PHP convet class to json data

/*********************************************************************
 *              PHP convet class to json data
 * 说明：
 *     突然想使用class自动转换为json数据，这样的代码可扩展性会好一点，
 * 只需要修改class的属性就能够达到最终json数据输出，不过有遇到class中
 * 初始化class变量需要在构造函数中初始化的的问题。
 *
 *                                   2017-8-11 深圳 龙华樟坑村 曾剑锋
 ********************************************************************/


一、参考文档：
    1. getting Parse error: syntax error, unexpected T_NEW [closed]
        https://stackoverflow.com/questions/15806981/getting-parse-error-syntax-error-unexpected-t-new

二、测试代码：
    <?php
        class Uart {
            public $port = "/dev/ttyO0";
            public $value = "OK";
        }

        class Context {
            public $uart = new Uart();;
            public $version = "v0.0.1";
        }

        $context = new Context;

        $context_json = json_encode($context);
        echo $context_json
    ?>

三、报错内容：
    Parse error: syntax error, unexpected 'new' (T_NEW) in /usr/share/web/time.php on line 8

四、最终代码：
    <?php
        class Uart {
            public $port = "/dev/ttyO0";
            public $value = "OK";
        }

        class Context {
            public $uart;
            public $version = "v0.0.1";

            public function __construct() {
                $this->uart = new Uart();
            }
        }

        $context = new Context;

        $context_json = json_encode($context);
        echo $context_json
    ?>

五、输出结果：
    {"uart":{"port":"/dev/ttyO0","value":"OK"},"version":"v0.0.1"}

六、原因：
    you must do initialize new objects in the __construct function;