如何解決這個scala編譯器擦除警告？

Question

我想編譯這個scala代碼並得到以下編譯器警告。

scala> val props: Map[String, _] = 
| x match { 
|  case t: Tuple2[String, _] => { 
|  val prop = 
|  t._2 match { 
|   case f: Function[_, _] => "hello" 
|   case s:Some[Function[_, _]] => "world" 
|   case _ => t._2 
|  } 
|  Map(t._1 -> prop) 
| } 
| case _ => null 
| } 
<console>:10: warning: non-variable type argument String in type pattern (String, _) is unchecked since it is eliminated by erasure 
     case t: Tuple2[String, _] => { 
      ^
<console>:14: warning: non-variable type argument _ => _ in type pattern Some[_ => _] is unchecked since it is eliminated by erasure 
      case s:Some[Function[_, _]] => "world" 
       ^

在How do I get around type erasure on Scala? Or, why can't I get the type parameter of my collections?上給出的答案似乎指向同樣的問題。但我無法在這個特定的背景下推斷出解決方案。

Answer 1

使用case t @ (_: String, _)而不是case t: Tuple2[String, _]和case s @ Some(_: Function[_, _])而不是case s:Some[Function[_, _]]。

關於類型參數，斯卡拉不能match。

Answer 2

我會改寫這樣的：

x match { 
    case (name, method) => { 
    val prop = 
     method match { 
     case f: Function[_, _] => "hello" 
     case Some(f: Function[_, _]) => "world" 
     case other => other 
     } 

    Map(name -> prop) 
    } 
    case _ => null 
}