Typescript:递增数字类型

我只想硬编码如下:

type Increment<N extends number> = [
  1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,
  21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,
  38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54, // as far as you need
  ...number[] // bail out with number
][N]

type Zero = 0
type One = Increment<Zero> // 1
type Two = Increment<One>  // 2

type WhoKnows = Increment<12345>; // number

正如我在其他评论中所说,目前还没有对这种自然递归类型的强大支持。我 would love it 如果它被支持,但它不在那里。在实践中,我发现,如果某个东西可以处理长度为20左右的元组,这就足够了,但您的体验可能会有所不同。

Increment<123456789> 将评估为 123456790 )我很想看看。也许将来有一天它会成为语言的一部分。

此解决方案不是硬编码的,但由于TypeScript的递归限制,它没有用处。当我测试它时,它无法处理 Number2Nat

type SomeNat = [...unknown[]];
type Zero = [];
type Succ<N extends SomeNat> = [...N, unknown];
type Nat2Number<N extends SomeNat> = N["length"];
type Dec<N extends SomeNat> = N extends [unknown, ...infer T] ? T : never;
type Add<N extends SomeNat, M extends SomeNat> = [...N, ...M];
type Sub<N extends SomeNat, M extends SomeNat> = M extends Zero ? N : Sub<Dec<N>, Dec<M>>;
type Number2Nat<I extends number, N extends SomeNat = Zero> = I extends Nat2Number<N> ? N : Number2Nat<I, Succ<N>>;