CODEWITHSUNDEEP
listscalaprintingrecursionfunctional-programming
koenieee
koenieee

Reputation: 35

print list functional using scala

I have a problem: I have a list in scala like this (List("Entry1", "Entry2", "Entry3")) and I want to print this list functional. So I don't want to use any kind of loops.

I know it is possible using some kind of this code:

def test(input:List[_])
input match
{
case //and what now?
case _ => //recursion I guess?

}

I want to print this list using this method, every element on a new line.

Could someone help me please?

Thanks!

Upvotes: 0

Views: 6158

Answers (3)

dhg
dhg

Reputation: 52681

The standard way would be:

val xs = List("Entry1", "Entry2", "Entry3")
xs.foreach(println)

Or, if you want the index with it:

xs.zipWithIndex.foreach { case (x,i) => println(i + ": " + x) }

But I take it from your question that this is an exercise in writing a recursive function, so knowing the built-in way doesn't really help you.

So, if you want to do it yourself, recursively, without the built-in foreach method, try this:

@tailrec
def printList[T](list: List[T]) {
  list match {
    case head :: tail =>
      println(head)
      printList(tail)
    case Nil =>
  }
}

printList(List("Entry1", "Entry2", "Entry3"))

UPDATE: Regarding your comment about having the list index, try this:

def printList[T](list: List[T]) {
  @tailrec
  def inner(list: List[T], i: Int) {
    list match {
      case head :: tail =>
        println(i + ": " + head)
        inner(tail, i + 1)
      case Nil =>
    }
  }
  inner(list, 0)
}

Upvotes: 8

sbenitezb
sbenitezb

Reputation: 535

List("LALA", "PEPE", "JOJO").map(println)

Upvotes: 0

jwinandy
jwinandy

Reputation: 1749

Another solution using the list API :

List("Entry1", "Entry2", "Entry3").zipWithIndex.foreach(t => println(t._2 + ":" + t._1))

Upvotes: 1

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