Alex Rivera | Logout

Why call dispose(false) in the destructor?

Asked 2009-03-10T02:49:04.510
71

What follows is a typical dispose pattern example:

 public bool IsDisposed { get; private set; }

  #region IDisposable Members

  public void Dispose()
  {
    Dispose(true);
    GC.SuppressFinalize(this);
  }

  protected virtual void Dispose(bool disposing)
  {
    if (!IsDisposed)
    {
      if (disposing)
      {
        //perform cleanup here
      }

      IsDisposed = true;
    }
  }

  ~MyObject()
  {
    Dispose(false);
  }

I understand what dispose does, but what I don't understand is why you would want to call dispose(false) in the destructor? If you look at the definition it would do absolutely nothing, so why would anyone write code like this? Wouldn't it make sense to just not call dispose from the destructor at all?

c# dispose
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3

I think the confusion is due to the fact that in your example you aren't releasing any unmanaged resources. These also need to be released when dispose is called via garbage collection and they would be released outside the check for disposing. See the MSDN example relating to releasing unmanaged resources. The other that that would/should happen outside the check is a call to any base class Dispose method.

From the quoted article:

   protected override void Dispose(bool disposing) 
   {
      if (disposing) 
      {
         // Release managed resources.
      }
      // Release unmanaged resources.
      // Set large fields to null.
      // Call Dispose on your base class.
      base.Dispose(disposing);
   }
answered 2009-03-10T03:10:42.913

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