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Call static methods from regular ES6 class methods

People also ask

Can I call a static method inside a regular method?

If you have no object but just call a static method and in that method you want to call another static method in the same class, you have to use self:: . So to avoid potential errors (and strict warnings) it is better to use self .

How do you call a static method from another method?

A static method is bound to the class and not the object of the class. Therefore, we can call it using the class name. A static method doesn't have access to the class and instance variables because it does not receive an implicit first argument like self and cls .

Can you call a static method from an instance JavaScript?

To declare a static method we can simply use static keyword with the method signature. The static method are not called on the instance of class they are made to call directly on the class. So we can say that JavaScript provides us a static method that belongs to the class but not with the instance of the class.

What is static method in es6?

Static methods are often used to create utility functions for an application.” In other words, static methods have no access to data stored in specific objects. Note that for static methods, the this keyword references the class. You can call a static method from another static method within the same class with this.


Both ways are viable, but they do different things when it comes to inheritance with an overridden static method. Choose the one whose behavior you expect:

class Super {
  static whoami() {
    return "Super";
  }
  lognameA() {
    console.log(Super.whoami());
  }
  lognameB() {
    console.log(this.constructor.whoami());
  }
}
class Sub extends Super {
  static whoami() {
    return "Sub";
  }
}
new Sub().lognameA(); // Super
new Sub().lognameB(); // Sub

Referring to the static property via the class will be actually static and constantly give the same value. Using this.constructor instead will use dynamic dispatch and refer to the class of the current instance, where the static property might have the inherited value but could also be overridden.

This matches the behavior of Python, where you can choose to refer to static properties either via the class name or the instance self.

If you expect static properties not to be overridden (and always refer to the one of the current class), like in Java, use the explicit reference.


I stumbled over this thread searching for answer to similar case. Basically all answers are found, but it's still hard to extract the essentials from them.

Kinds of Access

Assume a class Foo probably derived from some other class(es) with probably more classes derived from it.

Then accessing

  • from static method/getter of Foo
    • some probably overridden static method/getter:
      • this.method()
      • this.property
    • some probably overridden instance method/getter:
      • impossible by design
    • own non-overridden static method/getter:
      • Foo.method()
      • Foo.property
    • own non-overridden instance method/getter:
      • impossible by design
  • from instance method/getter of Foo
    • some probably overridden static method/getter:
      • this.constructor.method()
      • this.constructor.property
    • some probably overridden instance method/getter:
      • this.method()
      • this.property
    • own non-overridden static method/getter:
      • Foo.method()
      • Foo.property
    • own non-overridden instance method/getter:
      • not possible by intention unless using some workaround:
        • Foo.prototype.method.call( this )
        • Object.getOwnPropertyDescriptor( Foo.prototype,"property" ).get.call(this);

Keep in mind that using this isn't working this way when using arrow functions or invoking methods/getters explicitly bound to custom value.

Background

  • When in context of an instance's method or getter
    • this is referring to current instance.
    • super is basically referring to same instance, but somewhat addressing methods and getters written in context of some class current one is extending (by using the prototype of Foo's prototype).
    • definition of instance's class used on creating it is available per this.constructor.
  • When in context of a static method or getter there is no "current instance" by intention and so
    • this is available to refer to the definition of current class directly.
    • super is not referring to some instance either, but to static methods and getters written in context of some class current one is extending.

Conclusion

Try this code:

class A {
  constructor( input ) {
    this.loose = this.constructor.getResult( input );
    this.tight = A.getResult( input );
    console.log( this.scaledProperty, Object.getOwnPropertyDescriptor( A.prototype, "scaledProperty" ).get.call( this ) );
  }

  get scaledProperty() {
    return parseInt( this.loose ) * 100;
  }
  
  static getResult( input ) {
    return input * this.scale;
  }
  
  static get scale() {
    return 2;
  }
}

class B extends A {
  constructor( input ) {
    super( input );
    this.tight = B.getResult( input ) + " (of B)";
  }
  
  get scaledProperty() {
    return parseInt( this.loose ) * 10000;
  }

  static get scale() {
    return 4;
  }
}

class C extends B {
  constructor( input ) {
    super( input );
  }
  
  static get scale() {
    return 5;
  }
}

class D extends C {
  constructor( input ) {
    super( input );
  }
  
  static getResult( input ) {
    return super.getResult( input ) + " (overridden)";
  }
  
  static get scale() {
    return 10;
  }
}


let instanceA = new A( 4 );
console.log( "A.loose", instanceA.loose );
console.log( "A.tight", instanceA.tight );

let instanceB = new B( 4 );
console.log( "B.loose", instanceB.loose );
console.log( "B.tight", instanceB.tight );

let instanceC = new C( 4 );
console.log( "C.loose", instanceC.loose );
console.log( "C.tight", instanceC.tight );

let instanceD = new D( 4 );
console.log( "D.loose", instanceD.loose );
console.log( "D.tight", instanceD.tight );

If you are planning on doing any kind of inheritance, then I would recommend this.constructor. This simple example should illustrate why:

class ConstructorSuper {
  constructor(n){
    this.n = n;
  }

  static print(n){
    console.log(this.name, n);
  }

  callPrint(){
    this.constructor.print(this.n);
  }
}

class ConstructorSub extends ConstructorSuper {
  constructor(n){
    this.n = n;
  }
}

let test1 = new ConstructorSuper("Hello ConstructorSuper!");
console.log(test1.callPrint());

let test2 = new ConstructorSub("Hello ConstructorSub!");
console.log(test2.callPrint());
  • test1.callPrint() will log ConstructorSuper Hello ConstructorSuper! to the console
  • test2.callPrint() will log ConstructorSub Hello ConstructorSub! to the console

The named class will not deal with inheritance nicely unless you explicitly redefine every function that makes a reference to the named Class. Here is an example:

class NamedSuper {
  constructor(n){
    this.n = n;
  }

  static print(n){
    console.log(NamedSuper.name, n);
  }

  callPrint(){
    NamedSuper.print(this.n);
  }
}

class NamedSub extends NamedSuper {
  constructor(n){
    this.n = n;
  }
}

let test3 = new NamedSuper("Hello NamedSuper!");
console.log(test3.callPrint());

let test4 = new NamedSub("Hello NamedSub!");
console.log(test4.callPrint());
  • test3.callPrint() will log NamedSuper Hello NamedSuper! to the console
  • test4.callPrint() will log NamedSuper Hello NamedSub! to the console

See all the above running in Babel REPL.

You can see from this that test4 still thinks it's in the super class; in this example it might not seem like a huge deal, but if you are trying to reference member functions that have been overridden or new member variables, you'll find yourself in trouble.