如何检测Windows服务中从睡眠模式唤醒？

Question

Good day! I write a windows service, which detects USB Flash drive connection and do something with connected drive. And now after tests i need to reprocess devices after wake up from sleep mode. I solve this problem when service work as windows program. To solve i additionally process DBT_DEVNODES_CHANGED , but service don't receive this message, it receives only DBT_DEVICE_ARRIVAL and DBT_DEVICEREMOVECOMPLETE . I also tried to check all messages handled by service except SERVICE_CONTROL_DEVICEEVENT and SERVICE_CONTROL_STOP when system is going to sleep and i don't get any message.

How can I determine when the system is waking up?

Answer 1

A normal application receives a WM_POWERBROADCAST:PBT_APMRESUMEAUTOMATIC window message, but a service will not get this message if it does not have a window. However, a service can receive SERVICE_CONTROL_POWEREVENT in its HandlerEx callback function, with the same parameters as WM_POWERBROADCAST

Notifies a service of system power events. The dwEventType parameter contains additional information. If dwEventType is PBT_POWERSETTINGCHANGE, the lpEventData parameter also contains additional information.

On Windows 8 and later, you can also use the PowerRegisterSuspendResumeNotification() function:

Registers to receive notification when the system is suspended or resumed.

The service equivalent of WM_DEVICECHANGE is SERVICE_CONTROL_DEVICEEVENT :

Notifies a service of device events. (The service must have registered to receive these notifications using the RegisterDeviceNotification function.) The dwEventType and lpEventData parameters contain additional information.

Answer 2

The best way to detect that system is waking up is process power messages. Windows must process WM_POWERBROADCAST Services must add SERVICE_ACCEPT_POWEREVENT to last call to SetServiceStatus and process SERVICE_CONTROL_POWEREVENT

My error was very stupid. I forget to add SERVICE_ACCEPT_POWEREVENT .