Typescrip类型体操 - IndexOf

题目

中文

实现类型版本的 Array.indexOf, indexOf<T, U> 接受一个数组T和any类型的U作为参数, 返回T中第一个U的索引

type Res = IndexOf<[1, 2, 3], 2>; // expected to be 1
type Res1 = IndexOf<[2,6, 3,8,4,1,7, 3,9], 3>; // expected to be 2
type Res2 = IndexOf<[0, 0, 0], 2>; // expected to be -1

English

Implement the type version of Array.indexOf, indexOf<T, U> takes an Array T, any U and returns the index of the first U in Array T.

type Res = IndexOf<[1, 2, 3], 2>; // expected to be 1
type Res1 = IndexOf<[2,6, 3,8,4,1,7, 3,9], 3>; // expected to be 2
type Res2 = IndexOf<[0, 0, 0], 2>; // expected to be -1

答案

注释中包含了部分解析

// 使用`[]`可以防止发生[`distribute`](https://www.typescriptlang.org/docs/handbook/2/conditional-types.html#distributive-conditional-types)
type IsEqual<L, R> = [L extends R ? 1 : 0, R extends L ? 1 : 0] extends [1, 1] ? true : false;
// `IsEqual`的错误写法
// `IsEqualMistake<any, string>`期望返回`true`, 实际是`boolean`, 发生了`distribute`
type IsEqualMistake<L, R> = L extends R ? R extends L ? true : false : false;
// IsEqual的另一种写法
type IsEqual2<L, R> = [Exclude<L, R>, Exclude<R, L>] extends [never, never] ? true : false;
type IndexOf<T extends any[], U extends any, S extends any[] = []> = T extends [infer L, ...infer R]
  ? true extends IsEqual<L, U>
    ? S['length']
    : IndexOf<R, U, [...S, 0]>
  : -1

在线演示