title: When a PHP function is called and something is RETURNED, does it stop running the code below it?
body: <p>I am just studying other user's PHP code right now to understand and learn better.  In the code below, it is part of a user class.  When I code using if/else blocks I format them like this... </p>
  
  <pre><code>if(!$this-&gt;isLoggedIn()){
      //do stuff
  }
  </code></pre>
  
  <p>But in the code below it is more like this</p>
  
  <pre><code>if (! $this-&gt;isLoggedIn())
      return false;
  </code></pre>
  
  <p>Also in the function below you can see that there is a couple times that there can be a <strong>RETURN</strong> value.  SO my question here, when <strong>RETURN</strong> is called, does it not run any code after that?  Like does it end the script for that function there?  </p>
  
  <p>In this case if this is ran...</p>
  
  <pre><code>if (! $this-&gt;isLoggedIn())
          return false;
  </code></pre>
  
  <p>Does it continue to run the code below that?</p>
  
  <hr>
  
  <p><strong>Here is the function</strong>  </p>
  
  <pre><code>&lt;?PHP
  private function logout($redir=true)
  {
      if (! $this-&gt;isLoggedIn())
          return false;
  
      $this-&gt;obj-&gt;session-&gt;sess_destroy();
  
      if ($this-&gt;isCookieLoggedIn())
      {
          setcookie('user','', time()-36000, '/');
          setcookie('pass','', time()-36000, '/');
      }
      if (! $redir)
          return;
  
      header('location: '.$this-&gt;homePageUrl);
      die;
  }
  ?&gt;
  </code></pre>