Using TypeScript will result in the return type of a combined generic type always being unknown

I am designing a settings store system where users can add settings with values of any type and retrieve them with correct typing. To achieve this, I thought about utilizing connected generic type arguments. The idea is that the return type of the value should be inferred based on the type argument passed to the `getSetting()` method. By using the type as a key, the system should be able to fetch the correct information.

Currently, I have implemented a method `getSetting()` that takes a type as input and should infer the return type accordingly. However, I have encountered a problem where the return type of `getSetting` is always shown as `unknown`, even though I expected it to be `number` in a specific example.

abstract class Setting<T> {
  abstract value: T;
}
class SD extends Setting<number> {
  value = 0;
}

let getSetting = <T extends Setting<K>, K>(type: new () => T): K => {
  // return value somehow
};

let setting = getSetting(SD); // the type of setting is shown as unknown instead of number

I am seeking to understand where the error lies in my approach, or if there might be a better way to achieve the desired functionality. Could there be an alternative solution to this problem?

Answer №1

When it comes to type inference, it occurs from arguments rather than from other type parameters. This means that in order to achieve the correct type inference, you will need to reposition the K in the parameter list.

An easy solution would be to eliminate T entirely. You can simply specify that the constructor returns Setting<K>, allowing any derived class to work effectively. Since T is not utilized in any other context, this approach seems to be the most straightforward:

let getSetting = <K>(type: new () => Setting<K>): K => {
  return new type().value
};

Playground Link

If you find that T is necessary for something else outside of this example, there are alternative approaches available. One option is to utilize both T and Setting<K> in the constructor return type:

let getSetting = <T extends Setting<K>, K>(type: new () => T & Setting<K>): K => {
  return new type().value
};

Playground Link

Alternatively, you could eliminate K altogether and extract the value type using an index type query:


let getSetting = <T extends Setting<any>>(type: new () => T): T['value'] => {
  return new type().value
};

Playground Link

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