斯卡拉錯誤的前向參考

Question

我正在通過一些練習：斯卡拉函數式編程特別是問題5.2。問題是，我用答案關鍵字拼湊了以下代碼。

sealed trait Stream[+A] 
{ 
    def take(n: Int): Stream[A] = this match { 
    case Cons(hs, ts) if n > 1 => cons(h(), t().take(n - 1)) 
    case Cons(hs, _) if n == 1 => cons(h(), empty) 
    case _ => empty 
    } 

} 
case object Empty extends Stream[Nothing] 
case class Cons[+A](h:() => A, t:() => Stream[A]) extends Stream[A] 

object Stream{ 
    def cons[A](hd: => A, tl: => Stream[A]): Stream[A] = { 
    lazy val head = hd 
    lazy val tail = tl 
    Cons(() => head ,() => tail) 
    } 

    def empty[A]: Stream[A] = Empty 

    def apply[A](as: A*): Stream[A] = 
    if (as.isEmpty) empty 
    else cons(as.head, apply(as.tail: _*)) 

} 

我得到的REPL如下：

<console>:10: error: not found: type A 
        def take(n: Int): Stream[A] = this match { 
              ^
<console>:11: error: not found: value Cons 
         case Cons(hs, ts) if n > 1 => cons(h(), t().take(n - 1)) 
         ^
<console>:11: error: not found: value cons 
         case Cons(hs, ts) if n > 1 => cons(h(), t().take(n - 1)) 
                ^
<console>:12: error: not found: value Cons 
         case Cons(hs, _) if n == 1 => cons(h(), empty) 
         ^
<console>:12: error: not found: value cons 
         case Cons(hs, _) if n == 1 => cons(h(), empty) 
                ^
<console>:13: error: not found: value empty 
         case _ => empty 

          ^

Answer 1

你有這個代碼2個問題：

1. 沒有明確規定該empty和cons方法位於伴侶物體Stream

爲了解決這個問題，你需要或者import Stream._到類：

sealed trait Stream[+A] { 
    import Stream._ 
    def take(n: Int): Stream[A] = this match { 
    case Cons(hs, ts) if n > 1 => cons(hs(), ts().take(n - 1)) 
    case Cons(hs, _) if n == 1 => cons(hs(), empty) 
    case _ => empty 
    } 
} 

或者你需要明確指定：

sealed trait Stream[+A] { 
    def take(n: Int): Stream[A] = this match { 
    case Cons(hs, ts) if n > 1 => Stream.cons(hs(), ts().take(n - 1)) 
    case Cons(hs, _) if n == 1 => Stream.cons(hs(), Stream.empty) 
    case _ => Stream.empty 
    } 
} 

使用的t變量名和h，它們在case class Cons而不是hs和ts的綁定變量中。

當你這樣做：

case Cons(hs, ts) if n > 1 => Stream.cons(hs(), ts().take(n - 1)) 
case Cons(hs, _) if n == 1 => Stream.cons(hs(), Stream.empty) 

你是說你要提取的情況下類參數hs和ts分別在接下來的代碼塊中使用它們。無論他們在案例類別中是否被稱爲h和t，它們都將被分配您在匹配中指定的名稱。

修復這兩個問題，你的代碼應該編譯（我個人使用Scala 2.11.5和Java 1.7測試，但我不認爲它應該的問題）：

sealed trait Stream[+A] { 
    def take(n: Int): Stream[A] = this match { 
    case Cons(hs, ts) if n > 1 => Stream.cons(hs(), ts().take(n - 1)) 
    case Cons(hs, _) if n == 1 => Stream.cons(hs(), Stream.empty) 
    case _ => Stream.empty 
    } 
} 
case object Empty extends Stream[Nothing] 
case class Cons[+A](h:() => A, t:() => Stream[A]) extends Stream[A] 

object Stream{ 
    def cons[A](hd: => A, tl: => Stream[A]): Stream[A] = { 
    lazy val head = hd 
    lazy val tail = tl 
    Cons(() => head ,() => tail) 
    } 

    def empty[A]: Stream[A] = Empty 

    def apply[A](as: A*): Stream[A] = 
    if (as.isEmpty) empty 
    else cons(as.head, apply(as.tail: _*)) 

}