How to select a type for a method based on the generic type provided in the class in TypeScript?

Question

I have the following class:

class Processor<TReq, TRes> {
    constructor(
        private handler: (input: TReq) => TRes
    ) {}

    public handleRequest(input: TReq): TRes {
       ...
       const res = this.handler(input);
       ...
    }
}

In this design, I can do this:

const myProcessor = new Processor<string, string>(
    (input: string) => `Hello ${input}`
);
myProcessor.handleRequest('Hello');

My issue is that I want to be able to pass never to TReq, and when that happens, I would like handleRequest to be of type () => TRes, and handler as well:

const myProcessor = new Processor<never, string>(
    () => 'Hi'
);
myProcessor.handleRequest();

How can I achieve this?

Answer 1

I think the closest you can get is the following

1. You need to define generic constraint for TReq
2. You need to make handleRequst argument optional

class Processor<TReq, TRes> {
  private handler: [TReq] extends [never] ? () => TRes : (input: TReq) => TRes;

  constructor(
    handler: [TReq] extends [never] ? () => TRes : (input: TReq) => TRes,
  ) {
    this.handler = handler;
  }

  public handleRequest(): TReq extends never ? TRes : never;
  public handleRequest(input: TReq): TReq extends never ? never : TRes;
  public handleRequest(input?: TReq): TRes {
    const res =
      typeof input === 'undefined'
        ? (this.handler as () => TRes)()
        : (this.handler as (input: TReq) => TRes)(input);
    return res;
  }
}

upd: added handleRequest overloads with generic constraint as suggested in comments. Still not perfect, but i'm afraid this is the best you can get from TS