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.

リンクされた TypeScript Playground の再現例から始め、リンクされた回避策と比較しながら、ジェネリックな mixin と継承の処理に焦点を当てます。元の例が回避策なしで型チェックに通り、UserRecord を通じて 'User' 型が保持されれば、issue は完了です。