[英]Generic type wrapping in TypeScript for tuples
我需要向函數添加一個類型聲明,該函數將元組[Foo<A>, Foo<B>, ...]
元素映射到函數() => [A, B, ...]
。 我怎樣才能在TypeScript中實現這一目標?
此示例在結構上與應用程序的相關部分類似:
interface SomethingWithAValue<T> { value: T; }
function collect(array) {
return () => array.map(a => a.value);
}
這將返回與每個對象關聯的值的元組。 collect
的類型聲明會是什么樣的?
偽代碼:
function collect<[SomethingWithAValue<T>...](array: [SomethingWithAValue<T>...]): () => [T...];
更新以回應jonrsharpe的建議:
interface SomethingWithAValue<T> { value: T; }
function collect<T>(array: SomethingWithAValue<T>[]): () => T[] {
return () => array.map(a => a.value);
}
type myTupleType = [string, number];
let somethings: [SomethingWithAValue<string>, SomethingWithAValue<number>];
somethings = [{ value: 'foo' }, { value: 5 }];
let fn: () => myTupleType = collect(somethings);
這不起作用:
Argument of type '[SomethingWithAValue<string>, SomethingWithAValue<number>]' is not assignable to parameter of type 'SomethingWithAValue<string>[]'.
Types of property 'pop' are incompatible.
Type '() => SomethingWithAValue<string> | SomethingWithAValue<number>' is not assignable to type '() => SomethingWithAValue<string>'.
Type 'SomethingWithAValue<string> | SomethingWithAValue<number>' is not assignable to type 'SomethingWithAValue<string>'.
Type 'SomethingWithAValue<number>' is not assignable to type 'SomethingWithAValue<string>'.
Type 'number' is not assignable to type 'string'.
更新:從TS3.1開始,下面的答案已經過時。 我相信你現在可以使用映射的元組類型和條件類型推斷來獲得你想要的行為:
type ExtractValue<T extends ReadonlyArray<SomethingWithAValue<any>>> =
{ [K in keyof T]: T[K] extends SomethingWithAValue<infer V> ? V : never };
function collect<T extends ReadonlyArray<SomethingWithAValue<any>>>(
array: T
): () => ExtractValue<T> {
return () => array.map(a => a.value) as any;
}
讓我們使用它......首先讓我們很容易得到一個帶有輔助函數tuple()
的元組類型,它接受一個可變數量的參數並輸出一個元組(這是從TS3.0開始實現的)
type Narrowable = string | number | boolean | undefined | null | void | {};
const tuple = <T extends Narrowable[]>(...t: T) => t;
讓我們看看它是否有效:
const result = collect(tuple({ value: 10 }, { value: "hey" }, { value: true }));
// const result: () => [number, string, boolean]
看起來不錯!
老答案:
TypeScript中沒有可變參數類型 ,因此無法輸入通用元組(沒有類似語法[T...]
存在)。
作為一種變通方法,您可以為任何長度的元組提供函數重載,直到達到一些合理的最大值:
function collect<A, B, C, D, E>(array: [SomethingWithAValue<A>, SomethingWithAValue<B>, SomethingWithAValue<C>, SomethingWithAValue<D>, SomethingWithAValue<E>]): () => [A, B, C, D, E];
function collect<A, B, C, D>(array: [SomethingWithAValue<A>, SomethingWithAValue<B>, SomethingWithAValue<C>, SomethingWithAValue<D>]): () => [A, B, C, D];
function collect<A, B, C>(array: [SomethingWithAValue<A>, SomethingWithAValue<B>, SomethingWithAValue<C>]): () => [A, B, C];
function collect<A, B>(array: [SomethingWithAValue<A>, SomethingWithAValue<B>]): () => [A, B];
function collect<A>(array: [SomethingWithAValue<A>]): () => [A];
function collect<T>(array: SomethingWithAValue<T>[]): () => T[] {
// implementation
}
這應該適用於長度為5的元組,並且您可以在頂部添加其他重載以實現您在實踐中所需的任何內容。 這是冗長而丑陋的,但應該有效。
希望有所幫助; 祝好運!
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