Getting the IP Address of a Remote Socket Endpoint

C#SocketsEndpoints

C# Problem Overview


How do I determine the remote IP Address of a connected socket?

I have a RemoteEndPoint object I can access and well as its AddressFamily member.

How do I utilize these to find the ip address?

Thanks!

Currently trying

IPAddress.Parse( testSocket.Address.Address.ToString() ).ToString();

and getting 1.0.0.127 instead of 127.0.0.1 for localhost end points. Is this normal?

C# Solutions


Solution 1 - C#

http://msdn.microsoft.com/en-us/library/system.net.sockets.socket.remoteendpoint.aspx

You can then call the IPEndPoint..::.Address method to retrieve the remote IPAddress, and the IPEndPoint..::.Port method to retrieve the remote port number.

More from the link (fixed up alot heh):

Socket s;
        
IPEndPoint remoteIpEndPoint = s.RemoteEndPoint as IPEndPoint;
IPEndPoint localIpEndPoint = s.LocalEndPoint as IPEndPoint;

if (remoteIpEndPoint != null)
{
    // Using the RemoteEndPoint property.
    Console.WriteLine("I am connected to " + remoteIpEndPoint.Address + " on port number " + remoteIpEndPoint.Port);
}

if (localIpEndPoint != null)
{
    // Using the LocalEndPoint property.
    Console.WriteLine("My local IpAddress is " + localIpEndPoint.Address + " connected on port number " + localIpEndPoint.Port);
}

Solution 2 - C#

string ip = ((IPEndPoint)(testsocket.RemoteEndPoint)).Address.ToString();

Solution 3 - C#

RemoteEndPoint is a property, its type is System.Net.EndPoint which inherits from System.Net.IPEndPoint.

If you take a look at IPEndPoint's members, you'll see that there's an Address property.

Solution 4 - C#

I've made this code in VB.NET but you can translate. Well pretend you have the variable Client as a TcpClient

Dim ClientRemoteIP As String = Client.Client.RemoteEndPoint.ToString.Remove(Client.Client.RemoteEndPoint.ToString.IndexOf(":"))

Hope it helps! Cheers.

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
Questionbobber205View Question on Stackoverflow
Solution 1 - C#Cory CharltonView Answer on Stackoverflow
Solution 2 - C#buzzard51View Answer on Stackoverflow
Solution 3 - C#Bertrand MarronView Answer on Stackoverflow
Solution 4 - C#MarshallView Answer on Stackoverflow