Type casting using type parameter

Question

Given is a Java method that returns java.lang.Objects for a given string. I\'d like to wrap this method in a Scala method that converts the returned instances t

Answer 1

You could try shapeless's Typeable,

scala> import shapeless._ ; import syntax.typeable._
import shapeless._
import syntax.typeable._

scala> def someJavaMethod(key: String): AnyRef =
     |   key match {
     |     case "keyToSomeInt" => 23.asInstanceOf[AnyRef]
     |     case "keyToSomeString" => "foo"
     |   }
someJavaMethod: (key: String)AnyRef

scala> def convert[T: Typeable](key: String): Option[T] =
     |   someJavaMethod(key).cast[T]
convert: [T](key: String)(implicit evidence$1: shapeless.Typeable[T])Option[T]

scala> convert[Int]("keyToSomeInt")
res0: Option[Int] = Some(23)

scala> convert[String]("keyToSomeString")
res1: Option[String] = Some(foo)

scala> convert[String]("keyToSomeInt")
res2: Option[String] = None

scala> convert[Int]("keyToSomeString")
res3: Option[Int] = None

Answer 2

That's what a ClassTag if for:

import reflect.ClassTag

def convert[T : ClassTag](key: String): Option[T] = {
  val ct = implicitly[ClassTag[T]]
  someJavaMethod(key) match {
    case ct(x) => Some(x)
    case _ => None
  }
}

It can be used as an extractor to test and cast to the proper type at the same time.

Example:

scala> def someJavaMethod(s: String): AnyRef = "e"
someJavaMethod: (s: String)AnyRef

[...]

scala> convert[Int]("key")
res4: Option[Int] = None

scala> convert[String]("key")
res5: Option[String] = Some(e)

Edit: note however that a ClassTag does not automatically unbox boxed primitives. So, for example, convert[Int]("a") would never work, because the java method returns AnyRef, it would have to be convert[java.lang.Integer]("a"), and so on for other primitive types.

Miles's answer with Typeable seems to take care of those edge cases automatically.