Mixin / Inheritance problem with Typescript - string not assignable to string|number|symbol

Bug Report

I have a typescript class. I'd like to subclass it and add a mixin. However, it keeps telling me that string not assignable to string|number|symbol. I don't understand why because the method signatures are the same.

🔎 Search Terms

typescript mixin prototype

🕗 Version & Regression Information

4.9.5

I just started using Typescript, so 4.9.5 is the version I was using since I started seeing it.

⏯ Playground Link

this shows the problem with UserRecord

this shows the way to work around the problem

💻 Code

This is as short as I can make the example and still show what's happening:

type TModel = {
    'User' : {
        username : string,
        password : string  | undefined,
    },
    'APIKey' : {
        privateKey: string,
    }
};
type TProps<T extends keyof TModel> = Partial<TModel[T]>;

type Constructor<T = {}> = new (...args: any[]) => T;

 class VingRecord<T extends keyof TModel> {
    constructor( private props: TProps<T>) { }

    public set<K extends keyof TProps<T>>(_key: K) {}
}

function RoleMixin<T extends Constructor>(Base: T) {
    return class RoleMixin extends Base {  };
}

class UserRecord extends RoleMixin(VingRecord<'User'>) {
    public set<K extends keyof TProps<'User'>>(_key: K) {}
}

and if you replace the RoleMixin with this it hides the problem with the funky prototype:

function RoleMixin<T extends IConstructable>(Base: T) {
    class RoleMixin extends Base {};
    return RoleMixin as {
        new (...args: any): RoleMixin;
        prototype: any;
    } & T;
}

🙁 Actual behavior

The code works, but typescript has a problem with the mixin thinking the type on VingRecord is any rather than 'User'. So it reports a type mismatch on the UserRecord.

More detail here with the guys who spotted the behavior

🙂 Expected behavior

I should be able to do the inheritance as shown in the code sample above without having to do the funky return from the mixin.

Inizia con la riproduzione collegata in TypeScript Playground e confrontala con la soluzione alternativa collegata, concentrandoti sulla gestione dei mixin generici e dell’ereditarietà. L’issue è completata quando l’esempio originale supera il controllo dei tipi senza la soluzione alternativa, preservando il tipo 'User' attraverso UserRecord.