2
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Would you consider this to be idiomatic Scala? If not, what would you do to improve it?

def fibonaccis(max: Int): List[Int] = { fibonaccis(max, 2, 1) }
private def fibonaccis(max: Int, prev: Int, prevPrev: Int): List[Int] = prev >= max match {
  case true => List[Int](prevPrev)
  case false => prevPrev :: fibonaccis(max, prev + prevPrev, prev)
}
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1
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Naming the parameters prev and prevPrev is weird. Why not current and prev?

You shouldn't need the curly braces here for a function that consists of a single expression:

def fibonaccis(max: Int): List[Int] = fibonaccis(max, 2, 1)

Conventionally, the Fibonacci sequence is said to start like

$$1, 1, 2, 3, 5, 8, \ldots$$

or

$$0, 1, 1, 2, 3, 5, \ldots$$

Your fibonaccis(13) would produce List(1, 2, 3, 5, 8), which I would consider to be missing the first element.


The Fibonacci sequence is an infinite sequence. It would be a shame to limit it to a max value. The most idiomatic way to model it in Scala would be using an infinite lazy stream, just as suggested in the documentation:

def fibFrom(a: Int, b: Int): Stream[Int] = a #:: fibFrom(b, a + b)
val fibonaccis = fibFrom(0, 1)

Alternative implementation:

val fibonaccis: Stream[Int] = 0 #:: 1 #:: fibonaccis.zip(fibonaccis.tail).map(
    n => n._1 + n._2
)

Then, you could do fibonaccis.takeWhile(_ < 13).toList to obtain List(0, 1, 1, 2, 3, 5, 8).

| improve this answer | |
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  • \$\begingroup\$ I'm not crazy about prev and prevPrev. Not current, though, since it's not the current value, it's the previous. Awkward either way. I like the stream approach - thought the answer might include that, but I haven't gotten that far yet. When is the next value generated? Is it just in time? \$\endgroup\$ – Don Branson Nov 3 '16 at 0:34
  • 1
    \$\begingroup\$ Streams are lazy. Each successive element is computed as needed, and memoized for the future. \$\endgroup\$ – 200_success Nov 3 '16 at 0:36

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