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I should have been clearer its the choice of ipaddress to return that's the issue. If I return all addresses it get it in a reverse order than a C# program (both are copied below). if I use Khen1950fx's first example I get the answer as 192.168.0.1 which is the wrong address to use. Code that picks the first address to use now breaks down on a multiadapter machines.

Outputs via C# and perl

>listaddrcsharp.exe 10.123.67.139 192.168.0.1

>perl d:\temp\in.pl 192.168.0.1 10.123.67.139

Perl code
use Socket; use IO::Socket; my $machineName = lc($ENV{COMPUTERNAME}); my ($name, $aliases, $addrtype, $length, @addrs) = gethostbyname($mach +ineName); foreach(@addrs) { ($a,$b,$c,$d)=unpack ('C4', $_); print "$a.$b.$c.$d\n"; }
C# Code
public static void Main(string[] args) { IPAddress[] ips; ips = Dns.GetHostAddresses(Environment.MachineName); foreach (IPAddress ip in ips) { if (ip.AddressFamily == AddressFamily.InterNetwork) Console.WriteLine(" {0}", ip); } }

In reply to Re^2: gethostbyname returns reverse order of ipaddress expected by anubhavd
in thread gethostbyname returns reverse order of ipaddress expected by anubhavd

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