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数值运算 ​

TypeScript 虽然图灵完备，但没有加减乘除运算符，但可以通过构造不同的数组 ( 元组 ) ，然后取 length 的方式完成数值运算，把数值的加减乘除转化为对数组的提取和构造。

构造数组 ​

ts
type ConstructTuple<
  T extends number,
  Res extends unknown[] = [],
> = Res["length"] extends T ? Res : ConstructTuple<T, [...Res, unknown]>

type Res = ConstructTuple<3> // [unknown, unknown, unknown]
type ConstructTuple<
  T extends number,
  Res extends unknown[] = [],
> = Res["length"] extends T ? Res : ConstructTuple<T, [...Res, unknown]>

type Res = ConstructTuple<3> // [unknown, unknown, unknown]

加减乘除 ​

Sum ​

加法：分别构造对应加数长度的两个元组，然后合并成一个

ts
type Sum<A extends number, B extends number> = [
  ...ConstructTuple<A>,
  ...ConstructTuple<B>,
]["length"]

type Res = Sum<1, 1> // 2
type Res1 = Sum<999, 999> // 1998

// eslint-disable-next-line @typescript-eslint/ban-ts-comment
// @ts-expect-error
type Res2 = Sum<999, 1000> // 报错：类型实例化过深，且可能无限。ts(2589)
// eslint-disable-next-line @typescript-eslint/ban-ts-comment
// @ts-expect-error
type Res3 = Sum<-1, 0> // 报错：类型实例化过深，且可能无限。ts(2589)
type Sum<A extends number, B extends number> = [
  ...ConstructTuple<A>,
  ...ConstructTuple<B>,
]["length"]

type Res = Sum<1, 1> // 2
type Res1 = Sum<999, 999> // 1998

// eslint-disable-next-line @typescript-eslint/ban-ts-comment
// @ts-expect-error
type Res2 = Sum<999, 1000> // 报错：类型实例化过深，且可能无限。ts(2589)
// eslint-disable-next-line @typescript-eslint/ban-ts-comment
// @ts-expect-error
type Res3 = Sum<-1, 0> // 报错：类型实例化过深，且可能无限。ts(2589)

受限于 TypeScript 递归次数的限制，构造数组的最大长度为 999 ；受限于数组 ( 元组 ) 的定义，构造数组的最小长度为 0 ，无法处理负数。

Subtract ​

减法：减数 = 被减数 + 差

减数构造数组的长度 = 被减数构造数组和差构造数组合并后构造数组的长度

ts
// M => minuend 被减数, S => subtrahend 减数
type Subtract<M extends number, S extends number> = ConstructTuple<M> extends [
  ...ConstructTuple<S>,
  ...infer Res,
]
  ? Res["length"]
  : never

type Res = Subtract<2, 1> // 1
type Res1 = Subtract<0, 1> // never
type Res2 = Subtract<999, 998> // 1

// eslint-disable-next-line @typescript-eslint/ban-ts-comment
// @ts-expect-error
type Res3 = Subtract<1000, 999> // 报错：类型实例化过深，且可能无限。ts(2589)
// M => minuend 被减数, S => subtrahend 减数
type Subtract<M extends number, S extends number> = ConstructTuple<M> extends [
  ...ConstructTuple<S>,
  ...infer Res,
]
  ? Res["length"]
  : never

type Res = Subtract<2, 1> // 1
type Res1 = Subtract<0, 1> // never
type Res2 = Subtract<999, 998> // 1

// eslint-disable-next-line @typescript-eslint/ban-ts-comment
// @ts-expect-error
type Res3 = Subtract<1000, 999> // 报错：类型实例化过深，且可能无限。ts(2589)

TypeScript 无法处理负数，当差为负数时，返回 never 。

Multiply ​

乘法：

2 x 5 = 10 可以理解为 2 + 2 + 2 + 2 + 2 = 10 或 5 + 5 = 10

ts
type Multiply<
  A extends number,
  B extends number,
  Res extends unknown[] = [],
> = A extends 0
  ? Res["length"]
  : Multiply<Subtract<A, 1>, B, [...Res, ...ConstructTuple<B>]>

type Res = Multiply<2, 5> // 10
type Multiply<
  A extends number,
  B extends number,
  Res extends unknown[] = [],
> = A extends 0
  ? Res["length"]
  : Multiply<Subtract<A, 1>, B, [...Res, ...ConstructTuple<B>]>

type Res = Multiply<2, 5> // 10

乘法运算同样收到递归次数限制，且大数乘法及其消耗性能，会很卡。

TypeScript 中的元组长度也有上限。

Divide ​

除法：被除数不断减去减数，直到为 0 ，记录的次数就是商。

ts
type Divide<
  A extends number,
  B extends number,
  Res extends unknown[] = [],
> = A extends 0 ? Res["length"] : Divide<Subtract<A, B>, B, [unknown, ...Res]>

type Res = Divide<30, 5> // 6
type Divide<
  A extends number,
  B extends number,
  Res extends unknown[] = [],
> = A extends 0 ? Res["length"] : Divide<Subtract<A, B>, B, [unknown, ...Res]>

type Res = Divide<30, 5> // 6

只能实现整除，且有边界条件没有判断。

数组长度实现计数 ​

StrLen ​

ts
// 尾递归
type StrLen<
  S extends string,
  Res extends unknown[] = [],
> = S extends `${string}${infer R}`
  ? StrLen<R, [...Res, unknown]>
  : Res["length"]

type Res = StrLen<"hello">
// 尾递归
type StrLen<
  S extends string,
  Res extends unknown[] = [],
> = S extends `${string}${infer R}`
  ? StrLen<R, [...Res, unknown]>
  : Res["length"]

type Res = StrLen<"hello">

GreaterThan ​

ts
type GreaterThan<
  A extends number,
  B extends number,
  Res extends unknown[] = [],
> = A extends B
  ? false
  : Res["length"] extends B
  ? true
  : Res["length"] extends A
  ? false
  : GreaterThan<A, B, [...Res, unknown]>

type Res = GreaterThan<2, 1> // true
type Res1 = GreaterThan<1, 2> // false
type GreaterThan<
  A extends number,
  B extends number,
  Res extends unknown[] = [],
> = A extends B
  ? false
  : Res["length"] extends B
  ? true
  : Res["length"] extends A
  ? false
  : GreaterThan<A, B, [...Res, unknown]>

type Res = GreaterThan<2, 1> // true
type Res1 = GreaterThan<1, 2> // false

Fibonacci ​

ts
type FibonacciLoop<
  PrevArr extends unknown[],
  CurrentArr extends unknown[],
  IndexArr extends unknown[] = [],
  Num extends number = 1,
> = IndexArr["length"] extends Num
  ? CurrentArr["length"]
  : FibonacciLoop<
      CurrentArr,
      [...PrevArr, ...CurrentArr],
      [...IndexArr, unknown],
      Num
    >

type Fibonacci<Num extends number> = FibonacciLoop<[1], [], [], Num>

type Res = Fibonaccci<8> // 21
type FibonacciLoop<
  PrevArr extends unknown[],
  CurrentArr extends unknown[],
  IndexArr extends unknown[] = [],
  Num extends number = 1,
> = IndexArr["length"] extends Num
  ? CurrentArr["length"]
  : FibonacciLoop<
      CurrentArr,
      [...PrevArr, ...CurrentArr],
      [...IndexArr, unknown],
      Num
    >

type Fibonacci<Num extends number> = FibonacciLoop<[1], [], [], Num>

type Res = Fibonaccci<8> // 21