Scala: Is there a default class if no class is defined?

Question

According to this, Scala methods belong to a class. However, if I define a method in REPL or in a script that I then execute using scala, what class does the method belong to ?

scala> def hoho(str:String) = {println("hoho " + str)}
hoho: (str: String)Unit

scala> hoho("rahul")
hoho rahul

In this example, what class does the method belong to ?

Answer 1

The REPL wraps all your statements (actually rewrites your statements) in objects automagically. You can see it in action if you print the intermediate code by using the -Xprint:typer option:

scala> def hoho(str:String) = {println("hoho " + str)}
[[syntax trees at end of typer]]// Scala source: <console>
package $line1 {
  final object $read extends java.lang.Object with ScalaObject {
    def this(): object $line1.$read = {
      $read.super.this();
      ()
    };
    final object $iw extends java.lang.Object with ScalaObject {
      def this(): object $line1.$read.$iw = {
        $iw.super.this();
        ()
      };
      final object $iw extends java.lang.Object with ScalaObject {
        def this(): object $line1.$read.$iw.$iw = {
          $iw.super.this();
          ()
        };
        def hoho(str: String): Unit = scala.this.Predef.println("hoho ".+(str))
      }
    }
  }
}

So your method hoho is really $line1.$read.$iw.$iw.hoho. Then when you use hoho("foo") later on, it'll rewrite to add the package and outer objects.

Additional notes: for scripts, -Xprint:typer (-Xprint:parser) reveals that the code is wrapped inside a code block in the main(args:Array[String]) of an object Main. You have access to the arguments as args or argv.