128

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))
Peter Mortensen
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Felix
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7 Answers7

253
scala> 10 to 1 by -1
res1: scala.collection.immutable.Range = Range(10, 9, 8, 7, 6, 5, 4, 3, 2, 1)
Randall Schulz
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    @Felix: You're welcome. I should have also pointed out that there is also `until` that you can use in place of `to` to exclude the right-hand end-point from the range. The left-hand endpoint is always included. – Randall Schulz Apr 13 '10 at 14:26
  • I already knew about the until, the until is also a function on Integers, however, "by" must be a function on the range/iterator whatever is returned from the "to" and "until" functions. Thanks anyway :) – Felix Apr 15 '10 at 12:59
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    Randall's answer is best, but I think `Range.inclusive(10, 1, -1)` deserves mention. – john sullivan Jul 15 '13 at 01:35
44

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.
Peter Mortensen
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Chirlo
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16

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)
}
Dipak Shaw
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6

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)
Lionel Parreaux
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  • Thank your for the answer. I'm having trouble using this solution. Here is my stacktrace: `Error:(57, 17) value class may not be a member of another class implicit class RichInt(val value: Int) extends AnyVal { ^` – robert Dec 22 '15 at 20:40
  • As the error message (not a stack trace) suggests, you cannot define the value class inside of another class. Either define it outside of it, oike in an object, or remove the `extends AnyVal` part (which only serves to remove some overhead). – Lionel Parreaux Jan 05 '16 at 14:40
2

You can use Range class:

val r1 = new Range(10, 0, -1)
for {
  i <- r1
} println(i)
KaaPex
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2

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

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

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

n1cula
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