Question

Groovy

Groovy comes with a compiler called groovyc. For each script, groovyc generates a class that extends groovy.lang.Script, which contains a main method so that Java can execute it. The name of the compiled class matches the name of the script being compiled.

For example, with this HelloWorld.groovy script:

println "Hello World"

That becomes something like this code:

class HelloWorld extends Script {
    public static void main(String[] args) {
        println "Hello World"
    }
}

Scala

Scala comes with a compiler called scalac.

For example, with the same HelloWorld.scala script:

println("Hello World")

The code is not valid for scalac, because the compiler expected class or object definition, but works in the Scala REPL Interpreter. How is possible? Is it wrapped in a class before the execution?

Answer 1

The code in a Scala-Script is first placed in a Scala object, then compiled to JVM-Bytecode and at last executed. You can see the generated Scala object by writing scala -Xprint:parser my_file.scala:

package <empty> {
  object Main extends scala.ScalaObject {
    def <init>() = {
      super.<init>();
      ()
    };
    def main(argv: Array[String]): scala.Unit = {
      val args = argv;
      {
        final class $anon extends scala.AnyRef {
          def <init>() = {
            super.<init>();
            ()
          };
          println("hello world")
        };
        new $anon()
      }
    }
  }
}