Calculating consecutive IP addresses

Question: consecutive IP addresses must be calculated.

For example, if the starting IP address is 192.168.2.2 and the total number of IP addresses is 3, all required IP addresses are 192.168.2.2, 192.168.2.3, and 192.168.2.4. For another example, if the starting IP address is 192.168.2.253 and the total number of IP addresses is 5, all the required IP addresses are 192.168.2.253, 192.168.2.254, 192.168.2.255, 192.168.3.0, and 192.168.3.1.

This can be done according to the traditional solution:

Static void Main (string [] args) {string ip = "192.168.2.253"; // start IP int count = 5; // calculate the number of consecutive IP addresses var ipValue = BitConverter. toUInt32 (IPAddress. parse (ip ). getAddressBytes (). reverse (). toArray (), 0); for (uint I = 0; I <count; I ++) {IPAddress newIp = IPAddress. parse (ipValue + I ). toString (); Console. writeLine (newIp );}}

If we use linq for a slight transformation, we can do this:

Static void Main (string [] args) {string ip = "192.168.2.253"; // start IP int count = 5; // calculate the number of consecutive IP addresses var ipValue = BitConverter. toUInt32 (IPAddress. parse (ip ). getAddressBytes (). reverse (). toArray (), 0); var newIps = from p in Enumerable. range (0, count) let newIp = ipValue + p select new {IP = IPAddress. parse (newIp. toString ()}; foreach (var newIp in newIps) {Console. writeLine (newIp. IP );}}

Answer:

192.168.2.253
192.168.2.254
192.168.2.255
192.168.3.0
192.168.3.1