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Currying 1

提出詳細

type Currying<T> = T extends (...arg: infer Arg) => infer Ret
  ? Arg["length"] extends 0 | 1
    ? T
  : Arg extends [infer L, ...infer R]
    ? (a: L) => Currying<(...a: R) => Ret>
  : Ret
  : never;
提出日時2023-08-30 08:05:34
問題Currying 1
ユーザーookkoouu
ステータスWrong Answer
テストケース
import type { Equal, Expect } from '@type-challenges/utils'

const curried1 = Currying((a: string, b: number, c: boolean) => true)
const curried2 = Currying((a: string, b: number, c: boolean, d: boolean, e: boolean, f: string, g: boolean) => true)

type cases = [
  Expect<Equal<
    typeof curried1, (a: string) => (b: number) => (c: boolean) => true
  >>,
  Expect<Equal<
    typeof curried2, (a: string) => (b: number) => (c: boolean) => (d: boolean) => (e: boolean) => (f: string) => (g: boolean) => true
  >>,
]