Scala: Copying case classes with trait

10
votes

I'm fairly new to Scala and I have a question about the best way to copy a case class while preserving data that comes from traits. For example, let's say I have the following:

trait Auditing {

  var createTime: Timestamp = new Timestamp(System.currentTimeMillis)
}

case class User(val userName: String, val email: String) extends Auditing

val user = User("Joe", "[email protected]")

Then I want to make a new copy with one parameter changed:

val user2 = user.copy(email = "[email protected]")

Now, in the example above, the property createTime does not get copied over because it is not defined in the constructor of the User case class. So my question is: assuming that moving createTime into the constructor is not an option, what is the best way for getting a copy of the User object that includes the value from the trait?

I'm using Scala 2.9.1

Thanks in advance! Joe

3
scala
There's not much options ahead of you: either you manually implement such a method that will produce the copy you want in the declaration of the User case class or you use Scala 2.10 macros feature to automate that. The second option definitely won't be an easy task for a beginner. - Nikita Volkov
@NikitaVolkov If you were willing to provide me/us with an example of how you would do it with macros, that would be more than great. - Malte Schwerhoff
@mhs I join the club. I'm a newbee in macros, that's why I didn't post it as answer. But here's how I solved a very similar task with Toolbox api. The accepted answer there is based on macros, but I think it doesn't support the latest Scala version. - Nikita Volkov
@Joe You can post it as another question specifically about macros. I know for sure that there are some experts on macros monitoring this site. I'll gladly upvote that kind of a question. - Nikita Volkov

3 Answers

6
votes

You can override the copy method with that behavior.

case class User(val userName: String, val email: String) extends Auditing
{
  def copy(userName = this.userName, email = this.email) {
   val copiedUser = User(userName, email)
   copiedUser.createTime = createTime
   copiedUser      
  }
}
3
votes

While I see no other solution than Reuben's, I don't understand the requirement to leave the constructor args untouched. This would be the most natural solution:

case class User(userName: String, email: String, 
   override val createTime:Timestamp = new Timestamp(System.currentTimeMillis)) 
      extends Auditing

If you don't want the user to be able to overwrite createTime, you can still use:

case class User private (userName: String, email: String, 
   override val createTime:Timestamp) extends Auditing {
   def this(userName: String, email: String) =
     this(userName, email, new Timestamp(System.currentTimeMillis))
}

The only drawback is that you need to write new User("Joe", "[email protected]"), as the primary constructor is now private.

0
votes

You might be better of not using a case class. You can easily implement the functionality you need yourself. The below code implements the copy method you wanted, a constructor without new, hides the original constructor, and creates an extractor so that you can use User in case statements.

class User private(val userName: String,
                   val email: String,
                   val timeStamp: Timestamp =
                   new Timestamp(System.currentTimeMillis)) {

    def copy(uName: String = userName,
             eMail: String = email) =
    new User(uName, eMail, timeStamp)
}

object User {
  def apply(userName: String, email: String) =
    new User(userName, email)

  def unapply(u: User) = Some((u.userName, u.email, u.timeStamp))
}