타입 스크립트 4버전으로 하길 추천한다.

/*
  15 - Last of Array
  -------
  by Anthony Fu (@antfu) #medium #array #4.0
  
  ### Question
  
  > TypeScript 4.0 is recommended in this challenge
  
  Implement a generic `Last<T>` that takes an Array `T` and returns it's last element's type.
  
  For example
  
  ```ts
  type arr1 = ['a', 'b', 'c']
  type arr2 = [3, 2, 1]
  
  type tail1 = Last<arr1> // expected to be 'c'
  type tail2 = Last<arr2> // expected to be 1

View on GitHub: https://tsch.js.org/15 */

/* _____________ Your Code Here _____________ */

type Last<T extends any[]> = any

/* _____________ Test Cases _____________ */ import { Equal, Expect } from '@type-challenges/utils'

type cases = [ Expect<Equal<Last<[3, 2, 1]>, 1>>, Expect<Equal<Last<[() => 123, { a: string }]>, { a: string }>>, ]

/* _____________ Further Steps _____________ / /

Share your solutions: https://tsch.js.org/15/answer View solutions: https://tsch.js.org/15/solutions More Challenges: https://tsch.js.org */

못 풀겠어서 답을 봤다.

기발한 솔루션이 많다.

```tsx
type Last<T extends any[]> = T extends [...infer _, infer L] ? L : never

• infer를 잘 활용하면 일반 변수 값처럼 쓸 수가 있구나!

많은 추천을 받은 방법

type Last<T extends any[]> = [any, ...T][T['length']]

• 배열의 갯수를 강제로 하나 늘려서 length로 접근했다!
• 뭔가 꼼수처럼 보이지만 간단하고 기발한데?

type Last<T extends any[]> = 
  ((...args: T) => any) extends (a: infer N, ...r: infer R) => any ? R['length'] extends 0 ? N : Last<R> : never