How to create Class dynamically in a proper way in Typescript?

Question

I am trying to migrate my app to typescript. I have kind of a base class that is my base library object. My classes created dependent on the base class. Below is a glimpse of my problem.

Below code works but autocomplete doesn't work. Couldn't figure out what should be defined for the type of Model.

const map = new WeakMap();
function weakRef<T>(context: T): T {
  // @ts-ignore
  if (!map.has(context)) { map.set(context, {}); }
  // @ts-ignore
  return map.get(context);
}
function getModel(provider: Provider) {
  return class Model {
    getSomething(key: string) {
      return weakRef(provider).configuration(key);
    }
  };
}

class Provider {
  configuration: (key: string) => string;
  constructor() {
    weakRef(this).configuration = (key: string) => {
      return key;
    };
    weakRef(this).Model = getModel(this);
  }
  get Model(): any {
    return weakRef(this).Model;
  }
  set Model(model: any) {
    weakRef(this).Model = model;
  }
}

const provider = new Provider();
const obj = new (provider.Model)();
console.log(obj.getSomething('test')); // This works, but autocomplete doesn't

I don't want to pass provider to the constructor in base model. Any help is appreciated, thanks.

Gerrit0 · Accepted Answer · 2019-04-20 21:03:43Z

2

You can get the return type of a function type with ReturnType<T>.

In this case, ReturnType<typeof getModel> should do the trick.

You can also fix the //@ts-ignore lines by constraining T to be an object type.

const map = new WeakMap();
function weakRef<T extends object>(context: T): T {
  if (!map.has(context)) { map.set(context, {}); }
  return map.get(context);
}

That said, I think this architecture is unnecessarily complicated. Why not just use this instead of a value tied to this?

answered Apr 20, 2019 at 21:03

Gerrit0

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2 Comments

trd3v3lop Over a year ago

Thanks it works. To answer your question, main reason is to make memory optimized library. you can look weakmap or weakset for more information.

Gerrit0 Over a year ago

@trd3v3lop that would make sense if multiple instances were sharing the same data... but they aren't, at least in your example. It is useless overhead.

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