title: Get disposed state of an object
body: <p>I don't know if this is the best way of doing this but I'm trying to create a form that would ne a parent form for all my other forms in my app and would check if an instance of it (actually its child) is open. The idea is to minimize the amount of code in owning form, when instantiating and opening new forms. The code in my parent form is working well, constructor checks if Application.OpenForms contains the instance of its type and closes itself it true, so I'm not going to post it at this time. 
  The problem is with executing Show() method in owning form.
  This code will not work (obviously) if the duplicate was found</p>
  
  <pre><code>FormA form = new FormA();
  form.Show();
  </code></pre>
  
  <p>because form object would have already been disposed.
  But this doesn't work also:</p>
  
  <pre><code>FormA form = new FormA();
  if(form != null)
  form.Show();
  </code></pre>
  
  <p>The correct way would be to check if form is disposed so I'm looking for property that would suffice this condition:</p>
  
  <pre><code>if( state of form is Disposed)
  </code></pre>
  
  <p>So the question: is there a property that would allow this test?</p>
  
  <p>Now, I could use event handler to exception handler tp deal with that but that would contradict principle statet at the beginning - to limit the code needed to open the form.</p>