title: Does OpenFileDialog InitialDirectory not accept relative path?
body: <p><code>dialog</code> is an <code>OpenFileDialog</code> class object, and I am using <code>ShowDialog()</code> method.</p>
  <p>When I use path containing relative path, like:</p>
  <pre><code>dialog.InitialDirectory = &quot;..\\abcd&quot;;
  dialog.InitialDirectory = Directory.GetCurrentDirectory() + &quot;..\\abcd&quot;;
  </code></pre>
  <p><code>ShowDialog()</code> crashes; what I only can do is giving a definite path, starting with a disk drive:</p>
  <pre><code>dialog.InitialDirectory = &quot;C:\\ABC\\DEF\\abcd&quot;;
  </code></pre>
  <p>In this case I want the path to be 1 level up of my .exe's current directory, and then downward to directory <code>abcd</code>.<br>
  The .exe's current path can be found by <code>Directory.GetCurrentDirectory()</code>, which is perfectly fine, but I cant go on with <code>&quot;..&quot;</code>)</p>
  <p>The directory hierarchy is like:</p>
  <pre><code>ABC
      DEF 
          abcd (where I want to go)
          defg (where .exe is at)
  </code></pre>
  <p>So, is there any method to use <code>&quot;..\\&quot;</code> with <code>InitialDirectory</code>?<br>
  Or I must use definite path with it?<br>
  Thanks!</p>