Scala中BigInt与Byte数组的内存和性能比较

Question

I have to use a type that can hole Ipv4 and Ipv6 addresses in memory efficient manner (in scala). As well as they should be performant. The two options I see are, using the scala BigInt type or a byte array. What the memory/perf hit in both cases?

Answer 1

BigInteger in Java takes 5 * 4 bytes for 4 int fields plus int array. BigInt from Scala is just wrapper over BigInteger so it would be similar. So using Byte array would definitely take less place.

I might also consider using type aliases with companion object and implicit extensions to keep type safety and rich API without additional overhead (it will still the same amount of space as Array[Byte] .

trait IP

type IPv4 = Array[Byte] with IP //adding "with IP" would make sure compiler won't accept plain Array[Byte] when type IPv4 is needed

type IPv6 = Array[Byte] with IP

object IPv4 { //create similar for IPv6

  def apply(ip: String) = {
    ip.split("\\.").map(_.toByte).asInstanceOf[IPv4] //you can create IPv4("127.0.0.1")
  }

  object Implicits {
    implicit class RichIPv4(ip: IPv4) {
        def show(): String = ip.map(_.toString).mkString(".") //add method to show as string
      def verify(): Boolean = ??? //additional methods
    }
  }

}

import IPV4.Implicits._

def printIP(ip: IPv4) = println(ip.show) //Will only accept arrays created by IPv4.apply

printIP(IPv4("127.0.0.1")) //ok
printIP(Array[Byte](127,0,0,1)) //won't compile

Alternatively, you should take a look at the great library scala-newtype , which does similar things but without additional boilerplate.