How Can I Check an Object to See its Type and Return A Casted Object

Question

I have method to which I pass an object. In this method I check it's type and depending on the type I do something with it and return a Long. I have tried every which way I can think of to do this and I always get several compiler errors telling me it expects a certain object but gets another. Can someone please explain to me what I am doing wrong and guide me in the right direction? What I have tried thus far is below:

  override def getInteger(obj:Object) = {
    if (obj.isInstanceOf[Object]) null
    else if (obj.isInstanceOf[Number]) 
      (obj:Number).longValue()
    else if (obj.isInstanceOf[Boolean]) 
      if (obj:Boolean) 1 else 0
    else if (obj.isInstanceOf[String]) 
      if ((obj:String).length == 0 | (obj:String) == "null") 
        null
      else
          try {
            Long.parse(obj:String)
          } catch {
            case e: Exception => throw new ValueConverterException("value \"" + obj.toString() + "\" of type " + obj.getClass().getName() + " is not convertible to Long")        
          }
  }

Answer 1

Pattern matching would make it much more nicer.

def getInteger(obj: Any) = obj match {
  case n: Number => n.longValue
  case b: Boolean => if(b) 1 else 0
  case s: String if s.length != 0 && s != "null" => s.toLong
  case _ => null
}

Answer 2

This code cries out for using a match:

obj match {
  case n: Number => n.longValue
  case b: Boolean => if (b) 1 else 0
  case s: String => if ((s eq null) || s.length == 0) null else {
            // try ... catch ... etc.
          }
  case o: Object => null
}

Followed my own advice from my comment to my original reply...

Answer 3

This might be a start:

def getInteger (o : Any) : Long = o match {
     case (o: Boolean) => if (o) 1 else 0       
     case (l: Long) => l                        
     case (s: String) => java.lang.Long.parseLong (s)   
     case _ => 0L 
}  

(I don't have something to override, and skipped the try/catch for brevity)