title: How to use a SymfonyStyle object to get options
body: <p>According to:</p>
  
  <p><a href="http://symfony.com/blog/new-in-symfony-2-8-console-style-guide" rel="nofollow">http://symfony.com/blog/new-in-symfony-2-8-console-style-guide</a></p>
  
  <blockquote>
    <p>...
    Using the new style guide in your commands is as easy as instantiating
    the SymfonyStyle class, passing to it the $input and $output of your
    command.
    ... 
    We recommend you to name this variable $io because you
    will use it both for input and output operations.
    ...</p>
  </blockquote>
  
  <p>A SymfonyStyle instance can be used to handle both input and output operations.</p>
  
  <p>So instead of having an <code>$input</code> variable (instance of <code>InputInterface</code>) and an <code>$output</code> variable (instance of <code>OutputInterface</code>), one can just use an $io (instance of <code>SymfonyStyle</code>) to handle <strong>both types of operations</strong>.</p>
  
  <p>That is very useful.</p>
  
  <p>But what about the <code>getOption</code> of <code>InputInterface</code> method used to retrieve the options submitted by the user?</p>
  
  <p>It looks like that <code>$input-&gt;getOption(...)</code> can't be simply replaced by <code>$io-&gt;getOption(...)</code> because <code>SymfonyStyle</code> does not have that method.</p>
  
  <p>Is there a way to retrieve an user-submitted option simply using an instance of <code>SymfonyStyle</code> ?</p>