openevent返回访问被拒绝 [英] openevent return access denied

问题描述

我创建了活动
HANDLE h = CreateEvent(NULL, FALSE, FALSE, "eventname");
然后在另一个过程中打开它
HANDLE h1 = OpenEvent(EVENT_MODIFY_STATE, FALSE, "eventname");

我正常运行时效果很好.
但是当我在Windows服务中创建事件时，打开事件时会给出错误访问被拒绝.
有人可以对此问题表示怀疑吗?

I have created Event
HANDLE h = CreateEvent(NULL, FALSE, FALSE, "eventname");
And open it in another process
HANDLE h1 = OpenEvent(EVENT_MODIFY_STATE, FALSE, "eventname");

It work fine when I run normally.
But when I have created event in windows service it gives error Access denied while opening event.
Can anybody give sujjection to this problem

推荐答案

以下是CreateEvent的签名

Here is the signature for CreateEvent

HANDLE WINAPI CreateEvent(
  __in_opt  LPSECURITY_ATTRIBUTES lpEventAttributes,
  __in      BOOL bManualReset,
  __in      BOOL bInitialState,
  __in_opt  LPCTSTR lpName
);

您必须传递SECURITY_ATTRIBUTES结构，以授予将调用OpenEvent的用户(或组)访问权限.

看看为C ++中的新对象创建安全描述符 [ ^ ]-它显示了如何使用SECURITY_ATTRIBUTES.

最好的问候
Espen Harlinn

You ave to pass a SECURITY_ATTRIBUTES structure granting access to the user (or a group) that will call OpenEvent.

Take a look at Creating a Security Descriptor for a New Object in C++[^] - it shows how to work with SECURITY_ATTRIBUTES.

Best regards
Espen Harlinn

我认为这与第一个参数 LPSECURITY_ATTRIBUTES 有关.在msdn中，建议如果此参数为NULL，则子进程不能继承该句柄.".

您最好尝试解决此问题.

MSDN链接

I think it has something to do with first parameter LPSECURITY_ATTRIBUTES. in msdn it is suggested "If this parameter is NULL, the handle cannot be inherited by child processes.".

You better try to fix this issue.

MSDN link

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