javascript - Typescript 中带有函数的映射类型

Question

我在使用映射类型时遇到了一些问题。

我正在尝试输入一个组成 redux“选择器”的函数(本质上只是来自 A => B 的函数)

我的目标 API 基本上是:

const bar = composeSelectors([
  (s: string) => s.length,
  (s: string) => s.trim(),
])

bar // should be (s: string) => [number, string]

这是我到目前为止所得到的:

type Selector<A, Z> = (a:A) => Z
type Selected<T> = T extends Selector<any, infer U> ? U : never;
type MappedSelector<S, T> = Selector<S, { [Y in keyof T]: Selected<T[Y]> }>

const selectedNumber: Selected<Selector<string, number>> = null as any
selectedNumber // is number

const baz: MappedSelector<string, [Selector<string, number>, Selector<string, string>]> = null as any
baz  // is Selector<string, [number, string]> !!!

这似乎工作得很好，直到我尝试在函数中使用MappedSelector:

type InferableMappedSelector<S> = <T extends any[]>(...values: T) => MappedSelector<S, T>
function createInferredSelector<S>(): InferableMappedSelector<S> {...}
const inferredSelectorCreator = createInferredSelector<string>()


const fooSelector: Selector<string, number> = s => s.length;
const barSelector: Selector<string, string> = s => s.trim();
const selectors: [Selector<string, string>, Selector<string, number>] = [barSelector, fooSelector]

const baz = inferredSelectorCreator(selectors)
baz // is Selector<string, [never]>, not Selector<string, [number, string]>

我也尝试过:

type InferableMappedSelector<S> = <T extends Selector<string, any>[]>(...values: T) => MappedSelector<S, T>

const baz = inferredSelectorCreator(selectors)
// doesn't typecheck due to: 
// Argument of type '[Selector<string, string>, Selector<string, number>]' is not assignable to // parameter of type 'Selector<string, any>'.
//   Type '[Selector<string, string>, Selector<string, number>]' provides no match for the signature // '(a: string): any'.

(最后一点似乎是 TS 中的错误)

Answer 1

你的问题其实很简单。 inferredSelectorCreator接受 T 类型的剩余参数，但是当您打电话时，您会调用 inferredSelectorCreator与整个数组，而不传播它( inferredSelectorCreator(selectors) )这意味着 T将被推断为 [[Selector<string, string>, Selector<string, number>]]而不是[Selector<string, string>, Selector<string, number>] .

如果您使用传播，您会得到预期的结果:

const bazz = inferredSelectorCreator(...selectors)