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强制数组在联合类型上是穷举的

typescript

0

Jamie Treworgy · 技术社区 · 7 年前

给定使用以下技术创建的强类型元组: here :

const tuple = <T extends string[]>(...args: T) => args;
const furniture = tuple('chair', 'table', 'lamp');

// typeof furniture[number] === 'chair' | 'table' | 'lamp'

我想在设计时断言,它比另一种联合类型更详尽:

type Furniture = 'chair' | 'table' | 'lamp' | 'ottoman'

我如何创建一个类型来确保 furniture 仅包含中的每个类型 Furniture 协会

我们的目标是能够在这样的设计阶段创建一个数组,并让它失败 家具 改变理想的语法可能如下所示:

const furniture = tuple<Furniture>('chair', 'table', 'lamp')

0 回复 | 直到 7 年前

1

9

jcalz 5 年前

TypeScript实际上并不直接支持“穷举数组”。您可以引导编译器检查这一点,但这对您来说可能有点混乱。一个绊脚石是缺少 部分类型参数推断 (按照 microsoft/TypeScript#26242 ).以下是我的解决方案:

type Furniture = 'chair' | 'table' | 'lamp' | 'ottoman';

type AtLeastOne<T> = [T, ...T[]];

const exhaustiveStringTuple = <T extends string>() =>
  <L extends AtLeastOne<T>>(
    ...x: L extends any ? (
      Exclude<T, L[number]> extends never ? 
      L : 
      Exclude<T, L[number]>[]
    ) : never
  ) => x;


const missingFurniture = exhaustiveStringTuple<Furniture>()('chair', 'table', 'lamp');
// error, Argument of type '"chair"' is not assignable to parameter of type '"ottoman"'

const extraFurniture = exhaustiveStringTuple<Furniture>()(
  'chair', 'table', 'lamp', 'ottoman', 'bidet');
// error, "bidet" is not assignable to a parameter of type 'Furniture'

const furniture = exhaustiveStringTuple<Furniture>()('chair', 'table', 'lamp', 'ottoman');
// okay

如你所见, exhaustiveStringTuple 是一个 curried 函数,其唯一目的是获取手动指定的类型参数 T 然后返回一个新函数,该函数接受类型受约束的参数 T 但根据电话推断。(如果我们有适当的部分类型参数推断,可以消除currying。)就你而言, T 将指定为 Furniture .如果你只关心 exhaustiveStringTuple<Furniture>() ,然后你可以用它来代替:

const furnitureTuple =
  <L extends AtLeastOne<Furniture>>(
    ...x: L extends any ? (
      Exclude<Furniture, L[number]> extends never ? L : Exclude<Furniture, L[number]>[]
    ) : never
  ) => x;

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2

0

Przemyslaw Jan Beigert 7 年前

我还有别的建议

type RemoveFirstFromTuple<T extends any[]> = 
  T extends [] ? undefined :
  (((...b: T) => void) extends (a: any, ...b: infer I) => void ? I : [])

const tuple = <T extends string[]>(...args: T) => args;

type FurnitureUnion = 'chair' | 'table' | 'lamp';
type FurnitureTuple = ['chair', 'table' , 'lamp'];

type Check<Union, Tuple extends Array<any>> = {
  "error": never,
  "next": Check<Union, RemoveFirstFromTuple<Tuple>>,
  "exit": true,
}[Tuple extends [] ? "exit" : Tuple[0] extends Union ? "next" : "error"];

type R = Check<FurnitureUnion, FurnitureTuple>; // true
type R1 = Check<'chair' | 'lamp' | 'table', FurnitureTuple>; // true
type R2 = Check<'chair' | 'lamp' | 'table', ['chair', 'table' , 'lamp', 'error']>; // nerver

Remove from tuple获取tuple并返回没有第一个元素的tuple(稍后需要)

Check将在元组上迭代。每一步都可以返回 never 当元组[0]不扩展并集时,当输入元组为空时退出,当元组[0]扩展并集时退出。在下一步中,我们将递归调用检查,但首先我们将通过前面的util从元组中删除第一个元素

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