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~~

ts 类型体操 github 我的解答

题目大纲

1. Easy Parameters
2. Medium Return Type
3. Medium Omit

01. Easy Parameters

题目要求

import { Equal, Expect, ExpectFalse, NotEqual } from '@type-challenges/utils'

const foo = (arg1: string, arg2: number): void => {}
const bar = (arg1: boolean, arg2: {a: 'A'}): void => {}
const baz = (): void => {}

type cases = [
  Expect<Equal<MyParameters<typeof foo>, [string, number]>>,
  Expect<Equal<MyParameters<typeof bar>, [boolean, {a: 'A'}]>>,
  Expect<Equal<MyParameters<typeof baz>, []>>,
]

我的答案

type MyParameters<T extends (...args: any[]) => any> = T extends (
  ...args: infer P
) => any
  ? P
  : [];

知识点

1. infer关键字的使用，推断出对应参数的类型，这里如果...args: infer P这个时候推断出的P其实是一个数组类型，如果是...args: infer P[]这个时候的P就是个联合类型了

2. Medium Return Type

题目要求

import { Equal, Expect } from '@type-challenges/utils'

type cases = [
  Expect<Equal<string, MyReturnType<() => string>>>,
  Expect<Equal<123, MyReturnType<() => 123>>>,
  Expect<Equal<ComplexObject, MyReturnType<() => ComplexObject>>>,
  Expect<Equal<Promise<boolean>, MyReturnType<() => Promise<boolean>>>>,
  Expect<Equal<() => 'foo', MyReturnType<() => () => 'foo'>>>,
  Expect<Equal<1 | 2, MyReturnType<typeof fn>>>,
  Expect<Equal<1 | 2, MyReturnType<typeof fn1>>>,
]

type ComplexObject = {
  a: [12, 'foo']
  bar: 'hello'
  prev(): number
}

const fn = (v: boolean) => v ? 1 : 2
const fn1 = (v: boolean, w: any) => v ? 1 : 2

我的解答

type MyReturnType<T extends (...args: any[]) => any> = T extends (
    ...args: any[]
) => infer P
  ? P
  : void;

知识点

1. infer的简单使用。

3. Medium Omit

题目要求

import { Equal, Expect } from '@type-challenges/utils'

type cases = [
  Expect<Equal<Expected1, MyOmit<Todo, 'description'>>>,
  Expect<Equal<Expected2, MyOmit<Todo, 'description' | 'completed'>>>
]

interface Todo {
  title: string
  description: string
  completed: boolean
}

interface Expected1 {
  title: string
  completed: boolean
}

interface Expected2 {
  title: string
}

我的解答

type MyEx<T, K> = T extends K ? never : T;

type MyOmit<T, K extends keyof T> = {
  [key in MyEx<keyof T, K>]: T[key];
};

知识点

1. 在Mapped的语法中可以使用Exclude，Extract等泛型，将原本的联合类型得到处理，得到我们自己想要的类型