[英]Is there a way to recursively unwrap a function type in TypeScript?
假設我們有以下類型I
:
type I = () => () => () => "a" | "b" | "c";
有沒有辦法創建一個通用類型Unwrap
使得Unwrap<I>
評估為"a" | "b" | "c"
"a" | "b" | "c"
"a" | "b" | "c"
?
type I = () => () => () => "a" | "b" | "c";
type Result = Unwrap<I>; // "a" | "b" | "c"
以下 (ofc) 會產生圓度誤差:
type Unwrap<
T extends (...args: any[]) => any,
R = ReturnType<T>
> = R extends (...args: any[]) => any
? Unwrap<R>
: R;
任何幫助將不勝感激。 謝謝!
好吧,這是在 TypeScript 3 中有效的 hack。實際上並沒有那么糟糕。
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";
如上面的評論中所列, Unwrap
類型適用於 TypeScript 4.1(即將發布)。
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