How to heal faulted WCF channels?

Question

When a single ClientBase instance is used for multiple WCF service calls, it can get a channel into a faulted state (ie. when the service is down).

Answer 1

This is what I'm currently doing, but I can't say this is the best option either.

I recreate the proxy when an exception is caught on the call.

try
{
    ListCurrentProcesses();
}
catch (TypeLoadException ex)
{
    Debug.Print("Oops: " + ex.Message);
    m_Proxy = new ProcessManagerProxy();
}
catch (EndpointNotFoundException endpointEX)
{
    Debug.Print("Oops: " + endpointEX.Message);
    m_Proxy = new ProcessManagerProxy();
}
catch (CommunicationException communicationEx)
{
    Debug.Print("Oops: " + communicationEx.Message);
    m_Proxy = new ProcessManagerProxy();
}

Answer 2

You can't. Once a channel is faulted, it's faulted for good. You must create a new channel. WCF channels are stateful (in a manner of speaking), so a faulted channel means the state may be corrupted.

What you can do is put the logic you're using into a utility method:

public static class Service<T> where T : class, ICommunicationObject, new()
{
    public static void AutoRepair(ref T co)
    {
        AutoRepair(ref co, () => new T());
    }

    public static void AutoRepair(ref T co, Func<T> createMethod)
    {
        if ((co != null) && (co.State == CommunicationState.Faulted))
        {
            co.Abort();
            co = null;
        }
        if (co == null)
        {
            co = createMethod();
        }
    }
}

Then you can invoke your service with the following:

Service<SampleServiceClient>.AutoRepair(ref service,
    () => new SampleServiceClient(someParameter));
service.SomeMethod();

Or if you want to use the default parameterless constructor, just:

Service<SampleServiceClient>.AutoRepair(ref service);
service.SomeMethod();

Since it also handles the case where the service is null, you don't need to initialize the service before calling it.

Pretty much the best I can offer. Maybe somebody else has a better way.