Coffeescript，我将如何编写这个排队函数示例，尤其是循环？

发布于 2024-10-03 13:42:33 字数 677 浏览 2 评论 0原文

我正在尝试获取一些示例，说明如何在 CoffeeScript 中以与 JavaScript 不同的方式执行某些操作。在这个排队函数的例子中，我对如何在 CoffeeScript 中处理这个问题感到困惑，

    wrapFunction = (fn, context, params) ->
            return ->
                fn.apply(context, params)        

    sayStuff = (str) ->
        alert(str)


    fun1 = wrapFunction(sayStuff, this, ['Hello Fun1'])
    fun2 = wrapFunction(sayStuff, this, ['Hello Fun2'])

    funqueue = []
    funqueue.push(fun1)
    funqueue.push(fun2)

    while (funqueue.length > 0) {
        (funqueue.shift())();   
    }

特别是我将如何在 CoffeeScript 中重写它？

while (Array.length > 0) {
    (Array.shift())();

原文

I'm trying to get some examples under my belt of how you would do something differently in CoffeeScript to JavaScript. In this example of queuing functions I'm confused at how you would do handle this in CoffeeScript

    wrapFunction = (fn, context, params) ->
            return ->
                fn.apply(context, params)        

    sayStuff = (str) ->
        alert(str)


    fun1 = wrapFunction(sayStuff, this, ['Hello Fun1'])
    fun2 = wrapFunction(sayStuff, this, ['Hello Fun2'])

    funqueue = []
    funqueue.push(fun1)
    funqueue.push(fun2)

    while (funqueue.length > 0) {
        (funqueue.shift())();   
    }

Especially how would I rewrite this in CoffeeScript?

while (Array.length > 0) {
    (Array.shift())();

分享到QQ

分享到微博

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

拿命拼未来 2024-10-10 13:42:33

fun1 = -> alert 'Hello Fun1'
fun2 = -> alert 'Hello Fun2'

funqueue = [fun1, fun2]

el() for el in funqueue

fun1 = -> alert 'Hello Fun1'
fun2 = -> alert 'Hello Fun2'

funqueue = [fun1, fun2]

el() for el in funqueue

回复收藏 0 原文

一曲琵琶半遮面シ 2024-10-10 13:42:33

f1 = (completeCallback) ->
  console.log('Waiting...')
  completeCallback()

funcs = [ f1, f2, f3 ]

next = ->
  if funcs.length > 0
    k = funcs.shift()
    k(next)

next()

f1 = (completeCallback) ->
  console.log('Waiting...')
  completeCallback()

funcs = [ f1, f2, f3 ]

next = ->
  if funcs.length > 0
    k = funcs.shift()
    k(next)

next()

回复收藏 0 原文

~没有更多了~