TypeScript 中泛型类型的子类型推断

Question

我想要一个函数，它接受一些对象并返回其x 属性。该对象需要限制为泛型类型Type<X>，并且我希望返回值的类型是属性x 的类型。

要将输入限制为Type<X>，我需要使用T extends Type<X>，但实际上我必须将X 设置为某种类型值，例如T extends Type<string>，它不适用于Type<number> 或T extends Type<any>丢弃x属性的类型信息。

我希望做类似<T extends Type<any>>(o: T) => T.X 或<T extends Type<???>>(o: T) => typeof o 之类的事情。

TypeScript 中有没有办法做到这一点？如果有，怎么做？

// Suppose I have this generic interface
interface Type<X> {
  readonly x: X
}

// I create one version for foo...
const foo: Type<string> = {
  x: 'abc',
}

// ...and another one for bar
const bar: Type<number> = {
  x: 123,
}

// I want this function to restrict the type of `o` to `Type`
// and infer the type of `o.x` depending on the inferred type of `o`,
// but to restrict to `Type` I must hardcode its X to `any`, which
// makes `typeof o.x` to evaluate to `any` and the type of `o.x` is lost.
function getX<T extends Type<any>> (o: T): typeof o.x {
  return o.x
}

// These are correctly typed
const okFooX: string = getX(foo)
const okBarX: number = getX(bar)

// These should result in error but it is OK due to `Type<any>`
const errorFooX: boolean = getX(foo)
const errorBarX: boolean = getX(bar)

Answer 1

如果我理解正确的话：

function getX<T>(o: Type<T>): T {
    return o.x;
}

然后：

const errorFooX: boolean = getX(foo); // error: Type 'string' is not assignable to type 'boolean'
const errorBarX: boolean = getX(bar); // error: Type 'number' is not assignable to type 'boolean'

(code in playground)