最近在看Redux的源码,发现Redux在使用中间件applyMiddleware.js的源码中,有一个对闭包非常巧妙的使用,解决了“鸡生蛋,蛋生鸡”的问题,特分享给大家。
Redux中间件的函数签名形式如下:
({dispatch, getState}) => next => action => {
// 函数体
}
applyMiddleware.js中的函数applyMiddleware(...middlewares)用于根据中间件生成action经过的中间件链。先来看一个错误版本的实现:
/*
* @param {...Function} middlewares The middleware chain to be applied.
* @returns {Function} A store enhancer applying the middleware.
*/
export default function applyMiddleware(...middlewares) {
return (createStore) => (reducer, initialState, enhancer) => {
var store = createStore(reducer, initialState, enhancer)
var chain = []
var middlewareAPI = {
getState: store.getState,
dispatch: store.dispatch
}
chain = middlewares.map(middleware => middleware(middlewareAPI))
var dispatch = compose(...chain)(store.dispatch) //compose(f, g, h) 等价于函数
//(...args)=>f(g(h(args)))
return {
...store,
dispatch
}
}
核心逻辑是chain = middlewares.map(middleware => middleware(middlewareAPI))和dispatch = compose(...chain)(store.dispatch)这两行。第1句代码是根据中间件生成一个数组chain,chain的元素是签名为next => action => {...}形式的函数,每个元素就是最终中间件链上的一环。第2句代码利用compose函数,将chain中的函数元素组成一个“洋葱式”的大函数,chain的每个函数元素相当于一层洋葱表皮。Redux发送的每一个action都会由外到内依次经过每一层函数的处理。假设有3层函数,从外到内依次是a,b,c,函数的实际调用过程是,a接收到action,在a函数体内会调用b(a的参数next,指向的就是b),并把action传递给b,然后b调用c(b的参数next指向的就是c),同时也把action传递给c,c的参数next指向的是原始的store.dispatch,因此是action dispatch的最后一环。这样分析下来,程序是没有问题的,但当我们的中间件需要直接使用dispatch函数时,问题就出来了。例如,常用于发送异步action的中间件redux-thunk,就需要在异步action中使用dispatch:
export function fetchPosts(subreddit) {
return function (dispatch) {
dispatch(requestPosts(subreddit))
return fetch(`https://www.reddit.com/r/${subreddit}.json`)
.then(
response => response.json(),
error => console.log('An error occured.', error)
)
.then(json =>
dispatch(receivePosts(subreddit, json))
)
}
}
fetchPosts使用的dispatch,是redux-thunk传递过来的,指向的是middlewareAPI对象中的dispatch,实际等于store.dispatch。当执行dispatch(requestPosts(subreddit))时,这个action直接就到了最后一环节的处理,跳过了redux-thunk中间件之后的其他中间件的处理,显然是不合适的。我们希望的方式是,这个action依然会从最外层的中间件开始,由外到内经过每一层中间件的处理。所以,这里使用的dispatch函数不能等于store.dispatch,应该等于compose(...chain)(store.dispatch),只有这样,发送的action才能经过每一层中间件的处理。现在问题出来了,chain = middlewares.map(middleware => middleware(middlewareAPI))需要使用dispatch = compose(...chain)(store.dispatch)返回的dispatch函数,而dispatch = compose(...chain)(store.dispatch)的执行又依赖于chain = middlewares.map(middleware => middleware(middlewareAPI))的执行结果,我们进入死循环了。
问题的解决方案就是闭包。当我们定义middlewareAPI的dispatch时,不直接把它指向store.dispatch,而是定义一个新的函数,在函数中引用外部的一个局部变量dispatch,这样就形成了一个闭包,外部dispatch变量的变化会同步反映到内部函数中。如下所示:
export default function applyMiddleware(...middlewares) {
return (createStore) => (reducer, initialState, enhancer) => {
var store = createStore(reducer, initialState, enhancer)
var dispatch = store.dispatch; // 需要有初始值,保证中间件在初始化过程中也可以正常使用dispatch
var chain = []
var middlewareAPI = {
getState: store.getState,
dispatch: (...args) => dispatch(...args) // 通过闭包引用外部的dispatch变量
}
chain = middlewares.map(middleware => middleware(middlewareAPI))
dispatch = compose(...chain)(store.dispatch) //compose(f, g, h) 等价于函数
//(...args)=>f(g(h(args)))
return {
...store,
dispatch
}
}
这样,“鸡生蛋,蛋生鸡”的问题就解决了。如果这个例子对你来说太复杂,可以用下面这个简化的例子帮助你理解:
const middleware = ({dispatch}) => (next) => (number) => {
console.log("in middleware");
if(number !== 0){
return dispatch(--number);
}
return next(number);
}
function test() {
var dispatch = (number) => {
console.log("original dispatch");
return number;
};
var middlewareAPI = {
dispatch
}
dispatch = middleware(middlewareAPI)(dispatch);
return {
dispatch
}
}
var {dispatch} = test();
dispatch(3);
//输出:
"in middleware"
"original dispatch"
const middleware = ({dispatch}) => (next) => (number) => {
console.log("in middleware");
if(number !== 0){
return dispatch(--number);
}
return next(number);
}
function test() {
var dispatch = (number) => {
console.log("original dispatch");
return number;
};
var middlewareAPI = {
dispatch: (number) => {dispatch(number);}
}
dispatch = middleware(middlewareAPI)(dispatch);
return {
dispatch
}
}
var {dispatch} = test();
dispatch(3);
//输出
"in middleware"
"in middleware"
"in middleware"
"in middleware"
"original dispatch"
第二种方式,middleware中dispatch的number会再次经历中间件的处理,当number=3,2,1,0时,都会进入一次middleware函数,当number=0时,next(0)调用的是test中定义的初始dispatch函数,因此不再经过middleware的处理。
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