[Typescript] 111. Hard – String Join

Create a type-safe string join utility which can be used like so:

const hyphenJoiner = join('-')
const result = hyphenJoiner('a', 'b', 'c'); // = 'a-b-c'

Or alternatively:

join('#')('a', 'b', 'c') // = 'a#b#c'

When we pass an empty delimiter (i.e ”) to join, we should concat the strings as they are, i.e:

join('')('a', 'b', 'c') // = 'abc'

When only one item is passed, we should get back the original item (without any delimiter added):

join('-')('a') // = 'a'

/* _____________ Your Code Here _____________ */
type Join<T extends unknown[], U extends string, ACC extends string = ''> = T extends [infer F extends string, ...infer RT] 
  ? ACC extends ''
    ?  Join<RT, U, `${F}`>
    : Join<RT, U, `${ACC}${U}${F}`>
  : ACC;
declare function join<J extends string>(delimiter: J): <Args extends string[]>(...parts: Args) => Join<Args, J>

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

// 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 cases = [
  Expect<Equal<typeof noCharsOutput, ''>>,
  Expect<Equal<typeof oneCharOutput, 'a'>>,
  Expect<Equal<typeof noDelimiterOutput, 'abc'>>,
  Expect<Equal<typeof twoCharOutput, 'a-b'>>,
  Expect<Equal<typeof hyphenOutput, 'a-b-c'>>,
  Expect<Equal<typeof hashOutput, 'a#b#c'>>,
  Expect<Equal<typeof longOutput, 'a-b-c-d-e-f-g-h'>>,
]

原文地址：http://www.cnblogs.com/Answer1215/p/16906933.html