Scala downwards or decreasing for loop?

Question

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))

Randall Schulz · Accepted Answer · 2010-04-11 15:37:24Z

255

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

answered Apr 11, 2010 at 15:37

Randall Schulz

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3 Comments

Randall Schulz Over a year ago

@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.

Felix Over a year ago

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 :)

john sullivan Over a year ago

Randall's answer is best, but I think Range.inclusive(10, 1, -1) deserves mention.

Peter Mortensen · Accepted Answer · 2013-05-03 17:19:47Z

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.

edited May 3, 2013 at 17:19

Peter Mortensen

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answered Apr 13, 2012 at 20:11

Chirlo

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3 Comments

om-nom-nom Over a year ago

+1 for first one one, but second one is evil -- less readable than by and IMO shouldn't be used under any circumstances

Zaheer Over a year ago

Second one is evil but builds intuition on what's available

Eric Jiang Over a year ago

Thanks, reverse is nice for exclusive start and inclusive end e.g. 1 until x reverse instead of x - 1 to 1 by -1.

Dipak Shaw · Accepted Answer · 2016-07-29 08:28:42Z

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)
}

edited Jul 29, 2016 at 8:28

answered Jul 29, 2016 at 7:54

Dipak Shaw

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Comments

Lionel Parreaux · Accepted Answer · 2014-05-13 21:54:36Z

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)

answered May 13, 2014 at 21:54

Lionel Parreaux

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2 Comments

robert Over a year ago

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 {                ^

Lionel Parreaux Over a year ago

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).

KaaPex · Accepted Answer · 2018-03-13 12:11:51Z

2

You can use Range class:

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

answered Mar 13, 2018 at 12:11

KaaPex

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Comments

Jonny · Accepted Answer · 2019-06-20 14:53:52Z

2

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

answered Jun 20, 2019 at 14:53

Jonny

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Comments

n1cula · Accepted Answer · 2020-04-01 09:36:09Z

1

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

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

edited Apr 1, 2020 at 9:36

answered Mar 20, 2020 at 14:06

n1cula

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