PHP静态类返回引用

Question

我有一个静态方法的类.static方法返回私有静态stdClass对象.

myclass::get() // returns stdClass object
myclass::get()->name // name is hardcoded into the class

我如何更改名称的值,如:

myclass::get()->name = 'bob';

并设置好吗？

我尝试返回对象,如:

return &self::$static_object;

但这会引发语法错误.

我能做什么？

EDIT发布了代码以便澄清

final class config {

    private static $configs = array();

    public static function get($config_name) {

        if (isset($configs[$config_name])) {

            return self::$configs[$config_name];
        }

        $file = __get_file_exists(M_CONFIGS . $config_name, 'conf.');

        if ($file) {

            $config = self::__scope_include($file);

            if (!is_array($config) && !$config instanceof stdClass) {
                /*
                 * 
                 * 
                 * FIX
                 * 
                 * 
                 * 
                 */
                die('ERROR config.php');
            }

            return self::$configs[$config_name] = self::__to_object($config);
        }
    }

    private static function __scope_include($file) {

        return include $file;
    }

    private static function __to_object($config) {

        $config = (object) $config;

        foreach ($config as &$value) {

            if (is_array($value)) {

                $value = self::__to_object($value);
            }
        }

        return $config;
    }
}

echo config::get('people')->name; //dave
config::get('people')->name = 'bob';
echo config::get('people')->name; // should be bob, is dave

Answer 1

在get()方法中通过引用返回应该做的诀窍:

public static function &get() {
    return self::$static_object;
}

但是,我认为您应该重新审视您的设计,因为这种编码非常不受欢迎,并且会导致维护和可测试性问题.