In Scala, how do I get the *name* of an `object` (not an instance of a class)?

Question

In Scala, I can declare an object like so:

class Thing

object Thingy extends Thing

How would I get "Thingy" (the name of the object) in Scala?

I've heard that Lift (the web framework for Scala) is capable of this.

Lappet

don't you already have the variable name? why do u want to do that at all? – Isometry

see this too: #5051182 – Isometry

@Isometry an object is not a variable. You're confused. When a variable points to an object, the name of the variable is not the same thing as the name of the object. – Faena

Answer 1

Just get the class object and then its name.

scala> Thingy.getClass.getName
res1: java.lang.String = Thingy$

All that's left is to remove the $.

EDIT:

To remove names of enclosing objects and the tailing $ it is sufficient to do

res1.split("\\$").last

Disconsolate

It also can be test.Main$Thingy$ or test.Main$Test1$2$ if object defined in other object or method. – Guff

@SergeyPassichenko: So then you have to do a bit more parsing to get the value out, but the basic idea is the same. – Disconsolate

@SergeyPassichenko: Included a solution in the answer. It doesn't cover objects defined in methods, but that's usually not needed. – Disconsolate

Here's another peculiarity. Thingy.getClass.getName => "Thingy$" but println(Thingy.getClass.getName) prints "$line2.$read$$iw$$iw$Thingy$".. Can anyone explain this? – Lappet

@Scoobie: That only happens in the REPL. – Disconsolate

In 2.10, you can do it like this: scala> object Foo defined module Foo scala> import scala.reflect.runtime.universe._ import scala.reflect.runtime.universe._ scala> typeOf[Foo.type].termSymbol.name res0: reflect.runtime.universe.Name = Foo – Gunpaper

Answer 2

If you declare it as a case object rather than just an object then it'll automatically extend the Product trait and you can call the productPrefix method to get the object's name:

scala> case object Thingy
defined module Thingy

scala> Thingy.productPrefix
res4: java.lang.String = Thingy

Overpraise

I like the fact that it's simple, and standard API. However, unfortunately Product does not guarantee that it will always just be the name of the object implementing it, AFAICT. – Trainer

Answer 3

I don't know which way is the proper way, but this could be achieved by Scala reflection:

implicitly[TypeTag[Thingy.type]].tpe.termSymbol.name.toString

Good

Answer 4

def nameOf[T](implicit ev: ClassTag[T]):String = ev.runtimeClass.getName.replace("$", "")

Shelf