pattern match on a list to check empty - scala

Question

I want to check if a list is empty using pattern matching
val simplelist: List[Char] = List('a', 'b', 'c', 'd')
def test(l:List[Char]):Unit= l.map(d => print(d))
simplelist match {
case x :: xs =>
test(xs)
case Nil=> log.info("empty")
}
Output is bcd and not abcd

Answer 1

You could try it as simple as:
scala> List(1,2).isEmpty
res0: Boolean = false
You do not need pattern matching for this.

Answer 2

For Empty List pattern match, you can do this.
val simplelist: List[Char] = List('a', 'b', 'c', 'd')
def test(l:List[Char]):Unit= l.map(d => print(d))
simplelist match {
case x :: xs =>
test(xs)
case List() => println("empty List") // this will pattern match as empty List
case Nil=> log.info("empty")
}
2) For the question, Output is bcd and not abcd
In case x :: xs, List('a', 'b', 'c', 'd') matches as
x = 'a' and
xs = List('b', 'c', 'd')
And you are passing only xs to test method. Hence it is printing only b,c,d as expected.

Answer 3

In the match expression, "x :: xs" matches Lists with a first element, which will be stored as "x", and a tail which will be stored as "xs". Then, you print the contents of "xs", so are passing only the tail of the list to test. It will match any list with a head element and a tail, including a -> Nil, a list of one element. In that example, "a" will be stored in x, and Nil will be stored in xs.
This will work if you try:
simplelist match {
case Nil => log.info("empty")
case x => test(x)
}
This will try to match the empty list, then match anything else and refer to it as x, and print the results. Because simplelist is definitely a List (and assuming you did not use null), the x will refer to a non-empty list, and you can call test(x).
It would also work to do
simplelist match {
case x :: xs => test (simplelist)
case Nil => log.info("empty")
}
which would match any list with a first element and a tail, but then pass simplelist to test.