TypeScript类型实现加减乘除详解

type A = Divide<1, 0> // never
type B = Divide<4, 2> // 2
type C = Divide<10, 3> // 3

type Tuple<T extends number, U extends any[] = []> =
    U['length'] extends T ? U : Tuple<T, [...U, any]>
type Subtract<
    A extends number,
    B extends number
> = Tuple<A> extends [...Tuple<B>, ...infer R] ? R['length'] : never
type SmallerThan<
    A extends number,
    B extends number,
    S extends any[] = []
> = S['length'] extends B
    ? false
    : S['length'] extends A
    ? true
    : SmallerThan<A, B, [...S,A]>
type Divide<A extends number, B extends number, S extends any[] = []> =
    B extends 0 ? never : SmallerThan<A, B> extends true ? S['length'] : Divide<Subtract<A, B>, B, [...S, any]>;
type res = Divide<200, 10> // 20

type Divide<A extends number, B extends number, S extends number[] = []> =
    B extends 0 ? never : SmallerThan<A, B> extends true ? S['length'] : Divide<Subtract<A, B>, B, [...S, any]>;

type res = SmallerThan<10,2>  // res为false
type res = SmallerThan<2,20> // res为true

type SmallerThan<
    A extends number,
    B extends number,
    S extends any[] = []
> = S['length'] extends B
    ? false
    : S['length'] extends A
    ? true
    : SmallerThan<A, B, [...S,any]>

type res = SmallerThan<3,2>  
// 首先S['length']=0 ,所以不等于A和B，此时将any推入数组，S数组变成[any]
// 接下来S['length'] =1，还是B等于A和B，此时继续将A放入数组，S数组变成[any,any]
// 此时S['length'] = 2，所以得出S的长度等于B，返回false

type Tuple<T extends number, U extends any[] = []> =
    U['length'] extends T ? U : Tuple<T, [...U, any]>
type res4 = Tuple<3> // [any, any, any]‘
// 基本上和上面的SmallerThan差不多，就是不够长度，就push一个any进去

type Subtract<
    A extends number,
    B extends number
> = Tuple<A> extends [...Tuple<B>, ...infer R] ? R : never
type res3 = Subtract<10,8> // [any, any]
type Subtract<
    A extends number,
    B extends number
> = Tuple<A> extends [...Tuple<B>, ...infer R] ? R['length'] : never
type res3 = Subtract<20,10> // 10
// 一开始把A转换长度为20的数组，B转换成长度为10的数组，然后让ts自己去infer，A的长度等于B的长度加上多少长度的数组，然后返回R的长度

type Add<A extends number, B extends number > = [...Tuple<A>,...Tuple<B>] extends [... infer T] ? T['length']:never

type Tuple<T extends number, U extends any[] = []> =
    U['length'] extends T ? U : Tuple<T, [...U, any]>
type Mutiply<A extends number, B extends number, S extends any[] = []> = B extends 0 ? S['length'] : Mutiply<A, Subtract<B, 1>, [...S, ...Tuple<A>]>
type res = Mutiply<5,6> //30

type Tuple<T extends number, S extends any[] = []> = S['length'] extends T ? S : Tuple<T, [...S, any]> // 不报错
// type Tuple<T extends number, S extends any[] = []> = T extends S['length'] ? S : Tuple<T, [...S, any]> 
// 不能写成T extends S['length']，ts会报递归可能是无限的