How does a for loop with multiple generators work in Scala?

Question

I'm new to Scala and coming at this from an imperative background but I'd like to understand it in functional terms. I'm very confused about the behavior of the following code snippets.

val res = for {
  i <- 1 to 2
  j <- 1 to 2
} yield (i, j)

println(s"res is $res")

This code will print res is Vector((1,1), (1,2), (2,1), (2,2)) as expected. However modifying the above slightly to

val res = for {
  i <- 1 to 2
  j <- i to 0
} yield (i, j)

println(s"res is $res")

prints res is Vector()

Why doesn't the second version of the loop yield Vector((1,1), (1,0), (2,2), (2,1), (2,0))?

This behavior has a significant impact on my ability to loop over 2D matrices using indices in Scala. In general, how could one loop over ONLY the upper triangular portion of a matrix in an idiomatic way while keeping track of the row and column indices?

also related - Decreasing for loop in Scala?

prayagupadhyay
– prayagupadhyay

2018-09-30 05:21:21 +00:00
Commented Sep 30, 2018 at 5:21 — prayagupadhyay
– prayagupadhyay, Commented Sep 30, 2018 at 5:21

jwvh · Accepted Answer · 2018-09-30 05:18:03Z

6

Let the REPL tell you.

scala> 1 to 2
res0: scala.collection.immutable.Range.Inclusive = Range 1 to 2

scala> 1 to 0
res1: scala.collection.immutable.Range.Inclusive = empty Range 1 to 0
                                                   ^^^^^

A Range won't go downward unless you tell it to.

val res = for {
  i <- 1 to 2
  j <- i to 0 by -1
} yield (i, j)
//res: IndexedSeq[(Int, Int)] = Vector((1,1), (1,0), (2,2), (2,1), (2,0))

answered Sep 30, 2018 at 5:18

jwvh

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1 Comment

milkbox Over a year ago

Tyvm the sidebar helped me realize this as well. For those who need more info: On ranges: stackoverflow.com/questions/2617513/… On Upper-triangular idioms: stackoverflow.com/questions/18707121/…

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