最近在看Redux的源码,发现Redux在使用中间件applyMiddleware.js的源码中,有一个对闭包很是巧妙的使用,解决了“鸡生蛋,蛋生鸡”的问题,特分享给你们。前端
Redux中间件的函数签名形式以下:json
({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通过的中间件链。先来看一个错误版本的实现:redux
/* * @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,每一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的每一个函数元素至关于一层洋葱表皮。Redux发送的每个action都会由外到内依次通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,而后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,所以是action dispatch的最后一环。这样分析下来,程序是没有问题的,但当咱们的中间件须要直接使用dispatch函数时,问题就出来了。例如,经常使用于发送异步action的中间件redux-thunk,就须要在异步action中使用dispatch:数组
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中间件以后的其余中间件的处理,显然是不合适的。咱们但愿的方式是,这个action依然会从最外层的中间件开始,由外到内通过每一层中间件的处理。因此,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才能通过每一层中间件的处理。如今问题出来了,chain = middlewares.map(middleware => middleware(middlewareAPI))
须要使用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,咱们进入死循环了。闭包
问题的解决方案就是闭包。当咱们定义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,这样就造成了一个闭包,外部dispatch变量的变化会同步反映到内部函数中。以下所示:app
export default function applyMiddleware(...middlewares) { return (createStore) => (reducer, initialState, enhancer) => { var store = createStore(reducer, initialState, enhancer) var dispatch = store.dispatch; // 须要有初始值,保证中间件在初始化过程当中也能够正常使用dispatch var chain = [] var middlewareAPI = { getState: store.getState, dispatch: (...args) => dispatch(...args) // 经过闭包引用外部的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。若是这个例子对你来讲太复杂,能够用下面这个简化的例子帮助你理解:异步
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,所以再也不通过middleware的处理。函数
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