Aha! That explains why there is possibility to have only one notification
event, as opposed to two. In this case the event will be signalled twice per
"loop". I've always wondered about this. Thanks Jeff 😊
Regards
/Robert Bielik
Dirac Research
-----Original Message-----
From: wdmaudiodev-bounce@xxxxxxxxxxxxx <wdmaudiodev-
bounce@xxxxxxxxxxxxx> On Behalf Of Jeff Pages
Sent: den 12 juli 2018 12:52
To: wdmaudiodev@xxxxxxxxxxxxx
Subject: [wdmaudiodev] Re: WaveRT event
In response to the notification, the Windows audio engine will query your
position register to figure out where in the buffer it can write to (render)
or
read from (capture), so at the driver level it doesn't matter.
Jeff
------ Original Message ------
From: "Gilbert Passador" <passador@xxxxxxxxxxxx
<mailto:passador@xxxxxxxxxxxx> >
To: wdmaudiodev@xxxxxxxxxxxxx <mailto:wdmaudiodev@xxxxxxxxxxxxx>
Sent: 12/07/2018 6:59:09 PM
Subject: [wdmaudiodev] WaveRT event
Hi,
I have a question regarding miniwavertstream.cpp sample
driver provided with Windows Driver Kit.
The RegisterNotificationEvent method appends the provided
event to m_NotificationList.
The TimerNotifyRT method notifies all events of
m_NotificationList.
What is the reason of such a behaviour ?
When two notifications are wanted (see
AllocateBufferWithNotification), it means the application cannot distinguish a
notification from the first part of the buffer from a notification of the
second
part.
On the other end, when dealing the wavecyclic or wavepci this
is possible with the overlapped structure.
Am I missing something ?
Thank you
Gilbert
