As far as I know, in Java I can
Object o = new String("abc")
String s = (String) o
But how to rewrite it in Scala?
val o: java.lang.Object = new java.lang.String("abc")
val s: String = // ??
A Java library I want to use returns java.lang.Object which I need to cast to a more specific type (also defined in this library). A Java example does it exactly the way like the first example of mine, but just using Scala's source: TargetType instead of Java's (TargetType)source doesn't work.
If you absolutely have to—for compatibility with a Java library, for example—you can use the following:
val s: String = o.asInstanceOf[String]
In general, though, asInstanceOf is a code smell and should be avoided.
Here's a safe way of doing it:
val o: java.lang.Object = new java.lang.String("abc")
o match { case s: String => /* do stuff with s */ }
If you need to return the result, this works as well:
import PartialFunction._
def onlyString(v: Any): Option[String] = condOpt(v) { case x: String => x }
val s: Option[String] /* it might not be a String! */ = onlyString(o)
asInstanceOf is just fine—e.g. deserializing with readObject, when you're stuck with libraries that use pre-generics Java collections, etc. I was assuming this was one of those cases.For the sake of future people having this issue, Travi's previous answer is correct and can be used for instance in Yamlbeans to map Desearialized objects to Maps doing something like:
val s: java.util.Map[String,String] = obj.asInstanceOf[java.util.Map[String,String]]
I hope this little comment comes handy for one in the future as a detail over the answer I found here.
