Typescript abstract optional method

Question

I have an abstract class with some abstract methods. Is there a way to mark some of these methods as optional?

abstract class Test {     protected abstract optionalMethod?(): JSX.Element[];     protected abstract requiredMethod(): string; } 

For some reasons I can add the ? as a suffix but it seems like it does nothing at all because I have to still implement the method in the derived class. For now I am using it like this to mark it that it can return null which is basically the poor mans optional.

protected abstract optionalMethod?(): JSX.Element[] | null; 

Answer 1

You can do this with class and interface merging:

interface Foo {    print?(l: string): void; }  abstract class Foo {   abstract baz(): void;    foo() {     this.print && this.print('foo');   } }  class Bar extends Foo {   baz(): void {     if (this.print) {       this.print('bar');     }   } } 

Link to the above code in the Typescript playground

Answer 2

I'm not sure if this changed at some point, but today (TS 4.3) you can simply make the base-class optional method non abstract:

abstract class Base {     protected abstract required(): void;     protected optional?(): string;      print(): void {         console.log(this.optional?.() ?? "Base");     } }  class Child1 extends Base {     protected required(): void { } }  class Child2 extends Base {     protected required(): void { }     protected optional(): string {         return "Child";     } }  const c1 = new Child1(); c1.print();  const c2 = new Child2(); c2.print(); 

Try it on the TS Playground.