Question

I’m a little confused regarding pattern matching on a list in Scala.

For example.

    val simplelist: List[Char] = List('a', 'b', 'c', 'd')

    //> simplelist  : List[Char] = List(a, b, c, d)

    def simple_fun(list: List[Char]) = list match {
           case (x:Char) :: (y:List[Char]) => println(x)
           case _ => Nil
     }                                                 
   //> simple_fun: (list: List[Char])Any

   simple_fun(simplelist)                            

   //> a
   //| res0: Any = ()

This currently prints only one line of output. Should it not run/pattern match on each element of the List ?

EDIT: I fixed the compile errors and copied the output from the REPL.

Answer 1

Unless you are repeatedly calling simple_fun in some way, what you have there will pattern match the first element and nothing more. To get it to match the whole list, you can get simple_fun to call itself recursively, like this:

val simplelist: List[Char] = List('a', 'b', 'c', 'd')

def simple_fun(list: List[Char]): List[Nothing] = list match {
  case x :: xs => {
    println(x)
    simple_fun(xs)
  }
  case _ => Nil 
}

Note I’ve also left out some of the types as the Scala compiler can infer them, leaving you with less cluttered, more readable code.

As a small side-note, calling println repeatedly inside the function like that is not particularly functional – as it is all about side effects. A more idiomatic approach would be to have the function construct a string describing the list, which is then output with a single call to println – so the side-effects are kept in a single well-defined place. Something like this would be one approach:

def simple_fun(list: List[Char]):String = list match {
  case x :: xs => x.toString + simple_fun(xs)
  case Nil => ""
}

println(simple_fun(simple_list))