目的
Github上的类型体操，让你写出更加优雅的TS类型定义，以及探寻TS中我们不熟悉的只是，让我们开始TS的类型挑战把~2022希望通过更文的方式来督促自己学习，每日三题，坚持日更不断~~
题目大纲
01. Medium Reverse
题目要求
import { Equal, Expect, ExpectFalse, NotEqual } from '@type-challenges/utils'
type cases = [
Expect<Equal<Reverse<['a', 'b']>, ['b', 'a']>>,
Expect<Equal<Reverse<['a', 'b', 'c']>, ['c', 'b', 'a']>>
]
我的答案
type Reverse<T extends any[]> = T extends [infer P, ...infer K]
? [...Reverse<K>, P]
: [];
2. Medium Filp Arguements
题目要求
import { Equal, Expect, ExpectFalse, NotEqual } from '@type-challenges/utils'
type cases = [
Expect<Equal<FlipArguments<() => boolean>, () => boolean>>,
Expect<Equal<FlipArguments<(foo: string) => number>, (foo: string) => number>>,
Expect<Equal<FlipArguments<(arg0: string, arg1: number, arg2: boolean) => void>, (arg0: boolean, arg1: number, arg2: string) => void>>
]
我的解答
type FlipArguments<T> = T extends (...args: infer P) => infer R
? (...args: Reverse<P>) => R
: never;
知识点
- 函数类型的推断
- 函数参数的参数用数组加上
...运算符，可以转换为,分割的参数
3. Medium Flatten Depth
题目要求
import { Equal, Expect, ExpectFalse, NotEqual } from '@type-challenges/utils'
type cases = [
Expect<Equal<FlattenDepth<[]>, []>>,
Expect<Equal<FlattenDepth<[1, 2, 3, 4]>, [1, 2, 3, 4]>>,
Expect<Equal<FlattenDepth<[1, [2]]>, [1, 2]>>,
Expect<Equal<FlattenDepth<[1, 2, [3, 4], [[[5]]]], 2>, [1, 2, 3, 4, [5]]>>,
Expect<Equal<FlattenDepth<[1, 2, [3, 4], [[[5]]]]>, [1, 2, 3, 4, [[5]]]>>,
Expect<Equal<FlattenDepth<[1, [2, [3, [4, [5]]]]], 3>, [1, 2, 3, 4, [5]]>>,
Expect<Equal<FlattenDepth<[1, [2, [3, [4, [5]]]]], 19260817>, [1, 2, 3, 4, 5]>>,
]
我的解答
type FlattenDepth<
T extends any[],
N = 1,
R extends any[] = [],
C extends any[] = []
> = T extends [infer P, ...infer K]
? C["length"] extends N
? [...R, ...T]
: P extends any[]
? FlattenDepth<K, N, [...R, ...FlattenDepth<P, N, [], [...C, number]>], C>
: FlattenDepth<K, N, [...R, P], C>
: R;
知识点
- 用数组的
length来进行记数 - 使用默认值来实现参数递归过程中叠加
