老师好，最终结果我最后可以实现，但是对FoodModule里面两个方法各自的属性的抓取，好像有点问题，还麻烦老师帮忙看一下

// 面试题：用keyof和infer处理复杂泛型问题
interface Action<T = any> {
  type: string;
  payload?: T;
}

class FoodModule {
  public static topic: string;
  public count!: number;
  delay(promise: Promise<number>) {
    return promise.then((second: number) => ({
      type: "delay",
      payload: `延迟${second}秒`
    }))
  }

  searchFoodByCity(action: Action<string>) {
    return {
      payload: action.payload,
      type: "searchFoodByCity"
    }
  }
}

// 要求1：实现方法名称
type AsyncMethodConnect<T, U> = (input: Promise<T>) => Promise<Action<U>>;
type SyncMethodConnect<T, U> = (action: Action<T>) => Action<U>;

// 要求2：把两个类型和原先的类合并成一个对象
// 所有方法名
type FoodModuleFuncKeys = Exclude<keyof FoodModule, "count">;
// 对比指定类型获取
type AsyncAndSyncFunc<T> =
  T extends AsyncMethodConnect<infer P, infer U> ? AsyncMethodConnect<P, U> :
  T extends SyncMethodConnect<infer P, infer U> ? SyncMethodConnect<P, U> : any;
// 综合输出结果
type Convert = {
  [P in FoodModuleFuncKeys]: AsyncAndSyncFunc<FoodModule[P]>
}

// TODO: 问题在这儿：分别抓两个方法的类型
type AsyncFunc<T> = T extends AsyncMethodConnect<infer P, infer U> ? AsyncMethodConnect<P, U> : any;
type SyncFunc<T> = T extends SyncMethodConnect<infer P, infer U> ? SyncMethodConnect<P, U> : any;

type FoodModuleDelay = AsyncFunc<FoodModule["delay"]>;
type FoodModuleSearch = SyncFunc<FoodModule["searchFoodByCity"]>;