How do I break out of a loop in Scala

Question

How do I break out a loop   var largest 0 for i lt -999 to 1 by -1        for  j lt -i to 1 by -1            val product i j         if  largest gt product                 I want to break out here         else            if product toString equals product toString reverse                 largest largest max product           How do I turn nested for loops into tail recursion   From Scala Talk at FOSDEM 2009 http   www slideshare net Odersky fosdem-2009-1013261 on the 22nd page      Break and continue   Scala does not have them  Why    They are a bit imperative  better use many smaller functions   Issue how to interact with closures    They are not needed    What is the explanation

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Ironically the Scala break in scala util control Breaks is an exception   def break    Nothing     throw breakException     The best advice is  DO NOT use break  continue and goto  IMO they are the same  bad practice and an evil source of all kind of problems  and hot discussions  and finally  considered be harmful   Code block structured  also in this example breaks are superfluous  Our Edsger W  Dijkstra    wrote      The quality of programmers is a decreasing function of the density of go to statements in the programs they produce

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import following package import scala util control       create a Breaks object as follows val loop   new Breaks      Keep the loop inside breakable as follows loop breakable     Loop will go here for                     Break will go here    loop break           use Break module http   www tutorialspoint com scala scala break statement htm

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An approach that generates the values over a range as we iterate  up to a breaking condition  instead of generating first a whole range and then iterating over it  using Iterator   inspired in  RexKerr use of Stream   var sum   0 for   i  lt - Iterator from 1  takeWhile      gt  sum  lt  1000    sum    i

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You have three  or so  options to break out of loops   Suppose you want to sum numbers until the total is greater than 1000   You try  var sum   0 for  i  lt - 0 to 1000  sum    i   except you want to stop when  sum   1000    What to do   There are several options    1a  Use some construct that includes a conditional that you test   var sum   0  0 to 1000  iterator takeWhile     gt  sum  lt  1000  foreach i   gt  sum  i     warning--this depends on details of how the takeWhile test and the foreach are interleaved during evaluation  and probably shouldn t be used in practice      1b  Use tail recursion instead of a for loop  taking advantage of how easy it is to write a new method in Scala   var sum   0 def addTo i  Int  max  Int      sum    i  if  sum  lt  max  addTo i 1 max    addTo 0 1000     1c  Fall back to using a while loop  var sum   0 var i   0 while  i  lt   1000  amp  amp  sum  lt   1000    sum    1  i    1      2  Throw an exception   object AllDone extends Exception     var sum   0 try     for  i  lt - 0 to 1000    sum    i  if  sum gt  1000  throw AllDone     catch     case AllDone   gt       2a  In Scala 2 8  this is already pre-packaged in scala util control Breaks using syntax that looks a lot like your familiar old break from C Java   import scala util control Breaks   var sum   0 breakable   for  i  lt - 0 to 1000      sum    i   if  sum  gt   1000  break        3  Put the code into a method and use return   var sum   0 def findSum   for  i  lt - 0 to 1000    sum    i  if  sum gt  1000  return     findSum   This is intentionally made not-too-easy for at least three reasons I can think of   First  in large code blocks  it s easy to overlook  continue  and  break  statements  or to think you re breaking out of more or less than you really are  or to need to break two loops which you can t do easily anyway--so the standard usage  while handy  has its problems  and thus you should try to structure your code a different way   Second  Scala has all sorts of nestings that you probably don t even notice  so if you could break out of things  you d probably be surprised by where the code flow ended up  especially with closures    Third  most of Scala s  loops  aren t actually normal loops--they re method calls that have their own loop  or they are recursion which may or may not actually be a loop--and although they act looplike  it s hard to come up with a consistent way to know what  break  and the like should do   So  to be consistent  the wiser thing to do is not to have a  break  at all   Note  There are functional equivalents of all of these where you return the value of sum rather than mutate it in place   These are more idiomatic Scala   However  the logic remains the same    return becomes return x  etc

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import scala util control    object demo brk 963       def main args  Array String               var a   0        var b   0        val numList1   List 1 2 3 4 5 6 7 8 9 10         val numList2   List 11 12 13          val outer   new Breaks    object for break       val inner   new Breaks    object for break        outer breakable    Outer Block                  for  a  lt - numList1                         println   Value of a      a                inner breakable    Inner Block                              for  b  lt - numList2                                     println   Value of b      b                      if  b    12                                             println   break-INNER                            inner break                                                        inner breakable             if  a    6                                 println   break-OUTER                     outer break                                      outer breakable           Basic method to break the loop  using Breaks class  By declaring the loop as breakable

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Below is code to break a loop in a simple way   import scala util control Breaks break  object RecurringCharacter     def main args  Array String         val str    nileshshinde        for  i  lt - 0 to str length   - 1          for  j  lt - i   1 to str length   - 1             if  str i     str j               println  First Repeted Character     str i             break         break method will exit the loop with an Exception  Exception in thread  main  scala util control BreakControl

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This has changed in Scala 2 8 which has a mechanism for using breaks  You can now do the following   import scala util control Breaks   var largest   0    pass a function to the breakable method breakable        for  i lt -999 to 1  by -1  j  lt - i to 1 by -1            val product   i   j         if  largest  gt  product                break     BREAK                     else if  product toString equals product toString reverse                 largest   largest max product

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Close to your solution would be this    var largest   0 for  i  lt - 999 to 1 by -1    j  lt - i to 1 by -1    product   i   j    if  largest  lt   product  amp  amp  product toString reverse equals  product toString reverse reverse        largest   product  println  largest    The j-iteration is made without a new scope  and the product-generation as well as the condition are done in the for-statement  not a good expression - I don t find a better one   The condition is reversed which is pretty fast for that problem size - maybe you gain something with a break for larger loops    String reverse implicitly converts to RichString  which is why I do 2 extra reverses     A more mathematical approach might be more elegant

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Clever use of find method for collection will do the trick for you    var largest   0 lazy val ij     for  i  lt - 999 to 1 by -1  j  lt - i to 1 by -1  yield  i  j   val largest ij   ij find   case i j    gt    val product   i   j   if  product toString    product toString reverse      largest   largest max product   largest  gt  product    println largest ij get  println largest

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Just use a while loop   var  i  sum     0  0  while  sum  lt  1000      sum    i   i    1

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Since there is no break in Scala yet  you could try to solve this problem with using a return-statement  Therefore you need to put your inner loop into a function  otherwise the return would skip the whole loop   Scala 2 8 however includes a way to break  http   www scala-lang org api rc scala util control Breaks html

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The third-party breakable package is one possible alternative  https   github com erikerlandson breakable  Example code   scala gt  import com manyangled breakable   import com manyangled breakable    scala gt  val bkb2   for             x  xLab   lt - Stream from 0  breakable      create breakable sequence with a method           y  yLab   lt - breakable Stream from 0       create with a function          if  x   2    1  continue xLab              continue to next in outer  x  loop          if  y   2    0  continue yLab              continue to next in inner  y  loop          if  x  gt  10  break xLab                     break the outer  x  loop          if  y  gt  x  break yLab                      break the inner  y  loop          yield  x  y  bkb2  com manyangled breakable Breakable  Int  Int     com manyangled breakable Breakable 34dc53d2  scala gt  bkb2 toVector res0  Vector  Int  Int     Vector  2 1    4 1    4 3    6 1    6 3    6 5    8 1    8 3    8 5    8 7    10 1    10 3    10 5    10 7    10 9

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I got a situation like the code below   for id lt -0 to 99        try         var symbol   ctx read    stocks     id      symbol   toString       var name   ctx read    stocks     id      name   toString       stocklist symbol    name      catch         case ex  com jayway jsonpath PathNotFoundException  gt  break              I am using a java lib and the mechanism is that ctx read throw a Exception when it can find nothing  I was trapped in the situation that  I have to break the loop when a Exception was thrown  but scala util control Breaks break using Exception to break the loop  and it was in the catch block thus it was caught   I got ugly way to solve this  do the loop for the first time and get the count of the real length   and use it for the second loop   take out break from Scala is not that good when you are using some java libs

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To add Rex Kerr answer another way     1c  You can also use a guard in your loop    var sum   0  for  i  lt - 0 to 1000   if sum lt 1000  sum    i

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Here is a tail recursive version  Compared to the for-comprehensions it is a bit cryptic  admittedly  but I d say its functional     def run start Int         tailrec   def tr i Int  largest Int  Int   tr1 i  i  largest  match       case x if i  gt  1   gt  tr i-1  x      case     gt  largest         tailrec   def tr1 i Int j Int  largest Int  Int   i j match       case x if x  lt  largest    j  lt  2   gt  largest     case x if x toString equals x toString reverse    gt  tr1 i  j-1  x      case     gt  tr1 i  j-1  largest         tr start  0      As you can see  the tr function is the counterpart of the outer for-comprehensions  and tr1 of the inner one  You re welcome if you know a way to optimize my version

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I don t know how much Scala style has changed in the past 9 years  but I found it interesting that most of the existing answers use vars  or hard to read recursion  The key to exiting early is to use a lazy collection to generate your possible candidates  then check for the condition separately  To generate the products   val products   for     i  lt -  999 to 1 by -1  view   j  lt -  i to 1 by -1  view   yield  i j    Then to find the first palindrome from that view without generating every combination   val palindromes   products filter  p   gt  p toString    p toString reverse  palindromes head   To find the largest palindrome  although the laziness doesn t buy you much because you have to check the entire list anyway    palindromes max   Your original code is actually checking for the first palindrome that is larger than a subsequent product  which is the same as checking for the first palindrome except in a weird boundary condition which I don t think you intended  The products are not strictly monotonically decreasing  For example  998 998 is greater than 999 997  but appears much later in the loops   Anyway  the advantage of the separated lazy generation and condition check is you write it pretty much like it is using the entire list  but it only generates as much as you need  You sort of get the best of both worlds

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It is never a good idea to break out of a for-loop  If you are using a for-loop it means that you know how many times you want to iterate  Use a while-loop with 2 conditions    for example  var done   false while  i  lt   length  amp  amp   done      if  sum  gt  1000         done   true

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I am new to Scala  but how about this to avoid throwing exceptions and repeating methods   object awhile   def apply condition       gt  Boolean  action       gt  breakwhen   Unit         while  condition              action   match               case breakwhen true       gt  return               case                      gt                         case class breakwhen break Boolean     use it like this   var i   0 awhile      gt  i  lt  20       gt        i   i   1     breakwhen i    5      println i    if you don   t want to break   awhile      gt  i  lt  20       gt        i   i   1     breakwhen false

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Simply We can do in scala is scala gt  import util control Breaks    scala gt  object TestBreak          def main args   Array String              breakable              for  i  lt - 1 to 10                 println i               if  i    5                 break                  output   scala gt  TestBreak main Array    1 2 3 4 5

[scala] How do I break out of a loop in Scala?

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