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Append Argument

提出詳細

type AppendArgument<T extends Function, K> = T extends (...args: infer R) => infer U ? (...args:[...R, K]) => U : never
提出日時2024-10-29 08:19:56
問題Append Argument
ユーザーbalckowl
ステータスAccepted
テストケース
import type { Equal, Expect } from '@type-challenges/utils'

type Case1 = AppendArgument<(a: number, b: string) => number, boolean>
type Result1 = (a: number, b: string, x: boolean) => number

type Case2 = AppendArgument<() => void, undefined>
type Result2 = (x: undefined) => void

type cases = [
  Expect<Equal<Case1, Result1>>,
  Expect<Equal<Case2, Result2>>,
]