CODEWITHSUNDEEP
scalacasting
Ivan
Ivan

Reputation: 64207

How to cast java.lang.Object to a specific type in Scala?

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.

Upvotes: 20

Views: 22569

Answers (3)

David Pelaez
David Pelaez

Reputation: 1384

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.

Upvotes: 5

Daniel C. Sobral
Daniel C. Sobral

Reputation: 297185

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)

Upvotes: 11

Travis Brown
Travis Brown

Reputation: 139038

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.

Upvotes: 26

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