有没有办法在 TypeScript 中递归解包函数类型？

Question

假设我们有以下类型 I:

type I = () => () => () => "a" | "b" | "c";

有没有一种方法可以创建泛型类型 Unwrap，使 Unwrap<I> 的计算结果为 "a" | "b" | "c"？

type I = () => () => () => "a" | "b" | "c";

type Result = Unwrap<I>; // "a" | "b" | "c"

以下 (ofc) 产生循环错误：

type Unwrap<
  T extends (...args: any[]) => any,
  R = ReturnType<T>
> = R extends (...args: any[]) => any
  ? Unwrap<R>
  : R;

如有任何帮助，我们将不胜感激。谢谢！

Answer 1

如上面评论中所列，Unwrap 类型适用于 TypeScript 4.1（soon-to-be 已发布）。

Answer 2

这是在 TypeScript 3 中有效的 hack。实际上并没有那么糟糕。

type I = () => (() => (() => "a" | "b" | "c")) | "e" | (() => "f" | "g");

type Unwrap<T> =
    T extends (...args: any[]) => infer R
        ? { 0: Unwrap<R> }[T extends any ? 0 : never] // Hack to recurse.
    : T;

type Result = Unwrap<I>;
// type Result = "e" | "a" | "b" | "c" | "f" | "g";

Playground Link

有没有办法在 TypeScript 中递归解包函数类型？

Is there a way to recursively unwrap a function type in TypeScript?

types

compiler-errors

inferred-type

typescript

unwrap