title: RabbitMq EasyNetQ Create message using Activator.CreateInstance
body: <p>I have an issue I don't fully understand.
  When I create message this way it works:</p>
  
  <pre><code>var message = new StartFakeJobCommand();
  using (var publishChannel = ApplicationSingleton.Instance.RabbitBus.OpenPublishChannel())
  {
      publishChannel.Publish(message);
  }
  </code></pre>
  
  <p>A message is put on the queue and my listener can consume it.
  But when I create a message using <strong>Activator.CreateInstance</strong> like this it does not work. Nothing is published to the queue.</p>
  
  <pre><code>var t = Type.GetType(string.Format("{0}.{1},{2}", job.CommandNamespace, job.Command, job.AssemblyName));
  if (t == null)
      throw new ArgumentException();
  
  var message = Activator.CreateInstance(t);
  using (var publishChannel = ApplicationSingleton.Instance.RabbitBus.OpenPublishChannel())
  {
      publishChannel.Publish(message);
  }
  </code></pre>
  
  <p>During debugging I can clearly see that the same type is created using both methods.
  Any idea why the second approach does not work? </p>
  
  <p>This is how i subscribe to the message:</p>
  
  <pre><code>bus.Subscribe&lt;StartFakeJobCommand&gt;("StartFakeJobCommand_ID", message =&gt; fakeJob.Handle(message));
  </code></pre>