Typescript anonymous class

Question

Is there a way in typescript to create anonymous class?

my code:

export abstract class Runnable {     public abstract run(); } 

and I'm trying to do something like this:

new Runnable {     runner() {         //implement     } } 

How can I do it?

Answer 1

Yes, here is the way.

The abstract class:

export abstract class Runnable {   public abstract run(); } 

An anonymous implementation:

const runnable = new class extends Runnable {   run() {     // implement here   } }(); 

Answer 2

Not quite, but you can do this:

abstract class Runnable {     public abstract run(); }  let runnable = new (class MyRunnable extends Runnable {     run() {         console.log("running...");     } })();  runnable.run(); // running... 

(code in playground)

The problem with this approach however is that the interpreter will evaluate the class every time it uses it, unlike with a compiled language (such as java) in which the compiler only evaluates it once.