这是我想要理解的代码(它来自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)()
}
}
}
代码的使用方法如下(我假设已经隐含了import io._
)
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
的定义开始。我是否正确地认为io.IO的apply
方法被调用了?那个apply
方法接受一个返回值的无参函数(是吗?)。我猜这就是我卡住的地方。有人能解释一下bufferFile
的定义是如何工作的吗?