Given:
case class FirstCC {
def name: String = ... // something that will give \"FirstCC\"
}
case class SecondCC extends FirstCC
val one = FirstCC()
val two =
def name = this.getClass.getName
def name = this.getClass.getName
Or if you want only the name without the package:
def name = this.getClass.getSimpleName
See the documentation of java.lang.Class for more information.
def name = getClass.getSimpleName.split('$').head
This will remove the $1
appearing at the end on some classes.
class Example {
private def className[A](a: A)(implicit m: Manifest[A]) = m.toString
override def toString = className(this)
}
Here is a Scala function that generates a human-readable string from any type, recursing on type parameters:
https://gist.github.com/erikerlandson/78d8c33419055b98d701
import scala.reflect.runtime.universe._
object TypeString {
// return a human-readable type string for type argument 'T'
// typeString[Int] returns "Int"
def typeString[T :TypeTag]: String = {
def work(t: Type): String = {
t match { case TypeRef(pre, sym, args) =>
val ss = sym.toString.stripPrefix("trait ").stripPrefix("class ").stripPrefix("type ")
val as = args.map(work)
if (ss.startsWith("Function")) {
val arity = args.length - 1
"(" + (as.take(arity).mkString(",")) + ")" + "=>" + as.drop(arity).head
} else {
if (args.length <= 0) ss else (ss + "[" + as.mkString(",") + "]")
}
}
}
work(typeOf[T])
}
// get the type string of an argument:
// typeString(2) returns "Int"
def typeString[T :TypeTag](x: T): String = typeString[T]
}
You can use the property productPrefix
of the case class:
case class FirstCC {
def name = productPrefix
}
case class SecondCC extends FirstCC
val one = FirstCC()
val two = SecondCC()
one.name
two.name
N.B.
If you pass to scala 2.8 extending a case class have been deprecated, and you have to not forget the left and right parent ()