Scala代码揭开神秘面纱 [英] Scala Code demystify

问题描述

有人可以揭开这个代码的神秘面纱，这是Play20框架中zentasks示例的一部分.我很好奇这是如何工作的，因为我是Java的Scala的新手，所以很多事情很难缠住我的头.

Could someone demystify this code which is part of the zentasks example in the Play20 framework. I'm curious how this works, granted I'm new to Scala from Java so a lot of things are tough to wrap my head around.

def IsAuthenticated(f: => String => Request[AnyContent] => Result) = 
  Security.Authenticated(username, onUnauthorized) { user =>
    Action(request => f(user)(request))
  }

推荐答案

您需要对签名进行一些拆分. f是接受尚未计算的字符串=> String并返回接受Request[AnyContent]并返回结果的另一个函数的函数.

You need to split up the signature a bit. f is a function that takes a not-yet-computed string => String and returns another function that accepts a Request[AnyContent] and returns a result.

Security.Authenticated调用接受两个参数列表.具有username和onUnauthorized的一个.第二个函数具有接受用户并返回操作的功能.

The Security.Authenticated call accepts two parameters lists. One that has username and onUnauthorized. The second takes a function accepting the user and returning an action.

Action.apply方法接受功能Request[AnyContent] => Result

因此，f以"curried"方式被调用.那是第一个函数被调用，然后结果函数立即被使用f(user)(request).

so, the f is called in 'curried' fashion. That is the first function is called, and then the resulting function is immediately used f(user)(request).

这是丑陋的(至少，尽我所能)和丑陋的东西:

Here's the same thing desugared (at least, as best I can) and ugly:

def isAuthenticated(f: => String => Request[AnyContent] => Result) =
  Security.Authenticated(username, onUnauthorized) { user: String =>
     Action.apply { request: Request[AnyContent] =>
       val hiddenTmp: Request[AnyContent] => Result = f(user)
       hiddenTemp.apply(request)
     }
  }

您可以看到编译器正在做一些工作来删除类型注释.希望这有助于解释它如何还原成原始scala.本质上，该功能执行很多功能组成.

You can see the compiler is doing a bit of work removing type annotations. Hopefully that helps explain how it desugars into raw scala. Essentially, the function does a lot of functional composition.

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