I have a method in Scala that checks if a game board is full or not. To speed this method up, I return inside the nested for loop to break out of both of them and stop execution. How would I achieve this in Scala in an idiomatic way?

def isFull: Boolean = {
  for (x <- getGrid.indices) {
    for (y <- getGrid(x).indices) {
      if (getGrid(x)(y).getState.isInstanceOf[Undecided]) {
        return false

You can apply forall to most containers to test if every element in the container satisfies some property. forall will only evaluate x + 1 if x is true. This type of evaluation is known as short-circuit evaluation. So basically we can use this property to perform the same behavior as the if-then statement in the original code.

Notice also that I've used pattern matching to get rid of the isInstanceOf[T].

def isFull: Boolean =
  getGrid.indices.forall(x =>
    getGrid(x).indices.forall(y =>
      getGrid(x)(y).getState match {
        case xyState: Undecided => False
        case _ => True
| improve this answer | |
  • \$\begingroup\$ How does this compare speed-wise to my solution? Scala 2.10 if it matters. \$\endgroup\$ – Martijn Apr 11 '16 at 11:39
  • \$\begingroup\$ Whoops I forgot to mention it, but yes, forall will return as soon as it encounters a false. \$\endgroup\$ – t. fochtman Apr 11 '16 at 18:08
  • \$\begingroup\$ So they are (or should be) equally fast. \$\endgroup\$ – t. fochtman Apr 11 '16 at 18:09
  • \$\begingroup\$ Could you elaborate on that pattern matching statement a bit? Undecided is a sealed trait and I want to check if it is a sub class or not. \$\endgroup\$ – Martijn Apr 11 '16 at 18:40
  • \$\begingroup\$ I've updated my answer \$\endgroup\$ – t. fochtman Apr 13 '16 at 8:17

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