I’m having some issues with using TypeScript and creating a custom type to pass to navigation.navigate(). I have a prop type called ScreenLink that looks like this:

interface ScreenLink {
  name: string,
  dest: keyof RootStackParamList
}

Where RootStackParamList is imported from another file that looks like this:

export type RootStackParamList = {
  Root: NavigatorScreenParams<RootTabParamList> | undefined;
  Home: undefined;
  Landing: undefined;
  LogIn: undefined;
  // ...etc.
}

export type RootTabParamList = {
  Home: undefined;
  Settings: undefined;
  // ...etc.
};

export type RootStackScreenProps<Screen extends keyof RootStackParamList> = NativeStackScreenProps<
  RootStackParamList,
  Screen
>;

I then use the custom type in a component like this:

  const LinkCard = (props: { index: number, screenLink: ScreenLink  }) => {

    const handleOnPress = () => {
      navigation.navigate(props.screenLink.dest)
    }

    return (<Pressable onPress={handleOnPress}> </Pressable>)
  }

This seems like it should work. When I mouseover the type of ‘dest’ it shows all of the potential keys I want to pass as a route. However, I get an error that “No overload matches this call”:

No overload matches this call.
  Argument of type '[keyof RootStackParamList]' is not assignable to parameter of type '[screen: "Root"] | [screen: "Root", params: NavigatorScreenParams<RootTabParamList> | undefined] | [screen: "Home"] | [screen: "Home", params: undefined] | [screen: "Landing"] | [screen: ...] | ... 53 more ... | [screen: ...]'.

Interestingly, the error goes away when I change keyof RootStackParamList to keyof RootStackParamList['Root'] and the code functions exactly like it should. But I have no idea why this works or what was wrong in the first place.

So far, I’ve been referencing the docs:
https://reactnavigation.org/docs/5.x/typescript/

And this answer:
Type of route name, route params in React Navigation Typescript

But I still can’t figure out what is wrong with the original code. Happy to provide more details if necessary. Thanks.