Best way to build and return a list from a method

Question

Is there a Scala construct for building and returning a list like so?

def getOutput(isValidInput: Boolean): List[Output] =
  if (isValidInput) {
    yield Output(1) // Pseudo-code... I know yield is not intended for this!
    yield Output(2)
  }

Instead of...

def getOutput(isValidInput: Boolean): List[Output] =
  if (isValidInput)
    List(Output(1), Output(2))
  else
    Nil

In C# the use of 'yield' allows you to return lazy-evaluated collections - is there something similar in Scala?

The question is not clear to me if you want lazyness or conciseness of code. The List won't be created in your example unless the boolean is true. — sksamuel, Jun 14 '13 at 20:34
What you want is a form of continuation. This is somewhat supported, but probably less readable than other options. See http://www.scala-lang.org/node/2096 — gzm0, Jun 14 '13 at 20:49

score 1 · Accepted Answer · edited May 23 '17 at 12:29

If you want a lazy sequence, use a Stream:

case class Output(value: Int)

def getOutput(isValidInput: Boolean):Stream[Output] = getOutput(isValidInput, 1)

def getOutput(isValidInput: Boolean, index: Int):Stream[Output] =
  if (isValidInput && index < 3) {
    println("Index: " + index)
    Output(index) #:: getOutput(isValidInput, index+1)
  } else {
    Stream.empty[Output]
  }

println("Calling getOutput")
val result: Stream[Output] = getOutput(true)
println("Finished getOutput")

result foreach println

This results in:

Calling getOutput
Index: 1
Finished getOutput
Output(1)
Index: 2
Output(2)

If you want to keep the return type as List[Output], using yield is a valid approach:

def getOutput(isValidInput: Boolean):List[Output] =
  if (isValidInput) {
    (for (i <- 1 until 3) yield Output(i))(collection.breakOut)
  } else {
    List.empty[Output]
  }

Also, using a Vector is often preferable over a List.

Related:

Best way to build and return a list from a method

1 Answers1