number 在类型体操中的运算

在TS类型体操运算中有时候涉及到number类型的操作比较麻烦
即不能直接进行字符串操作以及加减乘除等数字运算

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1、转成字符串

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假如现在有个题目 trunc 要求实现跟Math.trunc() 函数一样的功能，即取出数字整数部分，TS类型体操中应该怎么处理呢？

• 了解将number类型转成string类型， 通过${} 字符串模板转换

// 将number类型转成string
type NumberToString<T extends number> = `${T}`

type test = NumberToString<123> // "123"

• 通过TS推算infer 就可以进行拆分提取了

  // 类型函数Rrunc
  type Trunc<T extends string | number> = 
  `${T}` extends `${infer L}.${infer Rest}` 
  ? L extends '' 
    ? '0'
    : L
  : `${T}`
  
  // 测试用例
  import type { Equal, Expect } from '@type-challenges/utils'
  
  type cases = [
    Expect<Equal<Trunc<0.1>, '0'>>,
    Expect<Equal<Trunc<0.2>, '0'>>,
    Expect<Equal<Trunc<1.234>, '1'>>,
    Expect<Equal<Trunc<12.345>, '12'>>,
    Expect<Equal<Trunc<-5.1>, '-5'>>,
    Expect<Equal<Trunc<'.3'>, '0'>>,
    Expect<Equal<Trunc<'1.234'>, '1'>>,
    Expect<Equal<Trunc<'-10.234'>, '-10'>>,
    Expect<Equal<Trunc<10>, '10'>>,
  ]

2、加法运算

type BuildArray<Length extends number, Ele  = unknown, Arr extends unknown[] = []> =
    Arr['length'] extends Length
        ? Arr
        : BuildArray<Length, Ele, [...Arr, Ele]>

type Add<Num1 extends number, Num2 extends number> = 
    [...BuildArray<Num1>, ...BuildArray<Num2>]['length']

type Add2<Num1 extends 1000, Num2 extends 1> = 
    [...BuildArray<Num1>, ...BuildArray<Num2>]['length']

type AddResult1 = Add<999, 1> // 1000
type AddResult2 = Add<1000, 1> // Type instantiation is excessively deep and possibly infinite: 类型深度可能是无限的 

ts在递归超过1000次的时候会报 深度可能是无限的错误