React-Redux useSelector typescript type for state

This might not be the answer but I use it like so:

const isLoggedIn = useSelector<IRootState, boolean>(state => state.user.loggedIn);

EDIT: Or use Peter's answer which is shorter/cleaner

const isLoggedIn = useSelector((state: IRootState) => state.user.loggedIn);

You can create your custom typed useSelector like so:

import {
  useSelector as useReduxSelector,
  TypedUseSelectorHook,
} from 'react-redux'
import { RootState } from 'app/redux/store'

export const useSelector: TypedUseSelectorHook<RootState> = useReduxSelector

where RootState is the type of the store, usually defined as:

export type RootState = ReturnType<typeof rootReducer>

This is the method described in the definitely typed declaration.

Don't forget to install @types/react-redux.

Here is the suggestion (more or less) from the redux docs:

import { RootState } from 'app/redux/store';
const isLoggedIn = useSelector(state: RootState => state.user.loggedIn);

The advantage over @Federkun's answer is that it is much simpler. The advantage over @alextrastero's answer is that I don't have to specify isLoggedIn's type manually.

1. Create config.d.ts

2. Define your custom state

   import 'react-redux';
   import { ApplicationState } from '@store/index';
   declare module 'react-redux' {
     interface DefaultRootState extends ApplicationState {}
   }