Scala return a value from a for cycle

Question

I have the following piece of code

def my_function(condition: Int=>Boolean): Int = {
  for (i <- 0 to 10)
    if (condition(i)) return i

  return -1
}

The code is simple: if some condition is met for a number between 1 and 10, return the number, else return invalid result (-1).

It works perfectly fine, but it violates some of the functional programming principles, because of the return in the for cycle. How can I refactor this method (I got unit tests, too) and remove the return statements. I suppose that I must use yield, but it seems to produce list, I just need one value.

Answer 1

This is the functional translation of your code:

def my_function(condition: Int => Boolean): Int = {
  (0 to 10).find(i => condition(i)).getOrElse(-1)
}

Or more succinctly as:

def my_function(condition: Int => Boolean): Int = {
  (0 to 10).find(condition).getOrElse(-1)
}

I've used the find method which takes, as an argument, a function that produces a Boolean. The find method returns the first item in the collection that satisfies the condition. It returns the item as an Option so that if there is no satisfying item, then the result will be None. The getOrElse method of Option will return the found result if there is one, and return -1 if there isn't.

However, you should not use "error codes" like returning -1 in Scala programming. They are bad practice. Instead your should throw an exception or return an Option[Int] such that returning None indicates no value was found. The right way to do this would be something like:

def my_function(condition: Int => Boolean): Option[Int] = {
  (0 to 10).find(condition)
}

println(my_function(_ > 5))   // Some(6)
println(my_function(_ > 11))  // None

Answer 2

You can use the native collection method

def my_function(condition: Int => Boolean): Int =
  (1 to 10).find(condition).getOrElse(-1)

Usually in scala you should avoid "error code" using an Option return

def my_function(condition: Int => Boolean) : Option[Int] =
  (1 to 10).find(condition)

Similarly you can use a for-yield comprehension

def my_function(condition: Int => Boolean): Int =
  (for (i <- 1 to 10; if condition(i)) yield i).headOption.getOrElse(-1)

or with Option

def my_function(condition: Int => Boolean): Int =
  (for (i <- 1 to 10; if condition(i)) yield i).headOption

or use recursion as @Jan suggestion