在这种情况下，可以临时强制执行该操作，以保存麻烦：

interface Store {
  state: MyStateType,
  dispatch: Dispatch,
}

const StoreContext = createContext<Store>({} as Store);

const StoreProvider = ({children}) => {
  const {state, dispatch} = theseAreAlwaysPresent()

  return (
    <StoreContext.Provider value={{state, dispatch}}>
      {children}
    </StoreContext.Provider>
  )
}
...

因为我刚刚做了一个账户来回答这个问题，所以我不能在上面发表评论。Alex韦恩是正确的，这对Typescript来说是可以接受的。但是，useContext钩子不会真正起作用，因为就Typescript而言，你可能在那个上下文中有一个null值。
根据最初的回答......这里是我们的上下文，采用 typescript const StoreContext = createContext<{state: MyStateType, dispatch: Dispatch} | null>(null);
因此，我们需要创建一个新的钩子，我们将其命名为useContextAndErrorIfNull，如下所示：

const useContextAndErrorIfNull = <ItemType>(context: Context<ItemType | null>): ItemType => {
  const contextValue = useContext(context);
  if (contextValue === null) {
    throw Error("Context has not been Provided!");
  }
  return contextValue;
}

用这个钩子代替useContext，它应该都能很好地工作。

当你用null初始化一个上下文时，不能推断出想要的类型。在这种情况下，你必须显式地给予上下文的类型。
在这里，它看起来像这样：

const StoreContext = createContext<{
  state: MyStateType,
  dispatch: Dispact,
} | null>(null);

//So basically in my case you just have to assign an empty object that 
   will solve your problem.

 import React, { createContext, useReducer, FC, Dispatch } from 'react';
 import storeReducer, { initialState } from './reducer';
 import { Action, State } from './types';
 import { isNull } from 'util';

 // replaced null with {}
 // because assigning a value and any type of parameter will not be 
    acceptable in typescript
 const StoreContext = createContext({});

const StoreProvider:FC = ({ children }) => {
const [state, dispatch] = useReducer(storeReducer, initialState);
const value = {state, dispatch}
return (
<StoreContext.Provider value={value}>
  {children}
</StoreContext.Provider>
 )};

export { StoreProvider, StoreContext };

export interface IStoreContext {
dispatch: (action: Action<any>) => {};
state: State;
}