为什么$db PDO对象在构造函数参数中作为空传递

我想知道为什么在下面的PHP代码中，PDO对象$db被传递为NULL。即$db=NULL在构造函数参数。

class ColoredListsUsers
{
    /**
     * The database object
     * @var object
     */
    private $_db;
    /**
     * Checks for a database object and creates one if none is found
     * @param object $db
     * @return void
     */
    public function __construct($db=NULL) /* why is #db passed as null here ? */
    {
        if(is_object($db))
        {
            $this->_db = $db;
        }
        else
        {
            $dsn = "mysql:host=".DB_HOST.";dbname=".DB_NAME;
            $this->_db = new PDO($dsn, DB_USER, DB_PASS);
        }
    }
}

先前$db被声明为PDO对象:

// Create a database object
    try {
        $dsn = "mysql:host=".DB_HOST.";dbname=".DB_NAME;
        $db = new PDO($dsn, DB_USER, DB_PASS);
    } catch (PDOException $e) {
        echo 'Connection failed: ' . $e->getMessage();
        exit;
    }

只是似乎没有意义，使$db一个PDO对象，然后传递它作为空....代码来自http://www.copterlabs.com/blog/creating-an-app-from-scratch-part-5/