I am trying impersonation in remoting .. The program runs fine when i turn off impersonation. I am running client and server both on the same machine(VISTA). I have created the user account on the machine .. and it runs fine when i use secure channel and turn impersonation off.. I have also checked that when impersonation is off and using secure channel .. it allows only the users that i have created to logon .. otherwise it gives authentication failure. Following is the code : Server: using System; using System.Runtime.Remoting; using System.Runtime.Remoting.Channels; using System.Runtime.Remoting.Channels.Tcp; using System.Collections; using System.Threading; using System.Security.Principal; namespace ns_application { namespace ns_server { public class cls_server { static void Main(string[] args) { //TcpChannel channel = new TcpChannel(8080); IDictionary props = (IDictionary) new Hashtable(); IPrincipal threadPrincipal; props.Add("port", "8080"); props.Add("secure", "true"); props.Add("impersonate", "true"); props.Add("protectionLevel", "EncryptAndSign"); //props.Add("username", "Vishwjeet"); //props.Add("password", "pwd"); //props.Add("authorizationModule", "AuthorizeUser"); //BinaryClientFormatterSinkProvider cp = new BinaryClientFormatterSinkProvider(); //BinaryServerFormatterSinkProvider sp = new BinaryServerFormatterSinkProvider(); IChannel chan = new TcpChannel(props,null,null); ChannelServices.RegisterChannel(chan,true); RemotingConfiguration.RegisterWellKnownServiceType(typeof(ns_registeruser.cls_registeruser), "ApplicationMain", WellKnownObjectMode.Singleton); threadPrincipal = Thread.CurrentPrincipal; Console.WriteLine(threadPrincipal.Identity.Name); Console.WriteLine(threadPrincipal.Identity.IsAuthenticated); Console.WriteLine(threadPrincipal.Identity.AuthenticationType); Console.WriteLine("Server Started .. "); Console.ReadLine(); } } } } Methods : using System; using System.Runtime.Serialization; using System.Reflection; using System.Data; using System.Threading; using System.Security.Principal; namespace ns_application { namespace ns_registeruser { //[Serializable] public class cls_whereitruns:MarshalByRefObject { private int _IDNo; public int IDNo { get { return _IDNo; } set { _IDNo = value; } } public int FutureofID() { IDNo = IDNo * 5; Console.WriteLine("Calculated future of ID : {0} :::: {1} :::: {2}", IDNo,this.ToString(),AppDomain.CurrentDomain.FriendlyName); return(IDNo); } public cls_wher

Any help on this error?? This happens when I run the server with ID_A and try to run the client with ID_B .. this is an attempt to try impersonation System.TypeInitializationException was unhandled Message="The type initializer for 'System.Runtime.Remoting.Identity' threw an exception." Source="mscorlib" TypeName="System.Runtime.Remoting.Identity" StackTrace: Server stack trace: at System.Runtime.Remoting.Identity.RemoveAppNameOrAppGuidIfNecessary(String uri) at System.Runtime.Remoting.IdentityHolder.MakeURIKey(String uri) at System.Runtime.Remoting.IdentityHolder.ResolveIdentity(String URI) at System.Runtime.Remoting.RemotingServices.GetServerTypeForUri(String URI) at System.Runtime.Remoting.Channels.BinaryServerFormatterSink.ProcessMessage(IServerChannelSinkStack sinkStack, IMessage requestMsg, ITransportHeaders requestHeaders, Stream requestStream, IMessage& responseMsg, ITransportHeaders& responseHeaders, Stream& responseStream) Exception rethrown at [0]: at System.Runtime.Remoting.Proxies.RealProxy.HandleReturnMessage(IMessage reqMsg, IMessage retMsg) at System.Runtime.Remoting.Proxies.RealProxy.PrivateInvoke(MessageData& msgData, Int32 type) at System.Object.FieldGetter(String typeName, String fieldName, Object& val) at ns_application.ns_client.cls_client.Main(String[] args) in D:\Vishwjeet\Study\Remoting_tcp\ConsoleApplication2\client.cs:line 64 at System.AppDomain._nExecuteAssembly(Assembly assembly, String[] args) at System.AppDomain.ExecuteAssembly(String assemblyFile, Evidence assemblySecurity, String[] args) at Microsoft.VisualStudio.HostingProcess.HostProc.RunUsersAssembly() at System.Threading.ThreadHelper.ThreadStart_Context(Object state) at System.Threading.ExecutionContext.Run(ExecutionContext executionContext, ContextCallback callback, Object state) at System.Threading.ThreadHelper.ThreadStart() InnerException: System.Security.Cryptography.CryptographicException Message="Unknown error \"-2146893801\"." Source="mscorlib" StackTrace: at System.Security.Cryptography.CryptographicException.ThrowCryptogaphicException(Int32 hr) at System.Security.Cryptography.Utils._AcquireCSP(CspParameters param, SafeProvHandle& hProv) at System.Security.Cryptography.Utils.AcquireProvHandle(CspParameters parameters) at System.Security.Cryptography.Utils.get_StaticProvHan

No,but kudos for taking the time to find out. If you wanted to ask about how WCF worked, this would be the place. Depending on what you want to know, you should look at the C# (or VB.NET) forum, the .NET forum or the ASP.NET forum for pure web service help. "WPF has many lovers. It's a veritable porn star!" - Josh Smith As Braveheart once said, "You can take our freedom but you'll never take our Hobnobs!" - Martin Hughes. My blog | My articles | MoXAML PowerToys | Onyx

Here is a better link http://www.codeguru.com/columns/dotnet/article.php/c10253[^]

Have you installed SQL server in mixed mode? If so, then just exempt the sqlbrowser and sqlserv.exe files from your firewall. Create a new login to your sql server and also add that user to the database you want to connect. you will get those files here : drive:\Program Files\Microsoft SQL Server\MSSQL.XX\MSSQL\Binn\sqlservr.exe drive:\Program Files\Microsoft SQL Server\XX\Shared\sqlbrowser.exe Now connect using SqlAuthentication using the userid and passwornd, and you are done. :-D Abhishek Sur My Latest Articles **Create CLR objects in SQL Server 2005 C# Uncommon Keywords Read/Write Excel using OleDB **Don't forget to click "Good Answer" if you like to.

Yes it is .. unless there is a specific message board for .net remoting .

The stuff that ASP.NET ALWAYS uses, is there by default. I would guess this is ASP.NET AJAX, or some other library you have added. Christian Graus Driven to the arms of OSX by Vista. Read my blog to find out how I've worked around bugs in Microsoft tools and frameworks.

a small one will do .. but i need some insight on .. what are the most popular design patterns being used for some typical applications nowadays. like an application for having a custom form for data entry then reporting using rdlc or maybe a gridview then having data modification forms .. a very normal application .. but with crucial data the tutorials on net just tell me what a design pattern is and how to use it

Have a look at dofactoryhttp://dofactory.com/Default.aspx[^] for details of common patterns, along with code. "WPF has many lovers. It's a veritable porn star!" - Josh Smith As Braveheart once said, "You can take our freedom but you'll never take our Hobnobs!" - Martin Hughes. My blog | My articles | MoXAML PowerToys | Onyx

Iterator blocks are called when you want to fetch the data from the Enumerable. Actually if you call the the function, it will not be called. Rather if you do a foreach on the Enumerable object the function will be called and will return you 10. ;) Just change your code a bit as I did: IEnumerable arr = Documented.GetValuesUpto(10); MessageBox.Show(arr.ToList().Count.ToString()); public static IEnumerable GetValuesUpto(int n) { for (int ctr = 1; ctr <= n; ctr++) { yield return ctr; } } Put a breakpoint in the line IEnumerable arr = Documented.GetValuesUpto(10);. and also a breakpoint within GetValueUpto block. You will see the function is called when we try to call ToList() rather than the actual call to GetValueUpto. This is the main advantage of iterator blocks. :rose::rose: Abhishek Sur My Latest Articles **Create CLR objects in SQL Server 2005 C# Uncommon Keywords Read/Write Excel using OleDB **Don't forget to click "Good Answer" if you like t

hey, i have posted the output that Im getting but im not able to understand what it means. Can you explain that?

Thanks buddy!! I will try this one. I hope this is what I was looking for

Thanks .. I can work now :) !!