Use static property of extended class in super()-CodePudding

Suppose we have the code

class A {
  static foo = 1;
  constructor() {
    console.log(A.foo);
  }
}
class B extends A {
  static foo = 2;
  constructor() {
    super();
  }
}
new A(); // prints "1"
new B(); // also prints "1", but make it print "2"

Here the constructor of A is accessing A.foo. When A is extended to B with a new value of foo, the constructor still prints 1, the value of A.foo, when really I want it to print the value of B.foo.

How would I modify this code such that the constructor prints the foo property of the current class?

CodePudding user response：

Use the constructor property to get access to the class itself, then you can refer to the static property from that.

class A {
  static foo = 1;
  constructor() {
    console.log(this.constructor.foo);
  }
}
class B extends A {
  static foo = 2;
  constructor() {
    super();
  }
}
new A(); // prints "1"
new B(); // also prints "1", but make it print "2"

CodePudding user response：

the super calls the constructor of the super class.

What you can do is the following

class A {
      static foo = 1;
      constructor(...args) {
        if(args.length == 1){
          console.log(args[0].foo)
        }
        else{
        
        console.log(A.foo);
        }
      }
    }
    
class B extends A {
  static foo = 2;
  constructor(...args) {
    super(...args); // calls A.foo not B.foo
  }
}
new A(); // prints "1"
new B(B); // pass the Class to it self as an arg