Can a instance method in a Javascript subclass call it's parents static method?
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0
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This doesn't work. Is there a way for a child class in JS to have access to its parents static methods?
class Person {
constructor() {}
static isHuman() {
return 'yes I am';
}
}
class Brian extends Person {
constructor() {
super();
}
greeting() {
console.log(super.isHuman())
}
}
const b = new Brian();
b.greeting();
javascript
asked Nov 11 at 0:56
GN.
7331640
add a comment |
up vote
0
down vote
favorite
This doesn't work. Is there a way for a child class in JS to have access to its parents static methods?
class Person {
constructor() {}
static isHuman() {
return 'yes I am';
}
}
class Brian extends Person {
constructor() {
super();
}
greeting() {
console.log(super.isHuman())
}
}
const b = new Brian();
b.greeting();
javascript
asked Nov 11 at 0:56
GN.
7331640
No, is not possible. It doesn't make sense. You need to understand how the static members work. read more about it here
– Ele
Nov 11 at 0:59
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
This doesn't work. Is there a way for a child class in JS to have access to its parents static methods?
class Person {
constructor() {}
static isHuman() {
return 'yes I am';
}
}
class Brian extends Person {
constructor() {
super();
}
greeting() {
console.log(super.isHuman())
}
}
const b = new Brian();
b.greeting();
javascript
asked Nov 11 at 0:56
GN.
7331640
This doesn't work. Is there a way for a child class in JS to have access to its parents static methods?
class Person {
constructor() {}
static isHuman() {
return 'yes I am';
}
}
class Brian extends Person {
constructor() {
super();
}
greeting() {
console.log(super.isHuman())
}
}
const b = new Brian();
b.greeting();
javascript
javascript
asked Nov 11 at 0:56
GN.
7331640
asked Nov 11 at 0:56
GN.
7331640
asked Nov 11 at 0:56
GN.
7331640
asked Nov 11 at 0:56
GN.
7331640
asked Nov 11 at 0:56
GN.
7331640
7331640
No, is not possible. It doesn't make sense. You need to understand how the static members work. read more about it here
– Ele
Nov 11 at 0:59
add a comment |
No, is not possible. It doesn't make sense. You need to understand how the static members work. read more about it here
– Ele
Nov 11 at 0:59
No, is not possible. It doesn't make sense. You need to understand how the static members work. read more about it here
– Ele
Nov 11 at 0:59
No, is not possible. It doesn't make sense. You need to understand how the static members work. read more about it here
– Ele
Nov 11 at 0:59
add a comment |
2 Answers
2
active
oldest
votes
up vote
0
down vote
accepted
Yes you can. You can use super but to get at the static method you have to access the constructor property of it:
super.constructor.isHuman()
You can also use the class name directly:
Person.isHuman();
//this works because Brian inherits the static methods from Person
Brian.isHuman();
Or by going up the prototype chain
//would be referencing Brian
this.constructor.isHuman();
//would be referencing Person
this.__proto__.constructor.isHuman();
Object.getPrototypeOf(this).constructor.isHuman();
Demo
class Person {
constructor() {}
static isHuman() {
return 'yes I am';
}
}
class Brian extends Person {
constructor() {
super();
}
greeting() {
console.log("From super: ", super.constructor.isHuman())
console.log("From class name (Brian): ", Brian.isHuman())
console.log("From class name (Person): ", Person.isHuman())
console.log("From prototype chain: ", this.constructor.isHuman())
console.log("From prototype chain: ", Object.getPrototypeOf(this).constructor.isHuman())
}
}
const b = new Brian();
b.greeting();answered Nov 11 at 1:12
Patrick Evans
31.7k54370
Awesome! Thanks
– GN.
Nov 11 at 23:28
add a comment |
up vote
0
down vote
For static methods, use the class name rather than super.
class Person {
constructor() {}
static isHuman() {
return 'yes I am';
}
}
class Brian extends Person {
constructor() {
super();
}
greeting() {
console.log(Person.isHuman())
}
}
const b = new Brian();
b.greeting();answered Nov 11 at 1:01
Ryan C
678210
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Yes you can. You can use super but to get at the static method you have to access the constructor property of it:
super.constructor.isHuman()
You can also use the class name directly:
Person.isHuman();
//this works because Brian inherits the static methods from Person
Brian.isHuman();
Or by going up the prototype chain
//would be referencing Brian
this.constructor.isHuman();
//would be referencing Person
this.__proto__.constructor.isHuman();
Object.getPrototypeOf(this).constructor.isHuman();
Demo
class Person {
constructor() {}
static isHuman() {
return 'yes I am';
}
}
class Brian extends Person {
constructor() {
super();
}
greeting() {
console.log("From super: ", super.constructor.isHuman())
console.log("From class name (Brian): ", Brian.isHuman())
console.log("From class name (Person): ", Person.isHuman())
console.log("From prototype chain: ", this.constructor.isHuman())
console.log("From prototype chain: ", Object.getPrototypeOf(this).constructor.isHuman())
}
}
const b = new Brian();
b.greeting();answered Nov 11 at 1:12
Patrick Evans
31.7k54370
Awesome! Thanks
– GN.
Nov 11 at 23:28
add a comment |
up vote
0
down vote
accepted
Yes you can. You can use super but to get at the static method you have to access the constructor property of it:
super.constructor.isHuman()
You can also use the class name directly:
Person.isHuman();
//this works because Brian inherits the static methods from Person
Brian.isHuman();
Or by going up the prototype chain
//would be referencing Brian
this.constructor.isHuman();
//would be referencing Person
this.__proto__.constructor.isHuman();
Object.getPrototypeOf(this).constructor.isHuman();
Demo
class Person {
constructor() {}
static isHuman() {
return 'yes I am';
}
}
class Brian extends Person {
constructor() {
super();
}
greeting() {
console.log("From super: ", super.constructor.isHuman())
console.log("From class name (Brian): ", Brian.isHuman())
console.log("From class name (Person): ", Person.isHuman())
console.log("From prototype chain: ", this.constructor.isHuman())
console.log("From prototype chain: ", Object.getPrototypeOf(this).constructor.isHuman())
}
}
const b = new Brian();
b.greeting();answered Nov 11 at 1:12
Patrick Evans
31.7k54370
Awesome! Thanks
– GN.
Nov 11 at 23:28
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Yes you can. You can use super but to get at the static method you have to access the constructor property of it:
super.constructor.isHuman()
You can also use the class name directly:
Person.isHuman();
//this works because Brian inherits the static methods from Person
Brian.isHuman();
Or by going up the prototype chain
//would be referencing Brian
this.constructor.isHuman();
//would be referencing Person
this.__proto__.constructor.isHuman();
Object.getPrototypeOf(this).constructor.isHuman();
Demo
class Person {
constructor() {}
static isHuman() {
return 'yes I am';
}
}
class Brian extends Person {
constructor() {
super();
}
greeting() {
console.log("From super: ", super.constructor.isHuman())
console.log("From class name (Brian): ", Brian.isHuman())
console.log("From class name (Person): ", Person.isHuman())
console.log("From prototype chain: ", this.constructor.isHuman())
console.log("From prototype chain: ", Object.getPrototypeOf(this).constructor.isHuman())
}
}
const b = new Brian();
b.greeting();answered Nov 11 at 1:12
Patrick Evans
31.7k54370
Yes you can. You can use super but to get at the static method you have to access the constructor property of it:
super.constructor.isHuman()
You can also use the class name directly:
Person.isHuman();
//this works because Brian inherits the static methods from Person
Brian.isHuman();
Or by going up the prototype chain
//would be referencing Brian
this.constructor.isHuman();
//would be referencing Person
this.__proto__.constructor.isHuman();
Object.getPrototypeOf(this).constructor.isHuman();
Demo
class Person {
constructor() {}
static isHuman() {
return 'yes I am';
}
}
class Brian extends Person {
constructor() {
super();
}
greeting() {
console.log("From super: ", super.constructor.isHuman())
console.log("From class name (Brian): ", Brian.isHuman())
console.log("From class name (Person): ", Person.isHuman())
console.log("From prototype chain: ", this.constructor.isHuman())
console.log("From prototype chain: ", Object.getPrototypeOf(this).constructor.isHuman())
}
}
const b = new Brian();
b.greeting();class Person {
constructor() {}
static isHuman() {
return 'yes I am';
}
}
class Brian extends Person {
constructor() {
super();
}
greeting() {
console.log("From super: ", super.constructor.isHuman())
console.log("From class name (Brian): ", Brian.isHuman())
console.log("From class name (Person): ", Person.isHuman())
console.log("From prototype chain: ", this.constructor.isHuman())
console.log("From prototype chain: ", Object.getPrototypeOf(this).constructor.isHuman())
}
}
const b = new Brian();
b.greeting();class Person {
constructor() {}
static isHuman() {
return 'yes I am';
}
}
class Brian extends Person {
constructor() {
super();
}
greeting() {
console.log("From super: ", super.constructor.isHuman())
console.log("From class name (Brian): ", Brian.isHuman())
console.log("From class name (Person): ", Person.isHuman())
console.log("From prototype chain: ", this.constructor.isHuman())
console.log("From prototype chain: ", Object.getPrototypeOf(this).constructor.isHuman())
}
}
const b = new Brian();
b.greeting();answered Nov 11 at 1:12
Patrick Evans
31.7k54370
edited Nov 11 at 1:18
answered Nov 11 at 1:12
Patrick Evans
31.7k54370
answered Nov 11 at 1:12
Patrick Evans
31.7k54370
answered Nov 11 at 1:12
Patrick Evans
31.7k54370
31.7k54370
Awesome! Thanks
– GN.
Nov 11 at 23:28
add a comment |
Awesome! Thanks
– GN.
Nov 11 at 23:28
Awesome! Thanks
– GN.
Nov 11 at 23:28
Awesome! Thanks
– GN.
Nov 11 at 23:28
add a comment |
up vote
0
down vote
For static methods, use the class name rather than super.
class Person {
constructor() {}
static isHuman() {
return 'yes I am';
}
}
class Brian extends Person {
constructor() {
super();
}
greeting() {
console.log(Person.isHuman())
}
}
const b = new Brian();
b.greeting();answered Nov 11 at 1:01
Ryan C
678210
add a comment |
up vote
0
down vote
For static methods, use the class name rather than super.
class Person {
constructor() {}
static isHuman() {
return 'yes I am';
}
}
class Brian extends Person {
constructor() {
super();
}
greeting() {
console.log(Person.isHuman())
}
}
const b = new Brian();
b.greeting();answered Nov 11 at 1:01
Ryan C
678210
add a comment |
up vote
0
down vote
up vote
0
down vote
For static methods, use the class name rather than super.
class Person {
constructor() {}
static isHuman() {
return 'yes I am';
}
}
class Brian extends Person {
constructor() {
super();
}
greeting() {
console.log(Person.isHuman())
}
}
const b = new Brian();
b.greeting();answered Nov 11 at 1:01
Ryan C
678210
For static methods, use the class name rather than super.
class Person {
constructor() {}
static isHuman() {
return 'yes I am';
}
}
class Brian extends Person {
constructor() {
super();
}
greeting() {
console.log(Person.isHuman())
}
}
const b = new Brian();
b.greeting();class Person {
constructor() {}
static isHuman() {
return 'yes I am';
}
}
class Brian extends Person {
constructor() {
super();
}
greeting() {
console.log(Person.isHuman())
}
}
const b = new Brian();
b.greeting();class Person {
constructor() {}
static isHuman() {
return 'yes I am';
}
}
class Brian extends Person {
constructor() {
super();
}
greeting() {
console.log(Person.isHuman())
}
}
const b = new Brian();
b.greeting();answered Nov 11 at 1:01
Ryan C
678210
answered Nov 11 at 1:01
Ryan C
678210
answered Nov 11 at 1:01
Ryan C
678210
answered Nov 11 at 1:01
Ryan C
678210
678210
add a comment |
add a comment |
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No, is not possible. It doesn't make sense. You need to understand how the static members work. read more about it here
– Ele
Nov 11 at 0:59