I solved this prime number verifier problem:import scala.io.StdInobject Main extends App {  val testCases = StdIn.readInt  for (i <- 0 until testCases) {    val limits = StdIn.readLine.split( )    val m: Int = limits(0).toInt    val n: Int = limits(1).toInt    for (j <- m to n) {      var prime = false      if (isPrime(j))        println(j)    }    println()  }  def isPrime(number: Int) : Boolean = {    if (number == 1)       false    else {      for (h <- 2 to Math.sqrt(number).asInstanceOf[Int]) {        if (number % h == 0 || number == 1) {          return false        }      }      true    }  }}I wrote the code this way because Scala doesn't have break statements, and return statements aren't very functional either. How do I go about writing this code in a more functional and idiomatic manner?
Prime number verifier
functional programming;primes;scala
How about this?def isPrime(n:Int):Boolean = n match {  case 1 => false  case 2 => true  case 3 => true  case 5 => true  case 7 => true  case n if (n%2==0 || n%3==0 || n%5==0 || n%7==0) => false  case _ => (Stream.iterate(3, Math.sqrt(n).toInt+1){_ + 2} dropWhile((x:Int) => n%x != 0)).isEmpty}The code is straightforward. We know that the prime numbers below 10 are 2, 3, 5, 7. If a given number is the multiple of 2, 3, 5, or 7, is't not a prime. In other case (the last case branch) it generates stream from 3 to ceil(sqrt(n)) with step 2. We know that even numbers can't be a prime. And check that the argument is a multiple of any number in the stream. Basically the same logic with the original code. The reason I use Stream is to use lazy evaluation. With lazy evaluation, we can avoid unnecessary computation.