斯卡拉模式匹配混淆选项[任何]

Question

我有以下的Scala代码。

import scala.actors.Actor 

object Alice extends Actor { 
    this.start 
    def act{ 
    loop{ 
     react { 
     case "Hello" => sender ! "Hi" 
     case i:Int => sender ! 0 
     } 
    } 
    } 
} 
object Test { 
    def test = { 
    (Alice !? (100, "Hello")) match { 
     case i:Some[Int] => println ("Int received "+i) 
     case s:Some[String] => println ("String received "+s) 
     case _ => 
    } 
    (Alice !? (100, 1)) match { 
     case i:Some[Int] => println ("Int received "+i) 
     case s:Some[String] => println ("String received "+s) 
     case _ => 
    } 
    } 
} 

做Test.test后，我得到的输出：

scala> Test.test 
Int received Some(Hi) 
Int received Some(0) 

我期待输出

String received Some(Hi) 
Int received Some(0) 

有什么解释？

作为第二个问题，我得到unchecked警告上述如下：

C:\scalac -unchecked a.scala 
a.scala:17: warning: non variable type-argument Int in type pattern Some[Int] is unchecked since it is eliminated by erasure 
     case i:Some[Int] => println ("Int received "+i) 
      ^
a.scala:18: warning: non variable type-argument String in type pattern Some[String] is unchecked since it is eliminated by erasure 
     case s:Some[String] => println ("String received "+s) 
      ^
a.scala:22: warning: non variable type-argument Int in type pattern Some[Int] is unchecked since it is eliminated by erasure 
     case i:Some[Int] => println ("Int received "+i) 
      ^
a.scala:23: warning: non variable type-argument String in type pattern Some[String] is unchecked since it is eliminated by erasure 
     case s:Some[String] => println ("String received "+s) 
      ^
four warnings found 

我怎样才能避免警告？

编辑：谢谢你的建议。丹尼尔的想法是好的，但似乎并没有与泛型类型的工作，如下面

def test[T] = (Alice !? (100, "Hello")) match { 
    case Some(i: Int) => println ("Int received "+i) 
    case Some(t: T) => println ("T received ") 
    case _ => 
} 

的例子以下 错误遇到 警告：warning: abstract type T in type pattern T is unchecked since it is eliminated by erasure

Answer 1

这是因为类型擦除。除了数组之外，JVM不知道任何类型参数。因此，Scala代码无法检查Option是Option[Int]还是Option[String] - 该信息已被删除。

您可以治好你的代码这种方式，虽然：

object Test { 
    def test = { 
    (Alice !? (100, "Hello")) match { 
     case Some(i: Int) => println ("Int received "+i) 
     case Some(s: String) => println ("String received "+s) 
     case _ => 
    } 
    (Alice !? (100, 1)) match { 
     case Some(i: Int) => println ("Int received "+i) 
     case Some(s: String) => println ("String received "+s) 
     case _ => 
    } 
    } 
} 

这样，你是不是测试的Option类型是什么，但什么内容类型是 - 假如有任何内容。 A None将落入默认情况。

Answer 2

有关类型参数的任何信息，只适用于编译时间，而不是在运行时（这被称为类型擦除）。这意味着在运行时Option[String]和Option[Int]之间没有区别，所以类型Option[String]上的任何模式匹配也将匹配Option[Int]，因为在运行时，两者都只是Option。

由于这几乎总是不符合您的要求，所以您会收到警告。避免警告的唯一方法不是在运行时检查某些东西的泛型类型（这很好，因为它无法像你想要的那样工作）。

有没有办法来检查Option是否是Option[Int]或Option[String]在运行时（比如果它是一个Some检查内容等）。

Answer 3

如前所述，你在这里反对擦除。

对于解决方案......这是正常使用Scala的演员来定义你很可能会发送的每个信息类型的案件类别：

case class MessageTypeA(s : String) 
case class MessageTypeB(i : Int) 

object Alice extends Actor { 
    this.start 
    def act{ 
    loop{ 
     react { 
     case "Hello" => sender ! MessageTypeA("Hi") 
     case i:Int => sender ! MessageTypeB(0) 
     } 
    } 
    } 
} 
object Test { 
    def test = { 
    (Alice !? (100, "Hello")) match { 
     case Some(MessageTypeB(i)) => println ("Int received "+i) 
     case Some(MessageTypeA(s)) => println ("String received "+s) 
     case _ => 
    } 
    (Alice !? (100, 1)) match { 
     case Some(MessageTypeB(i)) => println ("Int received " + i) 
     case Some(MessageTypeA(s)) => println ("String received " + s) 
     case _ => 
    } 
    } 
}