Scala: unpacking an Object: how do I know the type

Question

Apache Avro GenericRecord.get(colName) did the (dirty?) trick of returning an Object so it can return whatever type it wants. The Object contains other types: String, Int, Date, etc. The values can be null.

Example:

val one = genericRecord.get("hello")     // "hello" resolves to an Int
val two = genericRecord.get("goodbye")   // "goodbye" resolves to a String

"one" and "two" are of type Object, but they are "actually" an Int and String, respectively.

I'm trying to write an implicit class method to return an Option(<actual_type>) with the result (to avoid dealing with the null case), so I can write

val myVal = genericRecord.getWithType("hello")

and have myVal be of type Option[Int].

Tried the following:

one match { 
    case i: Int => Option(i.asInstanceOf[Int])
    case s: String => Option(s.toString)
}

But I get

error: pattern type is incompatible with expected type;
  found   : Int
  required: Object
       case x: Int =>
               ^

Matching on

one.getClass 

doesn't work either. Is there a way to do this? If so, how?

Thanks!

Answer 1

When coming back from Java interoperability, Scala interprets Object as AnyRef. Since Int is not a subtype of AnyRef, the pattern matcher does not let you do a "stupid" thing.

You can force the interpretation as Any simply by ascribing the result of get with : Any:

val one = genericRecord.get("hello"): Any