在 TypeScript 中推断函数参数

我正在尝试创建一个类型安全的映射函数（不是下面的函数），但我坚持让函数参数正确推断。

    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？目前里面的功能mapped对象接受任何参数，但它应该只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。

Can use Parameters and ReturnType泛型类型获取函数的具体参数和返回类型：

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]>
}