title: bash shell tr -d -c options combination usage
body: <pre><code>echo $PATH | tr -d -c :
  </code></pre>
  
  <p>this output is:</p>
  
  <pre><code>::::
  </code></pre>
  
  <p>The value of <code>$PATH</code> is:</p>
  
  <pre><code>/import/adams/2/z1/bin-pc.i86.linux:/import/adams/2/z1/bin:/usr/local/bin:/usr/bin:/bin
  </code></pre>
  
  <p>why I get such an output <code>;;;;</code>? I cannot understand -d -c :. -c option needs two sets, but -d needs only one set. which option is executed first? how does this result be generated? </p>
  
  <p>Thanks.</p>