给定一个像这样的状态缩减程序:
const anObject = {
fruit: 'Apple',
today: new Date(),
}
function reducer(state, stateReducer) {
return stateReducer(state);
}
const fruit = reducer(anObject, state => state.fruit);
// fruit should be of type string.
const now = reducer(anObject, state => state.
我正在通过重新实现我的流型原型来学习打字稿。这个问题让我有点困惑。
error TS2322: Type '(state: State, action: NumberAppendAction) => State' is not assignable to type 'Reducer'.
Types of parameters 'action' and 'action' are incompatible.
Type 'Action' is not assignable to type 'Nu
我刚接触typescript,正在尝试将react应用程序转换为使用typescript。当在我的index.tsx文件中使用'redux‘中的createStore函数时,我得到以下错误。 Type '{}' is missing the following properties from type '{ customersLoading: boolean; customersLoadError: boolean; customers: {}; editCustomerInProgress: boolean; editCustomerSuccess: b
我正在测试一个简单的Swift Redux实现。有人能解释一下为什么调用store.dispatch(.test)会导致:
Could not cast value of type '(Test.AppAction) -> ()' to '(Test.Action) -> ()'.
尽管AppAction实现了操作协议,但为什么不能将AppAction转换为操作?
中间件接受(S, Action, (Action) -> Void),我将它作为第三个参数传递给dispatch(_ action: A)。这是一种((Action) -> V
我正在使用过滤列表。
我已经将代码简化到最基本的部分,以隔离正在发生的事情,但我不能理解哪里出了问题。
代码如下:
defmodule Servy.Reducer do
alias Servy.Wildthings
def index(%{} = bear_filter) do
Wildthings.list_bears()
|> Enum.filter(
fn(bear) ->
IO.inspect(bear_filter)
case bear do
bea
为什么新的JDK8 类只包含以下reduce方法:
T reduce(BinaryOperator<T> reducer)
T reduce(T identity, BinaryOperator<T> reducer)
U reduce(U identity, BiFunction<U, ? super T, U> reducer,
BinaryOperator<U> combiner)
但不是一个明显的方法,它对应于其他语言(例如Haskell foldl :: (a -> b -> a) -> a -> [b
我正在尝试使用类型来配置一个预加载状态的Redux存储。
的示例如下:
import { configureStore } from '@reduxjs/toolkit'
const store = configureStore({
reducer: {
one: oneSlice.reducer,
two: twoSlice.reducer
}
})
// Infer the `RootState` and `AppDispatch` types from the store itself
export type RootState = Retur
我想要创建一个对还原器做一些事情的函数(它不是重要的内容,只是一些类似于combineReducers的转换)。当我想为减速器抽象state时,我遇到了问题。以下是用于某些组件的简化减速器:
type someState = {
foo: number,
};
export const someSimpleReducer = (state: someState) => state;
type StateAbstract = {}; // this doesn't work
// type StateAbstract = any; // this works
在这里,具有抽象
信息:在Jetbrains骑手中使用类型记录3.3.3333
给出了Reducer<State, Action>的这种类型定义
* @template S The type of state consumed and produced by this reducer.
* @template A The type of actions the reducer can potentially respond to.
*/
export type Reducer<S = any, A extends Action = AnyAction> = (
state: S |
我在ReactNative中使用Redux,我想创建一个带有减速机的商店
而且,下面出现了错误,指向switchToTab()函数reducer.js中的行'switch (reducer.js)‘
undefined is not an object(evaluating 'action.type')
这是我的actions.js
export const SWITCH_TAB = 'switchTab'
export function switchTab(index) {
return {
type: SWITCH_TAB,
in
在更新"@reduxjs/toolkit": "^1.6.1"之后,我得到了一个getDefaultMiddleware已弃用的警告,那么我应该如何删除这个警告呢?我们有没有其他方法可以在configureStore函数中注入默认的中间件? import { configureStore, getDefaultMiddleware } from "@reduxjs/toolkit";
import reducer from "./reducer";
import api from "./middleware/api
我正在创建一个Redux存储增强器,它接受一个函数来序列化Redux状态。我将构造存储并为更改设置订阅--对于每个更改,我将序列化状态。对于这个MCVE,我忽略了订阅方面,只是立即调用序列化函数。
但是,由于函数的高阶性质,我无法将状态的泛型类型(序列化函数所需的)与存储创建者返回的泛型类型关联起来:
// Copied and reduced from Redux 4.0.1
type Reducer<S = any> = (
state: S | undefined,
) => S
type DeepPartial<T> = { [K in keyof
我在学习Redux的时候偶然发现了createStore函数。因此,据我所知,createStore接收3个参数:
reducer
initial state
enhancers (for simplicity we will use only middlewares)
但是当我们实际使用createStore时,我们不会传递初始状态作为第二个参数,而是传递带有默认状态的reducer,如下所示:
const initialState = {counter:0}
const reducer =(state=initialState, action)=>...
问题是,为什么我们不把初始状
我的减速机出了点问题。我可能遗漏了一些明显的东西,但我遇到了combineReducers函数的问题。
export default combineReducers<ConfigCategoryState>({
tree: treeReducer(),
});
这会产生以下错误:
No overload matches this call.
No overload matches this call.
Overload 1 of 3, '(reducers: ReducersMapObject<ConfigCategoryState, any>):