带有函数的 Typescript 中的映射类型

Question

我在使用映射类型时遇到了一些问题。

我正在尝试键入一个组成 redux "selectors" 的函数（本质上只是 A => B 中的函数）

我的API目标基本上是：

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

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

这是我目前得到的：

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]> !!!

这似乎运作良好，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 接受类型为 T 的剩余参数，但是当你调用时，你用整个数组调用 inferredSelectorCreator，而不展开它 (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>].

如果您使用价差，您会得到预期的结果：

const bazz = inferredSelectorCreator(...selectors)

带有函数的 Typescript 中的映射类型

Mapped Types in Typescript with functions

javascript

typescript

redux

reselect