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";

有没有办法创建一个通用类型Unwrap使得Unwrap<I>评估为"a" | "b" | "c" "a" | "b" | "c" "a" | "b" | "c" ？

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

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

以下 (ofc) 会产生圆度误差：

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

任何帮助将不胜感激。 谢谢！

Answer 1

好吧，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";

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Answer 2

如上面的评论中所列， Unwrap类型适用于 TypeScript 4.1（即将发布）。