javascript - typescript 泛型包装器 : Untyped function calls may not accept type arguments

Question

这段带有泛型的代码片段工作得很好( Link to Simple and Working Code )

const state: Record<string, any> = {
    isPending: false,
    results: ['a', 'b', 'c']
}

const useValue = <T extends {}>(name: string): [T, Function] => {
    const value: T = state[name]
    const setValue: Function = (value: T): void => {
        state[name] = value
    }
    return [value, setValue]
}

const [isPending, setIsPending] = useValue<boolean>('isPending')
const [results, setResults] = useValue<string[]>('results')

在这里我可以确定 isPending 是一个 bool 值并且 setIsPending 接收一个 bool 值作为参数。同样适用于 results 和 setResults 作为字符串数组。

然后我用另一种方法包装代码 useStore ( Link to Extended Broken Code )

interface UseStore {
    state: Record<string, any>
    useValue: Function
}

const state: Record<string, any> = {
    isPending: false,
    results: ['a', 'b', 'c']
} 

const useStore = (): UseStore => {
    const useValue = <T extends {}>(name: string): [T, Function] => {
        const value: T = state[name]
        const setValue: Function = (value: T): void => {
            state[name] = value
        }
        return [value, setValue]
    }

    return { state, useValue }
}

const { useValue } = useStore()
const [isPending, setIsPending] = useValue<boolean>('isPending')
const [results, setResults] = useValue<string[]>('results')

最后两行给我 typescript 错误:Untyped function calls may not accept type arguments.

我怀疑 useStore 接口(interface)有问题，但由于它的动态特性，我想不出更好的解决方案。

如何在使用泛型类型时消除错误以获得正确的类型提示和代码完成？

Answer 1

由于useValue 的类型是Function，所以传递泛型参数是没有意义的。他们对谁有利？运行时没有得到它们，它们在编译时被删除。编译器无法使用它们，因为 Function 只是一个未类型化的函数，因此没有任何好处。传递类型参数是无用的，并且可以说是一个错误(即用户不期望这是 Function 并且认为它们有一些效果而传递了类型参数)。

删除类型参数并放弃以任何方式键入的假装:

const { useValue } = useStore()
const [isPending, setIsPending] = useValue('isPending')
const [results, setResults] = useValue('results')

更有趣的问题是你为什么要写这样的代码，因为有一种方法可以完全键入这段代码中的所有内容:

const state = {
    isPending: false,
    results: ['a', 'b', 'c']
}
type State = typeof state;

const useStore = () => {
    const useValue = <K extends keyof State>(name: K) => {
        const value = state[name]
        const setValue = (value: State[K]): void => {
            state[name] = value
        }
        return [value, setValue] as const
    }

    return { state, useValue }
}
type UseStore = ReturnType<typeof useStore>;

const { useValue } = useStore()
const [isPending, setIsPending] = useValue('isPending')
const [results, setResults] = useValue('results')

上面的版本是完全类型安全的，不需要任何重复的名称或类型(您当然可以将其拆分为多个文件，但根据您的要求，可能需要进行一些重复)。如果这不适用于您的情况，我很想知道原因。

编辑

如果您只想让类型在最后几行运行并在那里有一些类型安全性，您只需要使用泛型指定函数的签名:

interface UseStore {
    state: Record<string, any>
    useValue: <T,>(name: string) => [T, (value: T)=> void]
}

const state: Record<string, any> = {
    isPending: false,
    results: ['a', 'b', 'c']
}

const useStore = (): UseStore => {
    const useValue = <T,>(name: string): [T, (value: T)=> void] => {
        const value: T = state[name]
        const setValue = (value: T): void => {
            state[name] = value
        }
        return [value, setValue]
    }

    return { state, useValue }
}

const { useValue } = useStore()
const [isPending, setIsPending] = useValue<boolean>('isPending')
const [results, setResults] = useValue<string[]>('results')

编辑 - 接口(interface)实现的开放式

您可以将State 定义为接口(interface)，因为接口(interface)是开放式的，您可以在需要时添加成员。好处是如果其他人定义了一个名称相同但类型不同的属性，你会得到一个错误

界面状态{

}
// Don't know what is in here, empty object for starters 
const state : State = {
} as State


const useStore = () => {
    const useValue = <K extends keyof State>(name: K) => {
        const value = state[name]
        const setValue = (value: State[K]): void => {
            state[name] = value
        }
        return [value, setValue] as const
    }

    return { state, useValue }
}
type UseStore = ReturnType<typeof useStore>;

const { useValue } = useStore()

interface State { isPending: boolean }
state.isPending = false; // no guarantee people actually intialize, maybe initial value can be passed to useValue ? 
const [isPending, setIsPending] = useValue('isPending')

interface State { results: string[] }
state.results = ['a', 'b', 'c'];
const [results, setResults] = useValue('results')


interface State { results: number[] } // some else wants to use results for something else, err 
const [results2, setResults2] = useValue('results')