在Scala中通过擦除消除了返回类型

Question

Consider the following Scala code snippet:

def func(param: Any): Int = param match {
  case f: (String => Int) => f("apple")
  case i: Int => i
}

println(func((s: String) => s.length))

Works as expected, however, at compilation I get the following warning:

<console>:11: warning: non-variable type argument String in type pattern String => Int is unchecked since it is eliminated by erasure
         case f: (String => Int) => f("apple")

How can I get rid of this warning message?

Thanks your help in advance!

Answer 1

The reason why you get the message is because of Java's generic type erasure . In this particular case, your function which is of type Function[String, Int] will be matched by any Function[A, B] .

In order to get rid of this warning you should use scala typetags which will allow you to differentiate between the different function types.

The code snippet is below,

import scala.reflect.runtime.universe._

object Answer {
 def function[A](param: A)(implicit tt: TypeTag[A]): String = param match {
   case f: (String => Int) @unchecked if typeOf[String => Int] =:= typeOf[A] => f("apple").toString
    case f: (Int => String) @unchecked if typeOf[Int => String] =:= typeOf[A] => f(32 + 1)
    case s: String => s"hello $s"
  }

  def main (args: Array[String]) {
     println(function((s: String) => s.length))
     println(function((i: Int) => i.toString))
     println(function("world"))
  }
}

The key part is to have an implicit TypeTag[A] which is added at compile time which includes the metadata that the function typeOf needs to check the types of A against anything else.