Typescript: instance of an abstract class

user_388506 0
Viewed 12709

Ok. this is very simple, just to remove an annoying error message.

I have an abstract class Router:

export abstract class Router {
}

And an interface like this

interface IModule {
    name: string;
    forms: Array<IForm>,
    route: typeof Router
}

Now I have a class which looks like this, and many others based on Router abstract

export class Module1 extends Router {
}

Now, I want to instantiate the route like this:

let module: IModule = { 
    name: "My Module",
    forms: [],
    route: Module1
}

let router = new module.route(); 
// TS2511: Cannot create an instance of an abstract class.

The code run just fine, and the instance of router = new Module1() is made properly, but I obviously doesn't write it properly in Typescript because I see TS2511: Cannot create an instance of an abstract class. when transpiling.

What is the correct way to define this?

Thank you

typescript
2 Answers

When you have an abstract class, it's constructor is only invokable from a derived class, since the class is abstract and should not be instantiated. This carries over when you type something as typeof AbstractClass, the constructor will not be callable.

The solution would to be to not use typeof Router but rather a constructor signature that returns a Router

interface IModule {
    name: string;
    forms: Array<IForm>,
    route: new () => Router
}

Another solution if you need to access static methods of Router is to create an interface derived from typeof Router which will erase the abstractness of the constructor:

export abstract class Router {
    static m(): void { console.log("Router"); }
}
type RouterClass = typeof Router;
interface RouterDerived extends RouterClass { }

interface IModule {
    name: string;
    route: RouterDerived
}

export class Module1 extends Router {
    static m(): void { console.log("Module1"); }
}

let module: IModule = { 
    name: "My Module",
    route: Module1
}

module.route.m();
let router = new module.route();
user_1257340

Note, second part of the accepted answer covers:

if you need to access static methods of Router ...

The proposed solution doesn't actually work. Just spent fair bit of time trying to figure this out so though I'd add: the way to do this is:

type RouterDerived = {new (): Router} & typeof Router;

Full example playground.

user_20383830
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