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are you setting the BS_BITMAP style when you create the button?
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no because with BS_BITMAP style show bitmap only without text in the button.
I need [Bitmap+Text]
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Member 2965471 wrote: I need [Bitmap+Text]
See here[^] for more details.
The best things in life are not things.
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I know but don't work in that way!
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Member 2965471 wrote: I know but don't work in that way!
I have just tested this and it works OK. You need to ensure that your bitmap is the same size or smaller than your button. Also if you need to show both bitmap and text, then you must ensure that your bitmap is small enough to leave space for the text.
The best things in life are not things.
modified on Thursday, June 16, 2011 9:35 AM
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What is the return value of LoadBitmap function?
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Hi,
I've been working with some code that uses a TCP connection for the past few weeks. I have a socket class derived from CAsyncSocket.
Everything works fine initially. I instantiate my manager class which wraps around the socket class and calls connect (via a member pointer to the socket class). I am then able to send a receive data. The problem comes when there is a socket error. For a currently unknown reason, part of my code is causing a socket error. I have yet to find where this is happening but the problem forced me to test out the part of my code which shoudl attempt to recover the connection. It doesn't work, and I can't get things going.
I have overriden OnClose, which calls a method on the manager class (its parent) called OnSocketClose. This calls socket.shutdown(), socket.close() and 'delete socket'. It then sets the pointer to 0. From here, the next time I attempt to send anything down the socket it looks for the socket pointer == 0, to indicate that it should create a pointer to a new socket.
This new socket is created but fails to connect. I have got it to appear connected on a couple of occasions by moving things around but OnConnect doesn't seem to get called and I am unable to send data. The returned value from socket.connect however reports that the socket is already connected if I call socket.connect again, after a short period of time from the first call to connect (socket error 10056).
Apologies if this sounds a little confusing, but I think I have managed to confuse myself... massively 
If it will help, I can post code? Just let me know which parts you need to see.
Thanks!
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Thankfully, I have managed to solve this issue so thought I would post the solution in the hope that it might help someone else with the same problem.
The problem (rather annoyingly) came because the socket was being created initially on the main thread, but was later being created again (after the socket was destroyed because of a connection error) on a worker thread. This meant that the On* callbacks weren't being implemented, and so I got no feedback from OnConnect and didn't receive any data because OnReceive was never being called.
As soon as I managed the socket entirely from the main thread, everything was fixed 
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Hey
I have a MFC dialog app that will process the Enter/Return key presses in the keyboard.
Now if the dialog is minimized or hidden behind some other window, it will not be able to process the Return key presses.
Which makes sense considering the keyboard messages are routed to the window with focus.
I want to route all Return/Enter key presses to my MFC dialog irrespective of whether its hidden or minimized.
Is there a way i can do that?
Thanks in advance.
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Yes i want to do this, as this is just a sample app.
What kinda keyboard hooks?
Any particular function that i can use?
Any example code?
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HTH KeyBoard Hooks[^]
on side note, you are making your system unstable by using hooks. and as inputs from others posters you are changing default functionality of other application.
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Are you sure you want to do this? ...this sounds like it would create a headache for your users...
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dipuks wrote: I want to route all Return/Enter key presses to my MFC dialog irrespective of whether its hidden or minimized.
Is there a way i can do that?
Why would you want/need to do this? After capturing them, are you planning on passing those keystrokes on so that the window with focus will behave as expected?
"One man's wage rise is another man's price increase." - Harold Wilson
"Fireproof doesn't mean the fire will never come. It means when the fire comes that you will be able to withstand it." - Michael Simmons
"Some people are making such thorough preparation for rainy days that they aren't enjoying today's sunshine." - William Feather
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Hello friends,
I am doing program in c++ . But I faced some problem to create dsn dynamically using SQLConfigDataSoure function. It does not create dsn...
Here is my following code
CString strAttribute ;
strAttribute.Format (_T("DSN=DSNPREVIOUSLOG\\0 FIL=MS Access\\0 UID= \\0 SafeTransactions=0 \\0DBQ=%s\0"),strDest);
SQLConfigDataSource(NULL,ODBC_ADD_SYS_DSN,_T("Microsoft Access Driver (*.mdb)"),strAttribute);
it returns false...
kindly help me ...
Thank u in advance
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shashankacharya wrote: it returns false...
So then call SQLInstallerError() to find out why.
"One man's wage rise is another man's price increase." - Harold Wilson
"Fireproof doesn't mean the fire will never come. It means when the fire comes that you will be able to withstand it." - Michael Simmons
"Some people are making such thorough preparation for rainy days that they aren't enjoying today's sunshine." - William Feather
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strAttribute is not correct. debug and watch it, it only contains "DSN=DSNPREVIOUSLOG"
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hi,
I wish to create a vpn server/client using winsock, I know how to create the winsock sever, just don't know the vpn side of it..
I wish to create a network device driver too, just start into WDK
Thanks in advance..
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Hi,
I'm trying to animate a box moving from point A to point B. I'm using a .Net timer to
animate but I'm not sure on the math to get the box moving in a straight line. So far
the box finishes moving down before it finishes moving to the right.
Point A
x=9,y=6
Point B
x=291,y=178
I'm not very good at math but I'm thinking the equation should be straight foreward?
Thanks in advance.
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1. Think about how fast you want the box to go - i.e. how many timer ticks should it take. For example:
NumberOfTicks = 10;
2. Look at the coordinates independently.
DistanceX = 291 - 9 = 282
DistanceY = 178 - 6 = 172
3. Calculate the step needed in each tick to get there in time. Careful! It may not be precise if you use ints - but that's not a problem:
StepX = 282 / NumberOfTicks ~= 28
StepY = 172 / NumberOfTicks ~= 17
4. Change the box position by the given step in each timer tick. Because the steps are rounded down (see step 3) you should also store the end position you want to get to and move the box there in the last step (not too elegant, but it will get the job done without anyone noticing).
Sure hope this helps, but seriously if this math is too complex to figure out you've got a long way to go my friend. Best of luck to you.
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Thanks for the reply. I still run into the problem in my first post. The horizontal movement finishes before the vertical.
I ran into this equation for a straight line.
y = m*x + b
but the speed of the box varies depending on the slope. Can't seem to fix this.
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Can you post the exact code which you use for the animation? Perhaps we can work from there
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This is what's called the explicit function for a straight line, meaning that you calculate one coordinate based on the value of the other coordinate. The explicit representation comes with a few shortcomings that can be confusing at times when you deal with 2D geometrical elements. For example you wouldn't be able to draw a vertical line with an equation like what you posted! Also, for near vertical lines you'd have to take care what values of x would produce a valid y - you'll find that most values of x that lie well within the screen corrdinate range will produce y values that are way outside the screen.
The implicit representation avoids this confusion, but it is often much easier to use the parametric representation: In this case, you have two points, A and B. The parametric representation of a line through these two points is:
P(t) = A + t*(B-A) . This equation is in fact a set of equations, referring to the two coordinates, x and y:
P(t).x = A.x + t*(B.x-A.x)
P(t).y = A.y + t*(B.y-A.y)
If you implement a function to calculate P(t), using the formulas above, each value of t will yield a point P(t) somewhere on this line. For t==0 you will find that P(0)==A, and for t==1 you will find that P(1)==B.
Of course, you have to be careful with these calculations when dealing with pixel coordinates: you have to convert these coordinates to float values first, or else the rounding effects may lead to some unexpected results.
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Thanks for the replies I finally got it working.
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