title: How can I preserve quotes in printing a bash script's arguments
body: <p>I am making a bash script that will print and pass complex arguments to another external program.</p>
  
  <pre><code>./script -m root@hostname,root@hostname -o -q -- 'uptime ; uname -a'
  </code></pre>
  
  <p>How do I print the raw arguments as such:</p>
  
  <pre><code>-m root@hostname,root@hostname -o -q -- 'uptime ; uname -a'
  </code></pre>
  
  <p>Using <code>$@</code> and <code>$*</code> removes the single quotes around <code>uptime ; uname -a</code> which could cause undesired results. My script does not need to parse each argument. I just need to print / log the argument string and pass them to another program exactly how they are given. </p>
  
  <p>I know I can escape the quotes with something like  <code>"'uptime ; uname -a'"</code> but I cannot guarantee the user will do that. </p>