[TypeScript] Type Challenges #298 - Length of String

题目描述

计算字符串的长度，类似于String#length。

题解

// ============= Test Cases =============
import type { Equal, Expect } from './test-utils'

type cases = [
  Expect<Equal<LengthOfString<''>, 0>>,
  Expect<Equal<LengthOfString<'kumiko'>, 6>>,
  Expect<Equal<LengthOfString<'reina'>, 5>>,
  Expect<Equal<LengthOfString<'Sound! Euphonium'>, 16>>,
]


// ============= Your Code Here =============
type LengthOfString<T extends string, Acc extends string[] = []> =
  T extends `${infer Head}${infer Tail}`
      ? LengthOfString<Tail, [Head, ...Acc]>
      : Acc['length'];

类型约束

使用T extends string对T进行约束，确保T是一个字符串

引入泛型

引入泛型Acc，Acc是一个累积器，用来存储分割的字符串，默认值为[]

条件类型

type LengthOfString<T extends string, Acc extends string[] = []> =
  T extends `${infer Head}${infer Tail}`
      ? LengthOfString<Tail, [Head, ...Acc]>
      : Acc['length'];

• 如果T不是空字符串

  • infer Head：将T的第一个字符推断为Head
  • infer Tail：将T的剩余字符推断为Tail
  • 递归调用LengthOfString<Tail, [Head, ...Acc]>，将Head添加到Acc开头，继续处理Tail

• 如果T是空字符串

  • 返回Acc的长度，即Acc['length']