Bitwise Left Shift - C / C++ / MFC Discussion Boards

sweep12320-May-04 6:36

20-May-04 6:36

Thanks, got it working, one other question, will the processing for protecting the data be under 100msec;

i.e. the time to:-
Create(m)
WaitForSingleObject(m)
ReleaseMutex(m)

I know is a very general question but an order of magnitude would be helpful.

I have a timer checking for new data, the rate could be 100msec to several seconds?

Re: Long Wait for a Mutex

valikac20-May-04 6:45

valikac

20-May-04 6:45

There is no need to create the mutex each time. Just create it once and close it when the process terminates or when it is no longer needed throughout the process lifetime.

Kuphryn

Re: Long Wait for a Mutex

Grahamfff20-May-04 10:18

Grahamfff

20-May-04 10:18

I am now confused!

If I create the Mutex, then when I go into my routine which I WaitForSingleObject, then I wait forever. This was my query.

But if I create it each time I go into my WaitForMutex routine, I get it; i.e. my program works. See updated WaitForMutex routine.

<big>bool WaitForMutex(void)<br />
{<br />
      /* This procedure will wait for a Mutex before <br />
allowing any write access to the shared area. */<br />
<br />
m_hMutex = ::CreateMutex(NULL,FALSE,"CSLSharedMemLock");<br />
if(m_hMutex)<br />
{<br />
      if (::WaitForSingleObject(m_hMutex,INFINITE) ==    WAIT_OBJECT_0)<br />
      {<br />
         return true;<br />
     ReleaseMutex(m_hMutex);<br />
  }<br />
      return false;<br />
    	 }<br />
}</big>

My EXE uses CSLSharedMemLock for the Mutex and the libray (DLL) also uses CSLSharedMemLock as the Mutex name.
Sweep123 working from home!

grahamfff

Re: Long Wait for a Mutex

Paul Ranson20-May-04 10:34

Paul Ranson

20-May-04 10:34

Your EXE and DLL both need to call CreateMutex,

hMutex = ::CreateMutex ( 0, FALSE, "CSLSharedMemLock" ) ;

Now nobody owns the mutex but everybody has a handle to it.

So when you need the mutex call,

void EnterMutex ()
{
   ::WaitForSingleObject ( hMutex, INFINITE ) ;
}

Now you own it, and when you're done with it you must call ReleaseMutex

void LeaveMutex ()
{
   ::ReleaseMutex ( hMutex ) ;
}

I think your problem is probably a confusion over ownership. The two functions above are naturally wrapped up into the constructor/destructor of a class, and error returns handled by throwing exceptions, but that's another discussion really.

Paul

Re: Long Wait for a Mutex

Grahamfff20-May-04 10:45

Grahamfff

20-May-04 10:45

What you say makes sence, but in my situation if I call EnterMutex() then I have a long wait!!

I have checked that the DLL has released the mutex, so it should be 'unowned'.

Still confused!?!

Graham.

grahamfff

Re: Long Wait for a Mutex

Paul Ranson20-May-04 11:34

Paul Ranson

20-May-04 11:34

So you must have some sort of deadlock, the question is where, and I don't think we can help you here without all the code.

I think that a single thread can 'Wait' for a Mutex multiple times and the wait will always complete immediately. But each wait must be accompanied with a matching ReleaseMutex. Perhaps there's something like that but with insufficient releases happening in your code flow?

Paul

Re: Long Wait for a Mutex

valikac20-May-04 11:24

valikac

20-May-04 11:24

You only check the WAIT_OBJECT_0 returned from WaitForSingleObject(). Check for other possible return values.

Post the code that creates the mutex in the EXE.

Kuphryn

Bitwise Left Shift

Anthony988720-May-04 5:08

Anthony9887

20-May-04 5:08

I noticed that whenever I use the "<<" bitwise left shift operator. The result value is incorrect. The bitwise right shift is correct. Whats up with this? Any ideas? ie 49<<3 should be 136, but I get 392.

Thanks

Re: Bitwise Left Shift

Michael Dunn20-May-04 5:13

Michael Dunn

20-May-04 5:13

49*8 is 392. How are you calculating 136?

--Mike--
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Re: Bitwise Left Shift

Antony M Kancidrowski20-May-04 5:26

Antony M Kancidrowski

20-May-04 5:26

I assume he is taking it as 8bits and therefore the top bit falls off the end! Wink | ;)

Ant.

Re: Bitwise Left Shift

mirex20-May-04 21:39

mirex

20-May-04 21:39

Then you should be shifting an 8bit variable, like 'unsigned char'. You are probably shifting 'int' or something. Or cut it to 8 bits with AND 255

Re: Bitwise Left Shift

Antony M Kancidrowski20-May-04 5:25

Antony M Kancidrowski

20-May-04 5:25

I see what you have done. You need to AND with 255 to get the answer.

Ant.

Re: Bitwise Left Shift

deniz79s20-May-04 9:01

deniz79s

20-May-04 9:01

one time right shift is like divide with 2 and one time left shift is like multiple wiht 2. So How can you find 136 result

It is illegal to call out while inside message filter

Member 53435720-May-04 4:55

Member 534357

20-May-04 4:55

I am running my application in Visual Studio .NET & am getting the following error "It is illegal to call out while inside message filter"
What can be causing this error to occur?

Thanks

Re: It is illegal to call out while inside message filter

Andrew Quinn AUS20-May-04 5:14

Andrew Quinn AUS

20-May-04 5:14

Hi,

From my experience - this is caused by COM re-entrancy.

For example, COM server has called back into your app (via eventing) and you are then calling back into the COM server.

COM's telling you its blocking the call because it hasn't finished processing the previous method call.

I did the following which still blocks but without the annoying message

class CBusyState
{
public:

	// Constructor - by default the busy state is entered
	CBusyState(BOOL bEnterBusyState = TRUE, 
		SERVERCALL scBusyReply = SERVERCALL_RETRYLATER)
		: m_bBusy(FALSE)
	{
		// get the MFC message filter implementation. This will return NULL if called from a DLL.
		m_pMessageFilter = AfxOleGetMessageFilter();
		if (bEnterBusyState)
		   Begin(scBusyReply); // enter the busy state
	}

	// Destructor - ends the busy state (if previously entered)
	~COasisBusyState()
	{	
		End(); // end the busy state
	};

	// Begins the busy state. By default, any incoming COM calls will be
	// rejected, but retried later
	BOOL Begin(SERVERCALL scBusyReply = SERVERCALL_RETRYLATER)
	{	// check that we have a message filter, and that we're not already
		// busy.
		if (m_pMessageFilter && !m_bBusy)
		{	// set the reply code for IMessageFilter::HandlingIncomingCall()
			m_pMessageFilter->SetBusyReply(scBusyReply);
			// enter the busy state
			m_pMessageFilter->BeginBusyState();
			m_bBusy = TRUE;
		}
		return m_bBusy;
	}

	// Ends the busy state, if previously entered
	void End()
	{	// check that we have a message filter, and that we're busy
		if (m_pMessageFilter && m_bBusy)
		{	// end the busy state
			m_pMessageFilter->EndBusyState();
			m_bBusy = FALSE;
		}
	}

private:
	BOOL			m_bBusy;
	COleMessageFilter*	m_pMessageFilter;
};

Then whenever we call the COM server, I'd put the object on the stack, e.g.

function xyzxyzxyz()
{
CBusyState blocker(TRUE);
MyCOMServer->method(xyz)
}

Hope this helps,
Andy

Re: It is illegal to call out while inside message filter

Member 53435720-May-04 6:27

Member 534357

20-May-04 6:27

Thanks alot for your response, I will try your idea, for now I put some workaround in my code.

Thanks again for the info.

How do I form the HTTP headers to DL a .CAB file?

Terry O'Nolley20-May-04 4:39

Terry O'Nolley

20-May-04 4:39

I am getting server errors when trying to download a CAB file from a webserver.

I think the problem is that the default headers don't allow for CAB file mime type.

What string do I need to use in my call to
CHttpFile.AddRequestheaders(); to get this file?

I believe this is the problem, because if I change the URL to point to an HTML page, but leave everything else the same, then the file downloads just fine.

Thanks!

(I am also posting this in the Web Develpment forum)

//placeholder for witty verbiage

Re: How do I form the HTTP headers to DL a .CAB file?

Alexander ^M.,20-May-04 5:26

Alexander ^M.,

20-May-04 5:26

does "GET /.../.../foo.cab" really not work?

Don't try it, just do it! Wink | ;-)

Re: How do I form the HTTP headers to DL a .CAB file?

Terry O'Nolley20-May-04 6:29

Terry O'Nolley

20-May-04 6:29

You won't believe this - when I read your response I changed my mindset from "Why does it work with an HTM file but not a CAB file" to "It is supposed to work with a CAB file, so what else could be wrong?" because of the way you worded it Smile | :)

Thanks!

When I changed the #define from my clients production server to my test server, I forgot to change the #define'd path from my client's path to the path on my server Blush | :O

Of course it works just fine now Smile | :)

Thanks for the push in the right direction!

//placeholder for witty verbiage

about MSChart

yingkou20-May-04 3:16

yingkou

20-May-04 3:16

I have set a MSChart control in my dialog,a error happened.

m_Chart.SetColumn(i);

when i=5,compiler shows :index given is out of bound and i=4,all right. Why?

Re: about MSChart

*Dreamz20-May-04 4:13

*Dreamz

20-May-04 4:13

I think the default column count in the control is 4.You can set the number of columns from the properties of the control.

Re: about MSChart

yingkou20-May-04 14:03

yingkou

20-May-04 14:03

thanks a lot,I have resulved this problem.

Idle time in VC++

Anonymous20-May-04 3:08