帮助我理解这段 Scala 代码：scalaz IO Monad

这是我试图理解的代码（它来自http://apocalisp.wordpress.com/2010/10/17/scalaz-tutorial-enumeration-based-io-with-iteratees/ http://apocalisp.wordpress.com/2010/10/17/scalaz-tutorial-enumeration-based-io-with-iteratees/):

object io {
  sealed trait IO[A] {
    def unsafePerformIO: A
  }

  object IO {
    def apply[A](a: => A): IO[A] = new IO[A] {
      def unsafePerformIO = a
    }
  }

  implicit val IOMonad = new Monad[IO] {
    def pure[A](a: => A): IO[A] = IO(a)
    def bind[A,B](a: IO[A], f: A => IO[B]): IO[B] = IO {
      implicitly[Monad[Function0]].bind(() => a.unsafePerformIO,
                                        (x:A) => () => f(x).unsafePerformIO)()
    }
  }
}

这段代码是这样使用的（我假设import io._是隐含的）

def bufferFile(f: File) = IO {   new BufferedReader(new FileReader(f)) }

def closeReader(r: Reader) = IO {   r.close }

def bracket[A,B,C](init: IO[A], fin: A => IO[B], body: A => IO[C]): IO[C] = for { a <- init
      c <- body(a)
      _ <- fin(a) }   yield c

def enumFile[A](f: File, i: IterV[String, A]): IO[IterV[String, A]] =  bracket(bufferFile(f),
          closeReader(_:BufferedReader),
          enumReader(_:BufferedReader, i))

让我们从bufferFile定义。我的想法是否正确applyio.IO 的方法被调用？那apply方法采用一个返回值的无参数函数（正确吗？）。我想这就是我被困住的地方。有人可以解释一下它的定义吗bufferFile works?

是的，你是对的，差不多了；io.IO.apply使用所谓的“按名称”参数调用，该参数基本上是一个不带任何内容的函数（Unit）并返回A。最酷的事情是当你传递一个实例时A直接喜欢new BufferedReader(new FileReader(f))，它将被转换为类似的东西() => new BufferedReader(new FileReader(f)).

后果apply你得到一个实例IO[BufferedReader]它定义了一个方法def unsafePerformIO它只是返回捕获的实例BufferedReader.