C / C++ / MFC

Re: WM_DEVICECHANGE Problem

6-Mar-11 22:23

Ozer Karaagac wrote:
dbcv_unitmask in case DBT_DEVICEREMOVECOMPLETE

is 245760.

and this is occurred only once.

Some Day I Will Prove MySelf :: GOLD

Ozer Karaagac6-Mar-11 22:43

Ozer Karaagac

6-Mar-11 22:43

Well, 245760 = 0x3c000 in hex = 111100000000000000 in binary.
111100000000000000
RQPONMLKJIHGFEDCBA
Which means that you got notifications for all 4 volumes.
You should test its bit flags.

DWORD dwTestMask = 1;
char archVols[32];
int nCount = 0;

for(int i = 0; i < 26; i++)
{
    if(pVol->dbcv_unitmask & dwTestMask)
       archVols[nCount++] = 'A' + i;

    dwTestMask <<= 1;
}

You will receive array of volume letters in archVols.

This notifications may or may not be occurred one per volume.

Re: WM_DEVICECHANGE Problem

goldenrose96-Mar-11 23:43

6-Mar-11 23:43

thanx, it worked Thumbs Up | :thumbsup:

Some Day I Will Prove MySelf :: GOLD

Sockets and Port number

sawerr5-Mar-11 23:09

sawerr

5-Mar-11 23:09

Hi
I have a conceptual question that relates with socket programming.

I know that socket number = IP number + Port number and
A TCP packet contains four numbers, source IP source port, destination IP destination port.

How am I gonna picture port number and applications while socket programming.

Is it problem to have two or more applications use same port? or must every application has distinct port number?

Is it problem to have two or more clients making communication over same port? or must every client has distinct port number?

Bill SerGio, The Infomercial King

Bill SerGio, The Infomercial King6-Mar-11 0:50

6-Mar-11 0:50

Your question is an excellent question. Most books and articles that talk about sockets FAIL to explain this and question goes to the heart of all real world examples of writing socket code.
I will answer your question with a real world example that I worked on recently.
Let's say that you want to show stock data about a stock or group of stocks that changes every second in a web page and that the source of this data is some TCP server that you would connect to and receive a callback from every second in the form of strings. So let's say the user select a stock symbol like "IBM" and wants to see the price and volume traded of IBM stock in their web browser every second. Another user might want to see the value of a different stock symbol like "ACL" for Alcon, Inc.
How many IP addresses and Ports we would need? As you know we must use web Sockets to connect from a web page to a TCP Server--BUT there is another problem--none of the stock alert data servers have implemented the Web Sockets protocol handshake so we will need a web socket proxy between the web page and the Data Server. There are 2 samples of web Sockets Servers posted here on CodeProject that we can use.
If we put the ip address and port INSIDE the web page and connect to a Web Sockets Server then that web sockets server must spin off a thread that will in side each thread open a TCP connection to Data Server on a given ip address and port. The data from the thread received from the data server will be sent to the web socket server and then back to the client web browser that requested the data and NOT to any other browsers. Each browser on connecting to the web socket server passes in a "settings string" like say the stock symbol so that 2,000 different web browsers would see different data for 2,000 different stock symbols where each browser would have a datagrid displaying say hundreds of rows of stock data per second for that one symbol they requested.
When you set up this scenario--trial and error are good ways to learn socket programming--you will quickly realize that the amount of data that the web sockets data has to return is huge and the Data Server is passing the data for all 2,000 connected web browsers back to one place, namely the web sockets server--you could write a queue manager but that can't handle this kind of volume.
Here is the solution and answer to you question. You create a web service to return different port numbers to the web page. The web page sits on the server along with your web sockets server so the web page will first send a request to the web service and the web service will launch a NEW INSTANCE of the web sockets server with the SAME IP Addresses BUT a new port number and the web service will send that port number corresponding to the instance of the web sockets server it launched to the web page page that will then make a web socket connection via TCP to the new instance of the web sockets server. In this way you will have a new instance of Web Sockets server which is just a console app running for EACH web browser that is connected where each web browser has a different port number specified. You can easily connect 2,000 web browsers this way.
In summary, the ip addresses can be the same but the port numbers must change. Cool | :cool:

http://www.KabbalahCode.com

Andrew Brock6-Mar-11 0:54

6-Mar-11 0:54

Your knowledge is correct from a simplistic point of view.
Perhaps you should get some first hand experience with Wireshark[^].
Sniff a session, such as a opening the google homepage.

For a TCP session to work the following must happen:
1. The server starts up and listens (binds) on a port.
No other programs can be listening for data on this port (although they can send from it which is a different session)
This port is almost always pre-determined (like 80 for HTTP) but if you specify 0 the OS will give a random port which the client must be sent by some other means (like an email) before connecting
2. The client connects to the IP of the server and the port from part 1.
If you looked at Wireshark, this does the SYN, SYN/ACK and ACK that starts the session
3. Send/receive from both server and client.
4. Shutdown the socket when done with it from both ends, both server and client must shutdown both send and receive directions when finished.
If you looked at Wireshark, this does the FIN/ACK and ACK once from each end.
5. Close the socket. This frees the system handle on the socket

sawerr6-Mar-11 1:36

sawerr

6-Mar-11 1:36

@TV Mogul, thanks for the explanation. But I think my knowledge doesn't let me to understand it completely.

In your answer, port numbers are used for clients.

@Andrew Brock thank you too.

We all know that, IP address is client specific. Every client has a distinct IP address.

- If I am right, port number is application specific. If one application grabs a port no other application can use it.
So, for example,that means one computer mustn't have 2 or more web server that listening port 80. Right?

- But also a lot of clients can connect same port. But sometimes we need to serve them from different ports, ( I think TV Mogul's definition is about that)

Which conditions should i make a decision that I need a new socket?

Andrew Brock6-Mar-11 2:01

6-Mar-11 2:01

sawerr wrote:
If one application grabs a port no other application can use it.

Application specific would imply that each application could have its own set of ports. I think you meant to say "System Specific".

sawerr wrote:
So, for example,that means one computer mustn't have 2 or more web server that listening port 80. Right?

That is correct. This is called System Specific. 1 per system.

sawerr wrote:
But also a lot of clients can connect same port.

Yes, me and you can both connect to www.google.com on port 80 at the same time. (ignore the fact that we would almost certainly get different servers altogether).
When a server is listening, it has the option of setting the maximum number of sessions.

When you look at some code for starting a server you have something like this:

SOCKET sConnection = socket(AF_UNSPEC, SOCK_STREAM, IPPROTO_TCP); //Create a TCP stream socket with any address type (IPv4, IPv6, ...)
//pAddress = get listening address, usually 0.0.0.0
bind(sConnection, pAddress->ai_addr, pAddress->ai_addrlen);
listen(sConnection, SOMAXCONN);
for (;;) {
    SOCKADDR saClient;
    DWORD nClientAddrLen = sizeof(saClient);
    SOCKET sClient = accept(sConnection, &saClient, &nClientAddrLen);
    CreateThread(); //Create a new thread to handle the new connection
    //Now wait for the next connection in the new loop
}

Bill SerGio, The Infomercial King

Bill SerGio, The Infomercial King6-Mar-11 2:14

6-Mar-11 2:14

Andrew Brock gave you a textbook answer. What I am trying to explain is that teh textbooks are WRONG and all the design model examples are wrong in that they only deal with a single client connecting or multiple clients where the data is the same for each client and only a little data is passed.
The simple answer is this: The ip address can be same BUT the port number must change for each client that connects if you want the system to be practical.
This means that you create a console app with a minimum amount of memory as the server and launch a new instance of this for each client that connects so that each client is connecting on a different port number.
For example, I wrote a camera wall recently for the U.S. government and their specs stated that you may NOT use threads--that you must create a new instance of the server for each client connection. The reason for this is so that if one client goes down it will not take the whole system down. Each client uses the same ip address but has a unique port number.

Andrew Brock's answer is a textbook answer which doesn't apply to teh real world. Get real Anrew!

http://www.KabbalahCode.com

help with tcl/tk in C

basel-bh5-Mar-11 20:53

basel-bh

5-Mar-11 20:53

hi everybody
I'm working in Fedora 13 using C
to bind tcl/tk GUI in C program you use this two headers tcl.h and tk.h
when you use Tcl_EvalFile you use after it Tk_MainLoop
but Tk_MainLoop make program stop until it return from the call

so I but it in another thread
but I couldn't to modify the window using Tcl_Eval

what I should do ??

thanx alot

case statement in DLGPROC [SOLVED]

goldenrose95-Mar-11 16:36

Re: case statement in DLGPROC

5-Mar-11 16:36

LRESULT CALLBACK DlgProc(HWND hwnd, UINT Msg, WPARAM wParam, LPARAM lParam)
{
  case WM_INITDIALOG:
  return FALSE;
  case WM_COMMAND:
  return FALSE;
  case WM_CLOSE:
  return FALSE;
}

LRESULT CALLBACK DlgProc(HWND hwnd, UINT Msg, WPARAM wParam, LPARAM lParam)
{
  case WM_INITDIALOG:
  return TRUE;
  case WM_COMMAND:
  return TRUE;
  case WM_CLOSE:
  return TRUE;
}

LRESULT CALLBACK DlgProc(HWND hwnd, UINT Msg, WPARAM wParam, LPARAM lParam)
{
  case WM_INITDIALOG:
  break;
  case WM_COMMAND:
  break;
  case WM_CLOSE:
  break;
}

LRESULT CALLBACK DlgProc(HWND hwnd, UINT Msg, WPARAM wParam, LPARAM lParam)
{
  case WM_INITDIALOG:
  return 0;
  case WM_COMMAND:
  return 0;
  case WM_CLOSE:
  return 0;
}

out of these four examples, which one is correct and more accurate, and why.

Some Day I Will Prove MySelf :: GOLD

modified on Sunday, March 6, 2011 12:19 PM

Andrew Brock5-Mar-11 17:55

5-Mar-11 17:55

None of them are correct.

The message determines what you must do.

At the bottom of the DlgProc function there would be a call to DefWndProc() or the original window procedure.

Read the MSDN pages for what the return should be.
WM_INITDIALOG[^]: "The dialog box procedure should return TRUE to direct the system to set the keyboard focus to the control specified by wParam. Otherwise, it should return FALSE to prevent the system from setting the default keyboard focus." Generally you would break and let the default procedure take care of this, but you can return either TRUE or FALSE.
WM_COMMAND[^]: "If an application processes this message, it should return zero." implies that you should break the switch and call the original procedure if you don't handle it.
WM_CLOSE[^]: "If an application processes this message, it should return zero."

LRESULT CALLBACK DlgProc(HWND hwnd, UINT Msg, WPARAM wParam, LPARAM lParam) {
    switch (Msg) {
        case WM_INITDIALOG:
            break; //You can do whatever here

        case WM_COMMAND:
            switch (LOWORD(wParam)) {
                case IDC_MYBUTTON:
                    return 0; //something happened to MYBUTTON
            }
            break; //We havn't handled it, use default handler

        case WM_CLOSE:
            //cleanup
            return 0;
    }
    return pOriginalProcedure(hwnd, Msg, wParam, lParam); //This is set as GetWindowLong(GWL_MSGPROC) before changing the procedure.
}

WM_COMMAND [SOLVED]

goldenrose95-Mar-11 5:08

Moving GDI+ objects question

5-Mar-11 5:08

case WM_COMMAND:
{
	if(LOWORD(wParam)==frmMainOK && HIWORD(wParam) == BN_CLICKED)
	{
		MessageBox(NULL,"OK Pressed","Demo",MB_OK);
	}
	return TRUE;
}

case WM_COMMAND:
{
   switch(wParam)
   {
      case frmMainOK:
	MessageBox(NULL,"OK Pressed","Demo",MB_OK);
        break;
   }
break;
}

Which is the correct and most appropriate way between the two examples given above, and why?

Some Day I Will Prove MySelf :: GOLD

modified on Saturday, March 5, 2011 10:37 PM

Re: WM_COMMAND

Michael Dunn5-Mar-11 6:58

Michael Dunn

5-Mar-11 6:58

The second bit of code doesn't check the notification code, so if the control sends other codes besides BN_CLICKED, you'll treat those notifications as if they were BN_CLICKED. For buttons, this isn't a problem, but other controls like the edit box send other types of notifications.

--Mike--
Dunder-Mifflin, this is Pam.

Re: WM_COMMAND

CPallini5-Mar-11 7:37

CPallini

5-Mar-11 7:37

The second one is logically wrong and (by accident) working just for buttons notifying BN_CLICKED.
Smile | :)

If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler.
-- Alfonso the Wise, 13th Century King of Castile.

This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong.
-- Iain Clarke

[My articles]

csrss5-Mar-11 2:13

5-Mar-11 2:13

How can i move gdi+ objects drawn on window with a mouse? Such objects like: elipse, rectangle. For example, i have an elipse, created with DrawEllipse(), then it filled with GraphicsPath which contains some background color. So for example i have painted a "blue ball" on main application window. Now i am trying to figure out, how can i move this ball within window area with a mouse. I can think of such scenario:

1. On WM_LBUTTONDOWN i am checking if mouse pointer is inside the ball and setting flag to true, if it is
2. And now on WM_MOUSEMOVE, if flag is sat to true, what should i do next? Should i for example, erase this ball and paint it in new position regarding mouse pointer? Or there is a method somewhere to, like, "grab" this ball and literally move it to new position?

I'm just learning GDI+ so this question may look lame, however...
Thanks

011011010110000101100011011010000110100101101110
0110010101110011

Niklas L5-Mar-11 2:42

Niklas L

5-Mar-11 2:42

1. Store your object in vector form, so you can repaint them whenever needed.
2. On left button down, you do hit-testing to figure out which object is selected if any.
3. On mouse move, adjust the coordinates in your vector data, and repaint the screen using double buffering to avoid flicker.

home

csrss5-Mar-11 3:36

5-Mar-11 3:36

Storing things in vector sounds interesting - never thought about such approach, thanks Smile | :)

011011010110000101100011011010000110100101101110
0110010101110011

Niklas L5-Mar-11 5:39

Niklas L

5-Mar-11 5:39

I was talking about a vector representation of the graphics, not about storing them in a vector type container, although that could also perform well.

Just to avoid misinterpretations.

home

csrss5-Mar-11 6:16

5-Mar-11 6:16

Oh, i see, ok. Yeah, some misunderstanding occurred. But, my misinterpretation of a part of your post gave me some new ideas! Wink | ;)

011011010110000101100011011010000110100101101110
0110010101110011

Andrew Brock5-Mar-11 2:42

5-Mar-11 2:42

The general solution is to just invalidate the window[^] or better yet, the old ball rectangle and the new ball rectangle and then draw it again in the WM_PAINT which is called automatically when you invalidate the area.

1 minor issue with this is that it will do what is called "flickering" where it flashes.
To fix this mute the WM_ERASEBKGND message and use a memory DC. Search codeproject or google for "memory DC" or "flicker free drawing" to see how this is done in code.

csrss5-Mar-11 3:04

5-Mar-11 3:04

Thanks guys Smile | :)

Looks like invalidating / redrawing window is the only way. This "ball" is actually a part of custom control i am writing (has its own window) - it is already double buffered.

case WM_PAINT: // custom control window paint handler
{
// some checks for need of repainting
Gdiplus::Graphics graphics(hdc);
Gdiplus::Bitmap *bmp = new Gdiplus::Bitmap(CtrlRect.right, CtrlRect.bottom);
Gdiplus::Graphics *temp = Gdiplus::Graphics::FromImage(bmp);
temp->SetSmoothingMode(Gdiplus::SmoothingMode::SmoothingModeAntiAlias);
// .. drawing goes here
// .. paint "ball" with new mouse coords
Gdiplus::CachedBitmap *cbmp = new Gdiplus::CachedBitmap(bmp, &graphics);
delete bmp;
delete temp;
graphics.DrawCachedBitmap(cbmp, 0, 0);
// done
}

Just thought it could be done a way smoother - without redrawing. Anyway, thanks Smile | :)

011011010110000101100011011010000110100101101110
0110010101110011

Andrew Brock5-Mar-11 3:09

5-Mar-11 3:09

Well, I know that scrollbars work by moving the screen data, rather than redrawing, in things like ListBox's and web browser controls.
Presumably this behaviour could be implemented into your solution too, but for something so simple, it would not be worth the complexity that I imagine it might take.

csrss5-Mar-11 3:16