CODEWITHSUNDEEP
scalatypesarguments
O.Chounry
O.Chounry

Reputation: 322

Handling long type argument in Scala

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

  }

  def start(until:Long){

    var all_prime_fac:Array[Int] = Array()
    var i = 2

(compile:compileIncremental) Compilation failed

integer number too large

Even though I changed the arg type to Long, it's still not fixed.

Upvotes: 1

Views: 2962

Answers (3)

ashburshui
ashburshui

Reputation: 1410

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

Upvotes: 0

puhlen
puhlen

Reputation: 8529

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

start(600851475143L)

Upvotes: 1

Jeffrey Chung
Jeffrey Chung

Reputation: 19507

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

start(600851475143L)
               // ^

Upvotes: 2

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