Call static methods from regular ES6 class methods

Question

What's the standard way to call static methods? I can think of using constructor or using the name of the class itself, I don't like the latter since it doesn't feel necessary. Is the former the recommended way, or is there something else?

Here's a (contrived) example:

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

  static print(n){
    console.log(n);
  }

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

Answer 1

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.

Answer 2

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 class:
  • 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 class:
  • 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/getter
  • this is referring to current instance.
  • super is basically referring to same instance, but somewhat addressing methods/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/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/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 );

Answer 3

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.