推断TypeScript中的函数参数

Question

我正在尝试制作一个类型安全的映射函数（而不是下面的函数），但是我一直坚持获取要正确推断的函数参数。

    export type Mapper<U extends Unmapped> = {
      mapped: Mapped<U>
    };

    export type Unmapped = {
      [name: string]: (...args: any[]) => any
    };

    export type Mapped<U extends Unmapped> = {
      [N in keyof U]: (...args: any[]) => Promise<any>
    };

    const map = <U extends Unmapped>(unmapped: U): Mapper<U> => ({
      mapped: Object.entries(unmapped).reduce(
        (previous, [key, value]) => ({
          ...previous,
          [key]: (...args: any[]) => new Promise((resolve) => resolve(value(...args)))
        }),
        {}
      ) as Mapped<U>
    });

    const mapped = map({ test: (test: number) => test });

    mapped.mapped.test('oh no');

是否可以让TypeScript推断它们？当前，mapped对象内部的函数可以接受任何参数，但是只能接受未映射对象中定义的参数。函数名称确实可以正确推断。

Answer 1

如果您(...args: any[]) => Promise<any>在映射类型中用作签名，您将丢失所有参数类型信息并返回类型信息。使用条件类型可以实现对您想要做的事情的不完美解决方案。此处描述了这些限制。

该解决方案需要创建一个条件类型，分别处理具有给定数量参数的每个函数。下面的解决方案最多适用于 10 个参数（对于大多数实际情况来说已经足够了）

export type Mapper<U extends Unmapped> = {
    mapped: Mapped<U>
};

export type Unmapped = {
    [name: string]: (...args: any[]) => any
};

type IsValidArg<T> = T extends object ? keyof T extends never ? false : true : true;

type Promisified<T extends Function> =
    T extends (...args: any[]) => Promise<any> ? T : (
        T extends (a: infer A, b: infer B, c: infer C, d: infer D, e: infer E, f: infer F, g: infer G, h: infer H, i: infer I, j: infer J) => infer R ? (
            IsValidArg<J> extends true ? (a: A, b: B, c: C, d: D, e: E, f: F, g: G, h: H, i: I, j: J) => Promise<R> :
            IsValidArg<I> extends true ? (a: A, b: B, c: C, d: D, e: E, f: F, g: G, h: H, i: I) => Promise<R> :
            IsValidArg<H> extends true ? (a: A, b: B, c: C, d: D, e: E, f: F, g: G, h: H) => Promise<R> :
            IsValidArg<G> extends true ? (a: A, b: B, c: C, d: D, e: E, f: F, g: G) => Promise<R> :
            IsValidArg<F> extends true ? (a: A, b: B, c: C, d: D, e: E, f: F) => Promise<R> :
            IsValidArg<E> extends true ? (a: A, b: B, c: C, d: D, e: E) => Promise<R> :
            IsValidArg<D> extends true ? (a: A, b: B, c: C, d: D) => Promise<R> :
            IsValidArg<C> extends true ? (a: A, b: B, c: C) => Promise<R> :
            IsValidArg<B> extends true ? (a: A, b: B) => Promise<R> :
            IsValidArg<A> extends true ? (a: A) => Promise<R> :
            () => Promise<R>
        ) : never
    );

export type Mapped<U extends Unmapped> = {
    [N in keyof U]: Promisified<U[N]>
}

const map = <U extends Unmapped>(unmapped: U): Mapper<U> => ({
    mapped: Object.entries(unmapped).reduce(
        (previous, [key, value]) => ({
            ...previous,
            [key]: (...args: any[]) => new Promise((resolve) => resolve(value(...args)))
        }),
        {}
    ) as Mapped<U>
});

const mapped = map({ test: (test: number) => test });

mapped.mapped.test('oh no'); 

Answer 2

可以使用Parameters和ReturnType泛型类型来获取函数的特定参数和返回类型：

type Promisified<T extends (...args: any[]) => any> = (...args: Parameters<T>) => Promise<ReturnType<T>>;

export type Mapped<U extends Unmapped> = {
    [N in keyof U]: Promisified<U[N]>
}