Get all IP addresses from a given IP address and subnet mask

6

In Java, I need to get list of all IP Addresses contained by a given IP network.

For e.g. let the netowork be: 192.168.5.0/24 then the output will be (192.168.5.0 ... 192.168.5.255).

I could think of the following way but it looks dirty, is there any elegant way? There is no function for the same in InetAddress class.

  1. Get Network Ip from the input Ip and subnet mask.

    mask = (long)(0xffffffff) << (32-subnetMask);
    Long netIp = getLongfromIp(Inputip)& mask;
    

The function 'getLongfromIp' contains code from - How to convert string (IP numbers) to Integer in Java

  1. get the number of hosts by Subnet Mask

    maxRange = (long)0x1<<(32-subnetMask);

  2. Get address of all hopes by adding i for i in (0 .. maxRange) in the netIp

  3. Convert the ip from above step to octet string.

Ps: I am sure the the IP Addresses will be in IPV4 only.

java
network-programming
ip
asked on Stack Overflow Nov 4, 2014 by Mangat Rai Modi • edited May 23, 2017 by Community

4 Answers

5

Answering my own question, solution is to use Apache commons.net library

import org.apache.commons.net.util.*;

SubnetUtils utils = new SubnetUtils("192.168.1.0/24");
String[] allIps = utils.getInfo().getAllAddresses();
//appIps will contain all the ip address in the subnet

Read more: Class SubnetUtils.SubnetInfo

answered on Stack Overflow Nov 5, 2014 by Mangat Rai Modi • edited May 17, 2017 by Mangat Rai Modi
1

To include NetworkAddress and BroadcastAddress

import org.apache.commons.net.util.*;

 SubnetUtils utils = new SubnetUtils("192.168.1.0/28");
       utils.setInclusiveHostCount(true);

       String[] allIps = utils.getInfo().getAllAddresses();
answered on Stack Overflow Oct 14, 2015 by Labeo
1

The IPAddress Java library supports both IPv4 and IPv6 subnets in a polymorphic manner. Disclaimer: I am the project manager.

Here is sample code to list the addresses for an IPv4 or Ipv6 subnet transparently. Subnets can get quite large, especially with IPv6, and it is not wise to attempt to iterate through a large subnet, so the code for iterateEdges shows how to iterate through just the beginning and ending addresses in the subnet.

show("192.168.10.0/24");
show("2001:db8:abcd:0012::/64");

static void show(String subnet) throws AddressStringException {
    IPAddressString addrString = new IPAddressString(subnet);
    IPAddress addr = addrString.toAddress();
    show(addr);
}

static void show(IPAddress subnet) {
    Integer prefixLength = subnet.getNetworkPrefixLength();
    if(prefixLength == null) {
        prefixLength = subnet.getBitCount();
    }
    IPAddress mask = subnet.getNetwork().getNetworkMask(prefixLength, false);
    BigInteger count = subnet.getCount();
    System.out.println("Subnet of size " + count + " with prefix length " + prefixLength + " and mask " + mask);
    System.out.println("Subnet ranges from " + subnet.getLower() + " to " + subnet.getUpper());
    int edgeCount = 3;
    if(count.compareTo(BigInteger.valueOf(256)) <= 0) {
        iterateAll(subnet, edgeCount);
    } else {
        iterateEdges(subnet, edgeCount);
    }
}

Iterates through entire subnet, use with caution:

static void iterateAll(IPAddress subnet, int edgeCount) {
    BigInteger count = subnet.getCount();
    BigInteger bigEdge = BigInteger.valueOf(edgeCount), currentCount = count;
    int i = 0;
    for(IPAddress addr: subnet.getIterable()) {
        currentCount = currentCount.subtract(BigInteger.ONE);
        if(i < edgeCount) {
            System.out.println(++i + ": " + addr);
        } else if(currentCount.compareTo(bigEdge) < 0) {
            System.out.println(count.subtract(currentCount) + ": " + addr);
        } else if(i == edgeCount) {
            System.out.println("...skipping...");
            i++;
        }
    }
}

Iterates through subnet edges:

static void iterateEdges(IPAddress subnet, int edgeCount) {
    for(int increment = 0; increment < edgeCount; increment++) {
        System.out.println((increment + 1) + ": " + subnet.getLower().increment(increment));
    }
    System.out.println("...skipping...");
    BigInteger count = subnet.getCount();
    for(int decrement = 1 - edgeCount; decrement <= 0; decrement++) {
        System.out.println(count.add(BigInteger.valueOf(decrement)) + ": " + subnet.getUpper().increment(decrement));
    }
}

Here is the output:

Subnet of size 256 with prefix length 24 and mask 255.255.255.0
Subnet ranges from 192.168.5.0/24 to 192.168.5.255/24
1: 192.168.5.0/24
2: 192.168.5.1/24
3: 192.168.5.2/24
...skipping...
254: 192.168.5.253/24
255: 192.168.5.254/24
256: 192.168.5.255/24

Subnet of size 18446744073709551616 with prefix length 64 and mask ffff:ffff:ffff:ffff::
Subnet ranges from 2001:db8:abcd:12::/64 to 2001:db8:abcd:12:ffff:ffff:ffff:ffff/64
1: 2001:db8:abcd:12::/64
2: 2001:db8:abcd:12::1/64
3: 2001:db8:abcd:12::2/64
...skipping...
18446744073709551614: 2001:db8:abcd:12:ffff:ffff:ffff:fffd/64
18446744073709551615: 2001:db8:abcd:12:ffff:ffff:ffff:fffe/64
18446744073709551616: 2001:db8:abcd:12:ffff:ffff:ffff:ffff/64
answered on Stack Overflow Mar 20, 2019 by Sean F • edited Oct 14, 2019 by Sean F
1

The following is the same as Sean's (really nice!) answer using https://seancfoley.github.io/IPAddress/ , it is only reducing the signal to noise ratio:

subnetToIps("192.168.10.0/28");

    public void subnetToIps(String ipOrCidr) {
        IPAddressString addrString = new IPAddressString(ipOrCidr, IPAddressString.DEFAULT_VALIDATION_OPTIONS);
        IPAddress subnet = addrString.toAddress();
        System.out.println("Subnet ranges from " + subnet.getLower() + " to " + subnet.getUpper());

        int i = 0;
        for (IPAddress addr : subnet.getIterable()) {
            System.out.println(++i + ": " + addr);
        }
    }

answered on Stack Overflow Jun 27, 2019 by Dirk Hoffmann

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