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";
}

[download]
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);
            }
        }

[download]

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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