Prose\UsingYamlFile::__construct PHP Метод

    public function __construct(StoryTeller $st, $args)
    {
        // call our parent constructor
        parent::__construct($st, $args);
        // $args[0] will be our filename
        if (!isset($args[0])) {
            throw new E5xx_ActionFailed(__METHOD__, "Param #0 needs to be the name of the file to work with");
        }
    }