我有一个选项卡导航器，它有Home和Profile作为屏幕。Home是一个堆栈导航器，它包含RestaurantDetails和RestaurantList作为屏幕。
我有一个Button在RestaurantList屏幕中。我想导航到RestaurantDetails。
Button.tsx

const Button = ({name}) => {
  const navigation = useNavigation()
  const onPress = () => {
    navigation.navigate("RestaurantDetails", {name})
  }
  return (
    <TouchableOpacity onPress={onPress}>
    </TouchableOpacity>
   
  )
}

问题是 typescript 抱怨

Argument of type '["RestaurantDetails", { name: string}]' is not assignable to parameter of type '[screen: "Home", params: NavigatorScreenParams<HomeStackParamList, Readonly<{ key: string; index: number; routeNames: string[]; history?: unknown[] | undefined; routes: NavigationRoute<ParamListBase, string>[]; type: string; stale: false; }>>] | [screen: ...] | [screen: ...] | [screen: ...] | [screen: ...]'.
  Type '["RestaurantDetails", { name: string}]' is not assignable to type '[screen: "Profile", params: undefined]'.
    Type at position 0 in source is not compatible with type at position 0 in target.
      Type '"RestaurantDetails"' is not assignable to type '"Profile"'.

我很确定问题是在RestaurantDetails屏幕上，typescript在TRootParamList而不是HomeStackParamList中查找，因为如果我将RestaurantDetails添加到TRootParamList中，typescript就不会再抱怨了。但是，我不确定如何真正阻止这个问题。
types.ts

import { NavigatorScreenParams } from '@react-navigation/native';

type QueueDetailsParams = {
  name: string
};

type HomeStackParamList = {
  RestaurantDetails: QueueDetailsParams | undefined;
  RestaurantList: undefined
};

type TRootParamList = {
  Home: NavigatorScreenParams<HomeStackParamList>;
  Profile: undefined
};

declare global {
  namespace ReactNavigation {
    interface RootParamList extends TRootParamList {}
  }
}

我知道我可以用下面的替换我的onPress，但我宁愿不这样做。

const onPress = () => {
    navigation.navigate("Home", {screen: "RestaurantDetails", params: {name}})
  }