Re: Windows Messaeg Queue Query
I think you don't show the actual code.
"John" <reply@newsgroup.com> wrote in message
news:eTwrl.34143$Ii4.28358@newsfe19.ams2...
Hi,
I have a problem with posting messages to a message queue. I've read
through the MSDN docs but they have not really answered the problem I am
expierencing.
The setup is as follows:
I have a thread that receives UDP messages from the network. When the
messages are received it posts a message like so
AtlTrace("Posting new message...");
PostMessage(WM_DATA_READY,6,0);
WM_DATA_READY is mapped to a function in the header using
BEGIN_MSG_MAP(CTestApp)
MESSAGE_HANDLER(WM_DATA_READY, OnDataReady)
END_MSG_MAP()
The function OnDataReady has the body
LRESULT CTestApp::OnDataReady(UINT uMsg, WPARAM wParam, LPARAM lParam,
BOOL& bHandled)
{
static int nCnt = 0;
AtlTrace("Entered OnDataReady function with nCnt = %d\n", nCnt);
DWORD dwThreadID = GetCurrentThreadId();
AtlTrace("Thread ID = 0x%x\n", dwThreadID);
for(int nIndex=0; nIndex<5; ++nIndex)
{
AtlTrace("Loop number is %d\n", nIndex);
}
}
When I run the code I get the following output:
Posting new message...
Posting new message...
Posting new message...
Entered OnDataReady function with nCnt = 0
Thread ID = 0x9eb
Loop number is 0
Entered OnDataReady function with nCnt = 1
Thread ID = 0x9eb
Entered OnDataReady function with nCnt = 2
Thread ID = 0x9eb
Loop number is 1
Loop number is 2
Loop number is 3
Loop number is 4
Loop number is 1
Loop number is 2
Loop number is 3
Loop number is 4
Loop number is 1
Loop number is 2
Loop number is 3
Loop number is 4
Either my code is wrong or my understanding of the message queue is wrong.
How is it possible that the same thread can be in the same function 3
times at the same time? As you can see from the output, the loop has been
run three times but for some reason the final two times it has missed out
the first interation???
I can't even set up CRITICAL_SECTIONS inside the function because the
thread ID is always the same.
Can anyone provide an explanation as to what is happening in the code?
Many thanks,
John.
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