函数返回类型中缺少类型

Question

我对函数 createRouting 有问题，它实际上应该返回与它得到的一样。
但目前它返回它得到的最简单的表示。
这是代码：

type ParamValue = number | string;
type Params<Keys extends string = string> = {
  [K in Keys]: (rawParams: Record<string, ParamValue>) => ParamValue;
};
type NoInfer<T> = [T][T extends any ? 0 : never];

class RouteConfig<RouteKey, P extends Params> {
  constructor(
    public config: {
      params?: P;
      name?: RouteKey;
      validate: (
        loc: string,
        params: NoInfer<P> extends Params<infer PKeys>
          ? { [K in PKeys]: ReturnType<P[K]> }
          : never,
      ) => boolean;
      path: (
        params: NoInfer<P> extends Params<infer PKeys>
          ? { [K in PKeys]: ReturnType<P[K]> }
          : never,
      ) => string;
    },
  ) {}
}

const enum RouteName {
  Main,
}

export const createRouting = <RouteKeys extends string | number, T = any>(
  routes: {
    [K in RouteKeys]: T;
  },
): {
  [K in RouteKeys]: T extends RouteConfig<infer RK, infer RP>
    ? RP extends Params<infer PK>
      ? RouteConfig<RK, Params<PK>>
      : never
    : never;
} => {
  return routes as any;
};

const routing = createRouting<RouteName>({
  [RouteName.Main]: new RouteConfig({
    name: RouteName.Main as const,
    params: {
      foo: (params) => (params.foo && +params.foo) || 0,
      bar: () => 66,
    } as const,
    validate: (location, params) => {
      const barExp = params.bar.toExponential(15);
      const fooExp = params.foo.toExponential(15);

      if (barExp + fooExp) return true;

      return true;
    },
    path: ({ foo }) => `/${foo}`,
  }),
});

routing[RouteName.Main].config.params.foo;

在最后一行代码中，params 不应是未定义的，并且有两个键 foo 和 bar