typescript type类型使用梳理总结

interface Parent {
  id: string;
}
// 符合里氏替换原则，不会报错
interface Child extends Parent {
  id: "childIdA"|"childIdB"; // "childIdA"|"childIdB"是string的子集
  key: number; // Parent中没有key，该属性是Parent的拓展
}
// 不符合里氏替换原则，报错
interface ChildError extends Parent {
  id: number; // number不是string的子集
}
type A = {a: string};
type B = {a: "A"|"a", b: number};
let a: A = {a: "xxx"};
let b: B = {a: "a", b: 2};
// 符合B是A的子集，该赋值符合里氏替换原则, 反之 b=a 会报错
a = b;

type Fn<T> = T; // 看作 const Fn = val => val;

type R = "A" | "B"
type T = {
  [k in R]: k
}

type MyPick<T, K extends keyof T> = {
  [k in K]: T[k]
}
type A = MyPick<{a: string, b: number, c: boolean}, "a"|"c">

type Equal5<T> = T extends 5 ? true : false;

// eq1: 实现`GetProp<Obj, K>`（获取Obj[K]类型）
type GetProp<Obj, K> = K extends keyof Obj ? Obj[K] : undefined;
// eq2: 实现getName<User>
type GetName<User> = GetProp<User, "name">

// 实现KeyOf
type KeyOf<T> = T extends {[k in infer U]: unknown} ? U : never;
// 实现ValueOf
type ValueOf<T> = T extends { [k in keyof T]: infer U } ? U : never;
// 实现Parameters和ReturnType
type FuncBase<F, C extends "params"|"return"> = F extends (...params: infer P) => infer R ? (C extends "params" ? P : R) : never;
type Params<F> = FuncBase<F, "params">;
type Return<F> = FuncBase<F, "return">;

// 需要实现类似js的const {name, ...extra} = user,求extra。（其实就是Omit方法）
type MyPick<Obj, T extends keyof Obj> = {[k in T]: Obj[k]};
type MyExclude<Obj, T extends keyof Obj> = keyof Obj extends T|(infer U extends keyof Obj) ? U : never;
type MyOmit<Obj, T extends keyof Obj> =  MyPick<Obj, MyExclude<Obj, T>>;
// 实现数组的解构赋值const [A, ...extra] = arr;
type GetArrBase<T extends unknown[], C extends "first"|"rest"> = T extends [infer First, ...infer Rest] ? (C extends "first"?First: Rest): never;
type GetFirst<T extends unknown[]> = GetArrBase<T, "first">;
type GetRest<T extends unknown[]> = GetArrBase<T, "rest">;

type Obj = {
  name: string;
  age: 20;
}
Obj["name"] // string;
Obj.age // Error
Obj["age"] // 20
Obj['name'|'age'] // string | 20 , 这个特性很重要！！！

interface Test {
  a: string;
  b: number;
  c: boolean;
}
// 之前不知道这个特性时，用infer也能达到同样的效果，但实现不如这个直观
type ValueOf<T> = T[keyof T];
type R = ValueOf<Test>;

type A = string[];
type B = [string, number, boolean];
// =========================分割线==========================
// 重点注意！！！
A[0]; // string
A[1]; // string
B[0]; // string
B[1]; // number;
B[0|2]; // string|boolean
// 注意以下写法，为什么可以这么写，因为number是所有数字的集合
A[number]; // string
B[number]; // string | number | boolean
A["length"]; // number
B["length"]; // 3
// ts数组同样可以像js数组那样展开
type Spread<T extends unknown[]> = [...T]
Spread<[1,2,3]> // [1,2,3]

// eq1: 实现 `ItemOf`方法（获取数组中项的类型）
type ItemOf<T extends unknown[]> = T[number];
// 之前不知道这个特性时，用infer实现的代码如下
type ItemOfByinfer<T> = T extends (infer N)[] ? N : never;
// eq2:实现`Append`方法
type Append<T extends unknown[], Ele> = [...T, Ele];
// eq3: 实现返回数组length+1
// ts虽然无法实现加减运算，但可以通过模拟生成对应新类型，返回其属性，从而模拟加减运算
type LengthAddOne<T extends unknown[]> = [unknown, ...T]["length"];

// 返回Push后的数组
type Append<T extends unknown[], E = unknown> = [...T, U];
// 同理，返回Pop后的数组代码如下，暂时用不到
// type RemoveTop<T extends unknown[]> = T extends [...(infer U), unknown] ? U : never;
type Tuple<N extends number, Arr extends unknown[] = []> = Arr["length"] extends N ? Arr : Tuple<N, Append<Arr>>

type Add<A extends number, B extends number> = [...Tuple<A>, ...Tuple<B>]["length"];
type Subtract<A extends number, B extends number> = Tuple<A> extends [...Tuple<B>, ...(infer U)] ? U["length"] : never;

type MultipleBase<A extends number, B extends number, S extends unknown[] = []> = 
    B extends 0
    ? S["length"]
    : MultipleBase<A, Subtract<B, 1>, [...S, ...Tuple<A>]>;

type BiggerThan<A extends number, B extends number> = Tuple<A> extends [...Tuple<B>, ...infer U] ? (U["length"] extends (never|0) ? false : true): false;
// 优化后的乘法如下
type Mutiple<A extends number, B extends number> = BiggerThan<A, B> extends true ? MultipleBase<A, B> : MultipleBase<B, A>;

type Divide<A extends number, B extends number, R extends unknown[] = []> = BiggerThan<B, A> extends true ? R["length"] : Divide<Subtract<A,B>, B, Append<R>>;