Note that there is already defined function:

scala> val x = "scala is awesome"
x: java.lang.String = scala is awesome

scala> x.reverse
res1: String = emosewa si alacs

But if you want to do that by yourself:

def reverse(s: String) : String =
(for(i <- s.length - 1 to 0 by -1) yield s(i)).mkString

or (sometimes it is better to use until, but probably not in that case)

def reverse(s: String) : String =
(for(i <- s.length until 0 by -1) yield s(i-1)).mkString

Also, note that if you use reversed counting (from bigger one to less one value) you should specify negative step or you will get an empty set:

scala> for(i <- x.length until 0) yield i
res2: scala.collection.immutable.IndexedSeq[Int] = Vector()

scala> for(i <- x.length until 0 by -1) yield i
res3: scala.collection.immutable.IndexedSeq[Int] = Vector(16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1)

s.head is never null. It'll throw an exception if the string is empty (or if it is null, which it should never be - nulls should never really appear in scala code). Also, you are returning too early - x.tail.isEmpty means you still have one char remaining to process. Finally, you always return "" instead of the actual result.

Something like this should work:

  str match {
     case "" => r
     case s => rev(s.tail, s.head + r)
  }

As mentioned in the comment, manipulating strings like this isn't very performant. So, in real life you'd probably want to convert it into a list, reverse the list, and then .mkstring to put it back together.

You can use the following definition:

def printNames(names: List[String]): String = {
  if(names.isEmpty) ""
  else printNames(names.tail) + " " + names.head
}

Note that this implementation is not tail recursive and cannot be optimized by scala compiler. I strongly advise using an accumulator for the resulting String.

Luis Miguel Mejía Suárez suggests using StringBuilder to avoid expensive String concatenation:

def printNames(names: List[String]): StringBuilder = {
  if(names.isEmpty) new StringBuilder("")
  else printNames(names.tail).append(" ").append(names.head)
}