbit extends Object"` does exactly the same as `"class Rabbit"`, or not?
Will it work?
```js
class Rabbit extends Object {
constructor(name) {
this.name = name;
}
}
let rabbit = new Rabbit("Rab");
alert( rabbit.hasOwnProperty('name') ); // true
```
If it won't please fix the code.