具有泛型和对象的TypeScript条件类型

Question

我有以下代码：

type Id = string | number;

type ObjectId<T extends Id> = {
  id: T;
};

type ValidObjectId = ObjectId<string> | ObjectId<number>

type ReverseId<T extends ValidObjectId> = T['id'] extends number ? string : number;

function encodeObjectId<T extends ObjectId<number>>(
  obj: T
): ReverseId<T> {
  const id = obj.id;

  // Why does the following does not accept a string type?
  // As it's the reverse of type number, because of the `ReverseId` type, it should be working.
  return id.toString();
}

^{Playground link.}

其中具有类型ReverseId，该类型将转换id属性的类型。例如，如果给定的对象id属性具有number类型，它将返回string；并且如果给定对象的id属性具有string类型，则将返回number类型。

这确实有效：

type Id = string | number;

type ObjectId<T extends Id> = {
  id: T;
};

type ValidObjectId = ObjectId<string> | ObjectId<number>

type ReverseId<T extends ValidObjectId> = T['id'] extends number ? string : number;

// The `t` constant has a `number` type, which is the "reverse" of the `string`
// type, used in the given `ObjectId`.
const t: ReverseId<ObjectId<string>> = 1;

^{Playground link.}

问题是：当ObjectId来自泛型时，如我们在第一个示例中的encodeObjectId函数中看到的那样，该类型不起作用。

是否可以使用条件类型解决此问题？