[英]Scala type erasure in pattern matching Map[String, Int]
在Scala 2.10
,编译器警告给定的代码:
private def getStrFromOpt[T](opt: Option[T]): String = opt match {
case Some(s: String) => s
case Some(i: Int) => i.toString()
case Some(l: Long) => l.toString()
case Some(m: Map[String, Int]) => m map ({ case (k, v) =>
"(" + k + ", " + v + ")" }) mkString ("(", ", ", ")")
case _ => ""
}
消息non-variable type argument String in type pattern Map[String,Int] is unchecked since it is eliminated by erasure: case Some(m: Map[String, Int]) ...
的消息non-variable type argument String in type pattern Map[String,Int] is unchecked since it is eliminated by erasure: case Some(m: Map[String, Int]) ...
的消息non-variable type argument String in type pattern Map[String,Int] is unchecked since it is eliminated by erasure: case Some(m: Map[String, Int]) ...
我如何摆脱这个警告? 如果我想在匹配中包含Map[String, MyObj]
,该怎么Map[String, MyObj]
-如何通过参数化Map区分这两种情况?
您可以使用Scala注释@unchecked禁止显示警告,对于第二个问题,建议您使用Scala反射-TypeTag。 我正在使用Scala 2.11.4,下面是示例代码FYI。
import scala.reflect.runtime.{ universe => ru }
import scala.reflect.runtime.universe.{ typeTag, TypeTag }
object MapType extends App {
def getTypeTag[T: TypeTag](t: T) = typeTag[T].tpe
def getTypeTag[T: TypeTag] = ru.typeOf[T]
// context bound T: ru.TypeTag cause typeOf
// requires implicit parameter
def getStrFromOpt[T: TypeTag](opt: Option[T]): String = {
opt match {
case Some(s: String) => s
case Some(i: Int) => i.toString()
case Some(l: Long) => l.toString()
case Some(m: Map[String, Int] @ unchecked)
if getTypeTag[T] =:= getTypeTag[Map[String, Int]] => "Int Map"
case Some(m: Map[String, String] @ unchecked)
if getTypeTag[T] =:= getTypeTag[Map[String, String]] => "String Map"
case _ => ""
}
}
// "Int Map"
println(getStrFromOpt(Some(Map("a" -> 2, "b" -> 3))))
// "String Map"
println(getStrFromOpt(Some(Map("a" -> "2", "b" -> "3"))))
}
实际上,Scala就像Java一样使用泛型的擦除模型。 因此,在运行时不会保留有关类型参数的信息。
def isIntMap(x: Any) = x match {
case m: Map[Int, Int] => true
case _ => false
}
对于上面的代码,Scala编译器无法确定m是否为Map [Int,Int]。 因此, isIntMap(Map("a"-> "b"))
返回true,这似乎是不直观的。 为了提醒您运行时行为,Scala编译器发出了该未经检查的消息。
但是,Array是一个异常,它的元素类型与元素值一起存储。
def isIntArray(x: Any) = x match {
case a: Array[String] => "yes"
case _ => "no"
}
scala> isIntArray(Array(3))
res1: String = no
scala> isIntArray(Array("1"))
res2: String = yes
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