Handling long type argument in Scala

Question

object LPrimeFactor {
  def main(arg:Array[String]):Unit = {
  start(13195)
  start(600851475143)

  }

  def start(until:Long){

    var all_prime_fac:Array[Int] = Arra         


        

Answer 1

Pass the argument as a Long (notice the L at the end of the number):

start(600851475143L)
               // ^

Answer 2

Please remember literals values, if you has not any type direct suffix, the compiler try to get your numeric type values, such as 600851475143 as type Int, which is 32-bit length, two complement representation

MIN_VALUE = -2147483648(- 2 ^ 31)  
MAX_VALUE = 2147483647(2 ^ 31 - 1)

So please add right suffix on the literal value, as 600851475143L

Answer 3

To create a Long literal you must add L to the end of it.

start(600851475143L)