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Nope. it does not help.I don't understand what's the problem 
i can resize it vertically only if it's type in properties is set to "Simple" , but thats not what i need i need "DropList" style
m0n0
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Go to the resource view and click on the dropdown button of the combo box. You can resize it vertically.
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I have a modeless dialog, when one of its button is pressed, I called a subroutine to load a file. Since the operation is lengthy, I want to give user some feedback that sth is going on. So I want to show a text string. I did the following:
<quote>
CStatic textProgress;
textProgress.Create(_T("Image loading..."), WS_CHILD | WS_VISIBLE|SS_CENTER,
CRect(CPoint(rectDlg.left+nOffset, rectDlg.CenterPoint().y-2*nOffset),
CSize(rectDlg.Width()-2*nOffset, nOffset)),
pParent);
but when the system runs, I saw the rectangle at the position specified, but no text. why?
Thanks for any help!
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Try this instead of pParent.
Haakon S.
A sure cure for seasickness is to sit under a tree.
Spike Milligan
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I don't have access to "this", since it's not a member function, but a global function.
I also have a CProgressCtrl in this global function, it's being showed ok.
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Did you try updating the static control?
Is this static control in another thread? If not then it needs to be repainted to show the text. like: mystatic.RedrawWindow() etc.
this is this.
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great! RedrawWindow() showed my text.
Thanks a lot! ![Rose | [Rose]](https://codeproject.global.ssl.fastly.net/script/Forums/Images/rose.gif)
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Running my program in debug mode I get the following three lines at the end of the output window:
The thread 0xFFEA33E3 has exited with code -1 (0xFFFFFFFF).<br />
The thread 0xFFEA76CF has exited with code 0 (0x0).<br />
The program 'D:\Prosjekter\MyProg\MyProg\Debug\MyProg.exe' has exited with code 0 (0x0).
So, what do I make of this? The last two codes indicates a succesfull run, but not the first. I just can't find out what it means.
Regards, Haakon.
A sure cure for seasickness is to sit under a tree.
Spike Milligan
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Did you create thread 0xFFEA33E3? If so, then you should be able to control its exit code.
"Ideas are a dime a dozen. People who put them into action are priceless." - Unknown
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No, I didn't. My program is single threaded. I guess one of the threads has to do with the debug routines.
Regards, Haakon.
A sure cure for seasickness is to sit under a tree.
Spike Milligan
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Sometimes threads also come and go if you use some of the common dialogs, such as the file open dialog (CFileDialog). IF it was not a thread you created, you usually don't have to worry about it.
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This may seem like a stupid question, but i simply don't know where to look. How can I execute a console program from another console program and catch the output so i can display it in my own program. Thing is I want to have a file compiled but i want to catch the output from the compiler binary in case of errors. The only functions I know to execute something are system() and ShellExecute() but I don't know how to catch the output after that.
Kuniva
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Kuniva wrote:
How can I execute a console program from another console program and catch the output so i can display it in my own program.
You can read here and here.
"Ideas are a dime a dozen. People who put them into action are priceless." - Unknown
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I have the following two functions:
unsigned PullTerminalId(unsigned long id)
{
return ((unsigned)(id >> 20));
}
unsigned long get_human_check_id(unsigned long id)
{
unsigned long terminalId = (unsigned long)PullTerminalId(id) * 10000;
unsigned long checkNbr = (unsigned)(id & 0xFFFF);
return terminalId + checkNbr;
}
As an example, if I call get_human_check_id with a value of 1048577, then the result is 10001
Could someone show me how to reverse this process?
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i'm not sure you can.
one major problem is that the algorithm doesn't preserve all of the information in the original id. it looks like at least 4 bits of the original data are thrown away completely: in effect, the ">> 20" throws away 20 bits, but the "& 0xffff" recovers 16 of them - the final number is still missing the information that was contained in bits 16..19.
that's before you even look at the arithmetic...
Cleek | Image Toolkits | Thumbnail maker
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[Tom Archer]
I think I got it. First, the 4 missing bits are not used in the system, which is why he's not accounting for them. Now that we know that the following function should work (I think)
unsigned long get_human_check_id_rev(unsigned long id)
{
unsigned long rev1 = id / 10000;
rev1 = rev1 << 20;
unsigned long rev2 = (unsigned)(id & 0x01);
return rev1 + rev2;
}
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i'm not sure that's gonna do it either.
in the encoding, the information in the high 12 bits is combined with the low 16 bits through addition. seperating them is equivalent to finding which two numbers were added to produce "1234".
just for a test:
<br />
UINT in = 0xffffffff;<br />
TRACE("%in: s\n", BinDump((BYTE *)&in, 4));<br />
UINT hid = get_human_check_id(in);<br />
UINT out = get_human_check_id_rev(hid);<br />
TRACE("out: %s\n", BinDump((BYTE *)&out, 4));<br />
<br />
outputs:<br />
in: 11111111 11111111 11111111 11111111 <br />
out: 00000001 00000000 01010000 00000000 <br />
Cleek | Image Toolkits | Thumbnail maker
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Except that we know *how* they were added. IOW, the get_human function basically isolates everything above 20 bits, multiplies it by 1000 and states that its the terminalID. Ex: 1048577 = terminalId of 1000.
It then takes bits 0-15 (hence the loss of 4 bits) and states that it's the checkNbr. Ex: 1048577 = checkNbr of 1
The two are added together. Ex: 1048577 == 1001
It seems like this should be reversible but I'm not seeing it.
Cheers,
Tom Archer - Archer Consulting Group
"Eat your brussel sprouts, Junior. There are starving Chinese children American programmers that would kill for that food!"
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I think I've got it now - proving that the get_human functoin can be reversed:
unsigned long get_human_check_id(unsigned long id)
{
unsigned long terminalId = (unsigned long)PullTerminalId(id) * 10000;
unsigned long checkNbr = (unsigned)(id & 0xFFFF);
return terminalId + checkNbr;
}
okay, lets break it down:
the terminal id part is in bits 20 and up, then scaled up by 10000
the checknumber is in the lower 16 bits.
So to reverse it you need to take :
terminalid / 10000, then shift it left 20 bits
plus
the checknumber.
My first attempt had roughly the right idea, except for that mask.
Cheers,
Tom Archer - Archer Consulting Group
"Eat your brussel sprouts, Junior. There are starving Chinese children American programmers that would kill for that food!"
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So to reverse it you need to take :
terminalid / 10000, then shift it left 20 bits
plus
the checknumber
where do you get the checknumber ?
Cleek | Image Toolkits | Thumbnail maker
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Check number is bits 0-15 (id & 0xFFFF)
Therefore,
* terminalId is bits 20 and above (multiplied by 1000)
* checkNbr is bits 0-15
The only bits lost are bits 16-19, which are never used.
Cheers,
Tom Archer - Archer Consulting Group
"Eat your brussel sprouts, Junior. There are starving Chinese children American programmers that would kill for that food!"
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Tom Archer wrote:
terminalId is bits 20 and above (multiplied by 1000)
but those bits are shift down 20 (which is equivalent to division by 1048576). they're now bits 0..12 of terminalId.
* checkNbr is bits 0-15
and those two groups of bits are combined by the addition step. they don't occupy distinct bits in the result.
Cleek | Image Toolkits | Thumbnail maker
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That's why the programmer multiplies terminalId by 1000; so that he can add the checkNbr without overlapping the terminalId digits.
Cheers,
Tom Archer - Archer Consulting Group
"Eat your brussel sprouts, Junior. There are starving Chinese children American programmers that would kill for that food!"
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