从 object 属性推断类型的通用接口

Question

I've got an interface I with the goal of making sure that the property type of val is the same as the return type of fn .

The type in question is restricted to string or number .

interface I<T extends string | number> {
  val: T;
  fn: () => T;
}

const x1: I<string> = {
  val: "hello",
  fn: () => "world",
};

const x2: I<number> = {
  val: 3,
  fn: () => 4,
};

Is it somehow possible to not having to explicitly set T and rather infer it? So that I could just do:

// accepted
const x1: I = {
  val: "hello",
  fn: () => "world",
};

// accepted
const x2: I = {
  val: 3,
  fn: () => 4,
};

// rejected
const x3: I = {
  val: "hello",
  fn: () => 4,
};

// rejected
const x4: I = {
  val: 3,
  fn: () => "world",
};

Answer 1

As far as I know, it is not possible for TS to infer the type for your example. However, there is a way to achieve similar effects using a helper function:

function make<T extends string|number>(input: I<T>) {
  return input;
}

// accepted
const x1 = make({
  val: "hello",
  fn: () => "world",
});

// accepted
const x2 = make({
  val: 3,
  fn: () => 4,
});

// rejected
const x3 = make({
  val: "hello",
  fn: () => 4,
});

// rejected
const x4 = make({
  val: 3,
  fn: () => "world",
});

Playground link