Scala downwards or decreasing for loop?

ScalaIteratorLoopsFor Loop

Scala Problem Overview


In Scala, you often use an iterator to do a for loop in an increasing order like:

for(i <- 1 to 10){ code }

How would you do it so it goes from 10 to 1? I guess 10 to 1 gives an empty iterator (like usual range mathematics)?

I made a Scala script which solves it by calling reverse on the iterator, but it's not nice in my opinion, is the following the way to go?

def nBeers(n:Int) = n match {

    case 0 => ("No more bottles of beer on the wall, no more bottles of beer." +
               "\nGo to the store and buy some more, " +
               "99 bottles of beer on the wall.\n")

    case _ => (n + " bottles of beer on the wall, " + n +
               " bottles of beer.\n" +
               "Take one down and pass it around, " +
              (if((n-1)==0)
                   "no more"
               else
                   (n-1)) +
                   " bottles of beer on the wall.\n")
}

for(b <- (0 to 99).reverse)
    println(nBeers(b))

Scala Solutions


Solution 1 - Scala

scala> 10 to 1 by -1
res1: scala.collection.immutable.Range = Range(10, 9, 8, 7, 6, 5, 4, 3, 2, 1)

Solution 2 - Scala

The answer from @Randall is good as gold, but for sake of completion I wanted to add a couple of variations:

scala> for (i <- (1 to 10).reverse) {code} //Will count in reverse.

scala> for (i <- 10 to(1,-1)) {code} //Same as with "by", just uglier.

Solution 3 - Scala

Scala provides many ways to work on downwards in loop.

1st Solution: with "to" and "by"

//It will print 10 to 0. Here by -1 means it will decremented by -1.     
for(i <- 10 to 0 by -1){
    println(i)
}

2nd Solution: With "to" and "reverse"

for(i <- (0 to 10).reverse){
    println(i)
}

3rd Solution: with "to" only

//Here (0,-1) means the loop will execute till value 0 and decremented by -1.
for(i <- 10 to (0,-1)){
    println(i)
}

Solution 4 - Scala

Having programmed in Pascal, I find this definition nice to use:

implicit class RichInt(val value: Int) extends AnyVal {
  def downto (n: Int) = value to n by -1
  def downtil (n: Int) = value until n by -1
}

Used this way:

for (i <- 10 downto 0) println(i)

Solution 5 - Scala

You can use Range class:

val r1 = new Range(10, 0, -1)
for {
  i <- r1
} println(i)

Solution 6 - Scala

You can use : for (i <- 0 to 10 reverse) println(i)

Solution 7 - Scala

for (i <- 10 to (0,-1))

The loop will execute till the value == 0, decremented each time by -1.

Attributions

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionFelixView Question on Stackoverflow
Solution 1 - ScalaRandall SchulzView Answer on Stackoverflow
Solution 2 - ScalaChirloView Answer on Stackoverflow
Solution 3 - ScalaDipak ShawView Answer on Stackoverflow
Solution 4 - ScalaLionel ParreauxView Answer on Stackoverflow
Solution 5 - ScalaKaaPexView Answer on Stackoverflow
Solution 6 - ScalaJonnyView Answer on Stackoverflow
Solution 7 - Scalan1culaView Answer on Stackoverflow