How do I break out of a loop in Scala?

Question

How do I break out a loop?

var largest=0
for(i<-999 to 1 by -1) {
    for (j<-i to 1 by -1) {
        val product=i*j
        if (largest>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?

Answer 1

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 <- 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(_ => sum < 1000).foreach(i => 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 < max) addTo(i+1,max)
}
addTo(0,1000)

(1c) Fall back to using a while loop

var sum = 0
var i = 0
while (i <= 1000 && sum <= 1000) { sum += 1; i += 1 }

(2) Throw an exception.

object AllDone extends Exception { }
var sum = 0
try {
  for (i <- 0 to 1000) { sum += i; if (sum>=1000) throw AllDone }
} catch {
  case AllDone =>
}

(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 <- 0 to 1000) {
  sum += i
  if (sum >= 1000) break
} }

(3) Put the code into a method and use return.

var sum = 0
def findSum { for (i <- 0 to 1000) { sum += i; if (sum>=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.).

Answer 2

Close to your solution would be this:

var largest = 0
for (i <- 999 to 1 by -1;
  j <- i to 1 by -1;
  product = i * j;
  if (largest <= product && 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.

Answer 3

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 <- 0 to str.length() - 1) {
      for (j <- 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"

        }
      }
    }
  }
}

Answer 4

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<-999 to 1  by -1; j <- i to 1 by -1) {
        val product = i * j
        if (largest > product) {
            break  // BREAK!!
        }
        else if (product.toString.equals(product.toString.reverse)) {
            largest = largest max product
        }
    }
}

Answer 5

I got a situation like the code below

 for(id<-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=>{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.

Answer 6

Clever use of find method for collection will do the trick for you.

var largest = 0
lazy val ij =
  for (i <- 999 to 1 by -1; j <- i to 1 by -1) yield (i, j)

val largest_ij = ij.find { case(i,j) =>
  val product = i * j
  if (product.toString == product.toString.reverse)
    largest = largest max product
  largest > product
}

println(largest_ij.get)
println(largest)