php - 为什么我不能直接访问带索引的数组？

Question

当我尝试直接使用其索引访问数组元素时，我感到很困惑。我想我可以在编码中更好地解释它:-

我有一个 Employee 类的对象，我将其类型转换为数组并尝试像这样显示它:

$arrOfObj = (array) $objEmployee;

$arrKeys = array_keys( $arrOfObj );

display( $arrOfObj ); // display() is a method in my library that prints an array in a mannered way.

这给了我以下输出:-

Array
(
   [*m_UserId] => 1155
   [*m_EmailPassword] => 
   [*m_IsAssignedToManagementCompany] => 
   [*m_ManagementCompanyId] => 
   [*m_DepartmentId] => 3
   [*m_DesignationId] => 4
   [*m_EmployeeCompletedMonth] => 
   [*m_EmployeeCompletedDay] => 
   [*m_EmailAddress] =>showket.mca@gmail.com
   ------
   ------
 )

现在我不明白这个 Star(*)。当我的成员变量很简单时，比如 m_UserId、m_EmialPassword 等等，它从哪里得到这颗星。当我尝试使用以下 2 个语句显示相同内容时，我收到了一个错误:-

display( $arrOfObj['*m_EmailAddress'] );

或

display( $arrOfObj['m_EmailAddress'] );

两者都给出错误消息 Undefined index: m_EmailAddress

当我尝试这样做时它工作正常:-

display( $arrOfObj[$arrKeys[8]] );

最后一个工作正常，任何人都可以向我解释问题。

            display( $arrOfObj[$arrKeys[11]] );
            display( $arrOfObj['m_strEmailAddress'] );

Answer 1

If an object is converted to an array, the result is an array whose elements are the object's properties. The keys are the member variable names, with a few notable exceptions: integer properties are unaccessible; private variables have the class name prepended to the variable name; protected variables have a '*' prepended to the variable name. These prepended values have null bytes on either side.

http://php.net/manual/en/language.types.array.php#language.types.array.casting

尝试 var_dump(bin2hex($arrKeys[8])) 获得启发。另请参阅上面链接的手册中的示例。