How to get the line separator of current os easily in Scala?

Question

I remember in Java, we do it like this:

System.getProperty("line.separator")

How to do the same in scala? Is there a better(easier) way?

Answer 1

scala> import util.Properties
import util.Properties

scala> Properties.lineSeparator
res14: java.lang.String =
"
"

Answer 2

Both scala.util.Properties.lineSeparator and System.lineSeparator will do the same job.

System.lineSeparator will directly call Java method which has to find property in systems props:

lineSeparator = props.getProperty("line.separator");

Here is the result:

scala> System.lineSeparator
res0: String = 
"
"

It falls back to default Java props if none found.

Similarly, Properties.lineSeparator will call:

def lineSeparator = propOrElse("line.separator", "\n")

which eventually calls:

System.getProperty(name, alt)

Result is the same:

scala> scala.util.Properties.lineSeparator
res2: String = 
"
"

So they both get line separator from Java props. The only difference is how they get defaults. I don't know why it's implemented this way :). It's like they don't trust that Java will have correct default in this case.