How to log in Scala *without* a reference to the logger in *every instance*? - scala

Question

I've looked at example of logging in Scala, and it usually looks like this:
import org.slf4j.LoggerFactory
trait Loggable {
private lazy val logger = LoggerFactory.getLogger(getClass)
protected def debug(msg: => AnyRef, t: => Throwable = null): Unit =
{...}
}
This seems independent of the concrete logging framework. While this does the job, it also introduces an extraneous lazy val in every instance that wants to do logging, which might well be every instance of the whole application. This seems much too heavy to me, in particular if you have many "small instances" of some specific type.
Is there a way of putting the logger in the object of the concrete class instead, just by using inheritance? If I have to explicitly declare the logger in the object of the class, and explicitly refer to it from the class/trait, then I have written almost as much code as if I had done no reuse at all.
Expressed in a non-logging specific context, the problem would be:
How do I declare in a trait that the implementing class must have a singleton object of type X, and that this singleton object must be accessible through method def x: X ?
I can't simply define an abstract method, because there could only be a single implementation in the class. I want that logging in a super-class gets me the super-class singleton, and logging in the sub-class gets me the sub-class singleton. Or put more simply, I want logging in Scala to work like traditional logging in Java, using static loggers specific to the class doing the logging. My current knowledge of Scala tells me that this is simply not possible without doing it exactly the same way you do in Java, without much if any benefits from using the "better" Scala.

Answer 1

Premature Optimization is the root of all evil
Let's be clear first about one thing: if your trait looks something like this:
trait Logger { lazy val log = Logger.getLogger }
Then what you have not done is as follows:
You have NOT created a logger instance per instance of your type
You have neither given yourself a memory nor a performance problem (unless you have)
What you have done is as follows:
You have an extra reference in each instance of your type
When you access the logger for the first time, you are probably doing some map lookup
Note that, even if you did create a separate logger for each instance of your type (which I frequently do, even if my program contains hundreds of thousands of these, so that I have very fine-grained control over my logging), you almost certainly still will neither have a performance nor a memory problem!
One "solution" is (of course), to make the companion object implement the logger interface:
object MyType extends Logger
class MyType {
import MyType._
log.info("Yay")
}

Answer 2

How do I declare in a trait that the
implementing class must have a
singleton object of type X, and that
this singleton object must be
accessible through method def x: X ?
Declare a trait that must be implemented by your companion objects.
trait Meta[Base] {
val logger = LoggerFactory.getLogger(getClass)
}
Create a base trait for your classes, sub-classes have to overwrite the meta method.
trait Base {
def meta: Meta[Base]
def logger = meta.logger
}
A class Whatever with a companion object:
object Whatever extends Meta[Base]
class Whatever extends Base {
def meta = Whatever
def doSomething = {
logger.log("oops")
}
}
In this way you only need to have a reference to the meta object.
We can use the Whatever class like this.
object Sample {
def main(args: Array[String]) {
val whatever = new Whatever
whatever.doSomething
}
}

Answer 3

I'm not sure I understand your question completely. So I apologize up front if this is not the answer you are looking for.
Define an object were you put your logger into, then create a companion trait.
object Loggable {
private val logger = "I'm a logger"
}
trait Loggable {
import Loggable._
def debug(msg: String) {
println(logger + ": " + msg)
}
}
So now you can use it like this:
scala> abstract class Abstraction
scala> class Implementation extends Abstraction with Loggable
scala> val test = new Implementation
scala> test.debug("error message")
I'm a logger: error message
Does this answer your question?

Answer 4

I think you cannot automatically get the corresponding singleton object of a class or require that such a singleton exists.
One reason is that you cannot know the type of the singleton before it is defined. Not sure, if this helps or if it is the best solution to your problem, but if you want to require some meta object to be defined with a specific trait, you could define something like:
trait HasSingleton[Traits] {
def meta: Traits
}
trait Log {
def classname: String
def log { println(classname) }
}
trait Debug {
def debug { print("Debug") }
}
class A extends HasSingleton[Log] {
def meta = A // Needs to be defined with a Singleton (or any object which inherits from Log}
def f {
meta.log
}
}
object A extends Log {
def classname = "A"
}
class B extends HasSingleton[Log with Debug] { // we want to use Log and Debug here
def meta = B
def g {
meta.log
meta.debug
}
}
object B extends Log with Debug {
def classname = "B"
}
(new A).f
// A
(new B).g
// B
// Debug