CODEWITHSUNDEEP
scalaipip-addressipv4
Garren S
Garren S

Reputation: 5782

How to Convert IPv4 Addresses to/from Long in Scala

I was looking for a basic utility with 2 functions to convert IPv4 Addresses to/from Long in Scala, such as "10.10.10.10" to its Long representation of 168430090 and back. A basic utility such as this exists in many languages (such as python), but appears to require re-writing the same code for everyone for the JVM.

What is the recommended approach on unifying IPv4ToLong and LongToIPv4 functions?

Upvotes: 4

Views: 5160

Answers (7)

papahugo
papahugo

Reputation: 51

I like @jwvh's comment on ipv4ToLong. As to longToIpv4, how about just simply:

def longToIpv4(v:Long):String = (for (i <- 0 to 3) yield (v >> (i * 8)) & 0x000000FF ).reverse.mkString(".")

Upvotes: 0

Pankaj
Pankaj

Reputation: 927

Try the ipaddr scala library. Create an IpAddress and get its long value like this:

val ip1: IpAddress = IpAddress("192.168.0.1")
val ip1Long = ip1.numerical // returns 3232235521L

Upvotes: 4

david hanley
david hanley

Reputation: 31

This is pretty straightforward for ipv4:

def ipToLong(ip:String) = ip.split("\\\\.").foldLeft(0L)((c,n)=>c*256+n.toLong)

def longToIP(ip:Long) = (for(a<-3 to 0 by -1) yield ((ip>>(a*8))&0xff).toString).mkString(".")

Upvotes: 3

tcat
tcat

Reputation: 361

Combining the ideas from leifbatterman and Elesin Olalekan Fuad and avoiding multiplication and power operations:

def ipv4ToLong(ip: String): Option[Long] = Try(
  ip.split('.').ensuring(_.length == 4)
    .map(_.toLong).ensuring(_.forall(x => x >= 0 && x < 256))
    .reverse.zip(List(0,8,16,24)).map(xi => xi._1 << xi._2).sum
).toOption

To convert Long to String in dotted format:

def longToipv4(ip: Long): Option[String] = if ( ip >= 0 && ip <= 4294967295L) {
  Some(List(0x000000ff, 0x0000ff00, 0x00ff0000, 0xff000000).zip(List(0,8,16,24))
    .map(mi => ((mi._1 & ip) >> mi._2)).reverse
    .map(_.toString).mkString("."))
} else None

Upvotes: 8

leifbattermann
leifbattermann

Reputation: 632

Adding to Elesin Olalekan Fuad's answer it can be made a little more robust like this:

def ipToLong(ip: String): Option[Long] = {
  Try(ip.split('.').ensuring(_.length == 4)
    .map(_.toLong).ensuring(_.forall(x => x >= 0 && x < 256))
    .zip(Array(256L * 256L * 256L, 256L * 256L, 256L, 1L))
    .map { case (x, y) => x * y }
    .sum).toOption
}

def longToIp(ip: Long): Option[String] = {
  if (ip >= 0 && ip <= 4294967295L)
    Some((0 until 4)
      .map(a => ip / math.pow(256, a).floor.toInt % 256)
      .reverse.mkString("."))
  else
    None
}

Upvotes: 0

Elesin Olalekan Fuad
Elesin Olalekan Fuad

Reputation: 3313

I have a GitHub gist that solves this. The gist contains code that converts from IP to Long likewise the reverse. Visit https://gist.github.com/OElesin/f0f2c69530a315177b9e0227a140f9c1

Here is the code:

def ipToLong(ipAddress: String): Long = {
   ipAddress.split("\\.").reverse.zipWithIndex.map(a=>a._1.toInt*math.pow(256,a._2).toLong).sum
}

def longToIP(long: Long): String = {
   (0 until 4).map(a=>long / math.pow(256, a).floor.toInt % 256).reverse.mkString(".")
}

Enjoy

Upvotes: 2

Garren S
Garren S

Reputation: 5782

import java.net.InetAddress
def IPv4ToLong(dottedIP: String): Long = {
  val addrArray: Array[String] = dottedIP.split("\\.")
  var num: Long = 0
  var i: Int = 0
  while (i < addrArray.length) {
    val power: Int = 3 - i
    num = num + ((addrArray(i).toInt % 256) * Math.pow(256, power)).toLong
    i += 1
  }
  num
}

def LongToIPv4 (ip : Long) : String = {
  val bytes: Array[Byte] = new Array[Byte](4)
  bytes(0) = ((ip & 0xff000000) >> 24).toByte
  bytes(1) = ((ip & 0x00ff0000) >> 16).toByte
  bytes(2) = ((ip & 0x0000ff00) >> 8).toByte
  bytes(3) = (ip & 0x000000ff).toByte
  InetAddress.getByAddress(bytes).getHostAddress()
}

scala> IPv4ToLong("10.10.10.10")
res0: Long = 168430090

scala> LongToIPv4(168430090L)
res1: String = 10.10.10.10

Upvotes: 4

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