打字稿泛型：从第二个参数推断类型以在第一个参数中使用

Question

我有这样的代码：

export type Subscribe<T extends object> = <U>(
  listener: (slice: U) => void,
  selector: (state: T) => U,
) => void

// implementation doesn't matter
const subscribe = {} as Subscribe<{ value: number }>

subscribe(
  (value) => value * 2, // ts error: object is of type unknown
  (state) => state.value
)

似乎不可能用 typescript 以这种方式推断类型

但如果我改变参数的顺序，它就可以正常工作：

export type Subscribe<T extends object> = <U>(
  selector: (state: T) => U,
  listener: (slice: U) => void,
) => void

const subscribe = {} as Subscribe<{ value: number }>

subscribe(
  (state) => state.value,
  (value) => value * 2 // now no error! number type is inferred
)

我希望selector 成为第二个参数以使其成为可选，请告诉我这是不可能的或者你是否知道方法

Answer 1

考虑这个例子：

const subscribe = <
    T extends Record<PropertyKey, unknown>,
    ReturnValue,
    >(
        listener: (slice: ReturnValue) => void,
        selector: (state: T) => ReturnValue,
) => {}

subscribe(
    (value) => value * 2,
    (state: { value: number }) => state.value
)

Playgroud

我不确定您是否需要单独的类型。您可以从函数级别推断所有泛型