Visibility of properties in scala class

Question

I defined a property in the constructor of my class the following way:

class Step(val message:String = "")

When I try access to message value from Java code y get a visbility error. Why?

Answer 1

Have you tried using getMessage()? Maybe scala is generating the accessor.

Answer 2

If you add the @scala.reflect.BeanProperty annontation you get "automatic" get and set methods

See http://www.scala-lang.org/docu/files/api/scala/reflect/BeanProperty.html

scala> class Step(@scala.reflect.BeanProperty val  message:String )
defined class Step

scala> val s = new Step("asdf")
s: Step = Step@71e13a2c

scala> s.message
res6: String = asdf

scala> s.getMessage
res10: String = asdf

Answer 3

The code is correct, message should be public in this case, but for some reason it is not. So, as a WO you could make it private (just drop the "val") and find a way to produce a getter for this value:

class Step(message: String = ""){
  def getMessage() = message  
}

Or:

class Step(@scala.reflect.BeanProperty message: String = "")

And compile:

> scalac -cp . Step.scala

Then create the calling Java class:

public class SomeClass{
  public static void main(String[] args) {
    Step step = new Step("hello");
    System.out.println(" " + step.getMessage());
  }
}

Then compile and run:

> javac -cp . SomeClass.java
> java -cp "/home/olle/scala-2.8.0.Beta1-prerelease/lib/scala-library.jar:." SomeClass
hello
>

Answer 4

I guess that in the Java code you're trying to access the field with step.message. Indeed, there is such a field, and it is private. That is why you get the visibility error. When you declare 'val' fields in Scala, the compiler generates a field and accessor method. So in java you should use step.message()