String Join

题目来源(type-challenges)

问题 #

实现一个拼接函数的定义。

// Edge cases
const noCharsOutput = join('-')()
const oneCharOutput = join('-')('a')
const noDelimiterOutput = join('')('a', 'b', 'c')

// Regular cases
const hyphenOutput = join('-')('a', 'b', 'c')
const hashOutput = join('#')('a', 'b', 'c')
const twoCharOutput = join('-')('a', 'b')
const longOutput = join('-')('a', 'b', 'c', 'd', 'e', 'f', 'g', 'h')

type case1 = typeof noCharsOutput // ''
type case2 = typeof oneCharOutput // 'a'
type case3 = typeof noDelimiterOutput // 'abc'
type case4 = typeof twoCharOutput // 'a-b'
type case5 = typeof hyphenOutput // 'a-b-c'
type case6 = typeof hashOutput // 'a#b#c'
type case7 = typeof longOutput // 'a-b-c-d-e-f-g-h'

解答 #

type Join<T, U extends string | number> =
  T extends [infer F, ...infer R]
  ? R extends []
    ? F
    : F extends string | number
      ? `${F}${U}${Join<R, U>}`
      : ''
  : ''

declare function join<D extends string>(delimiter: D): <T extends string[]>(...parts: T) => Join<T, D>

拆分 #

• 每个函数的开头加上泛型，不能跨函数声明泛型