ts怎么生成一个40长度的元组？

Question

let tuple:[string,string] = ['hh','123']

上面是一个长度为2的元组，如果我想要一个长度为40，里面都是string类型的元组，我该怎么做？

Answer 1

这可以通过递归来实现,如下:

type GeneraterTuple<
  T,
  N extends number,
  R extends Array<T> = []
> = R['length'] extends N ? R : GeneraterTuple<T, N, [T, ...R]>

type StringTuple40 = GeneraterTuple<string, 40>

TS Playground

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• Recursive Conditional Types

Answer 2

这里提供一种思路，可以通过扩展运算符来简化这种实现：

type MergeTuple<T extends any[], P extends any[]> = [...T, ...P];

type DoubleTuple<T> = T extends any[] ? MergeTuple<T, T> : MergeTuple<[T], [T]>;
type FourfoldTuple<T> = DoubleTuple<DoubleTuple<T>>;
type EightfoldTuple<T> = DoubleTuple<FourfoldTuple<T>>;
type TenfoldTuple<T> = MergeTuple<EightfoldTuple<T>, DoubleTuple<T>>;

type FourtyTuple<T> = MergeTuple< MergeTuple<TenfoldTuple<T>, TenfoldTuple<T> >, MergeTuple< TenfoldTuple<T>, TenfoldTuple<T>>>;

const a: FourtyTuple<number> = [0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39] // ok
const b: FourtyTuple<number> = [0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38] // error
const c: FourtyTuple<number> = [0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,''] // error

Answer 3

type M = {

length: 3;
0: string;

}

Answer 4

从概念上讲，元组是类型不同的元素线性集合。在typescript中好像确实没有限定数组长度的约束（毕竟JavaScript不太需要提前确定数组所需的内存空间大小）。

根据你的需求，尽量还是使用逻辑去控制，不能过分依赖类型系统。

Answer 5

const ls = [1, 2, 3, 4, 5, 6 ] as const;

Answer 6

type TempArr<T, N extends number> = [T, ...T[]] & {length: N};