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Blake Miller wrote:
There is typically (or used to be) limit of 255 controls in one window.
Ummm nope, there never was. Visual C++'s dialog editor has a limit on how many controls it supports on a dialog (255), but the limit as to the number of controls Windows supports is limited only by physical memory and the number of available window handler.
Ryan "Punctuality is only a virtue for those who aren't smart enough to think of good excuses for being late" John Nichol "Point Of Impact"
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I beg to differ.
A BYTE was used, previously, to store the control's identifier, thus limiting the number of child windows to 255 in the resource template. Since most people used dialog templates and not in-memory dynamic resource tempaltes, they were 'essentially' limited to 25 controls on a single dialog.
Of course, if you are just building windows on the fly, you are limited only by memory or GDI resources, or some other commodity you will eventually deplete.
Another thing is, even though the current windows systems are typically entirely 32 bit environments, there are still many 'window identifiers' passed around as the upper or lower 16-bits of various window messages, so you are, in some cases, limited to 65535 windows, depending upon your circumstances.
As far as 'child controls' the window control child identifier is a 16 bit number, so even if you made more than 65535 child control windows, you would only be able to distinguich 65535 of them from each other. And you would also run into the messaging LOWOR/HIWORD problem. Just see the wParam of the WM_COMMAND message, and I quote MSDN:
The high-order word specifies the notification code if the message is from a control. If the message is from an accelerator, this value is 1. If the message is from a menu, this value is zero.
The low-order word specifies the identifier of the menu item, control, or accelerator.
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Blake Miller wrote:
I beg to differ.
How is that different from what I said?
Ryan "Punctuality is only a virtue for those who aren't smart enough to think of good excuses for being late" John Nichol "Point Of Impact"
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"...the limit as to the number of controls Windows supports is limited only by physical memory and the number of available window handler."
The number of window handles is not the same as the number of available control identifiers. The limit of HWND is not the same as the numerical limit of UINT control identifiers. That is the difference I perceived between what I had said and what you had said.
AND since the original poster was asking about dialogs and controls on dialogs, it made sense to be very specific as to what limitations exist and what issues he might run into.
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Blake Miller wrote:
The limit of HWND is not the same as the numerical limit of UINT control identifiers.
While we're being pedantic, I'll mention that it is for Win95, 98 and ME - where only 65535 unique HWNDs are available (even though an HWND is 32 bits). Besides, control identifiers only have to be unique within a single parent window. Two different parent windows can have controls with the same identifiers. So if the dialog includes child windows that contain other controls, the limit on identifiers is not really relevant.
Ryan "Punctuality is only a virtue for those who aren't smart enough to think of good excuses for being late" John Nichol "Point Of Impact"
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Blake Miller wrote:
There is typically (or used to be) limit of 255 controls in one window.
That limit was imposed by the resource editor, not dynamically-created controls.
"Opinions are neither right nor wrong. I cannot change your opinion. I can, however, change what influences your opinion." - David Crow
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I have following code in a cpp file:
//test.cpp
int __cdecl calc1 (int a, int b, int c) {
return (a+b-c);
}
int __cdecl calc2 (int a, int b) {
return (a+b);
}
void main() {
int r = calc1(10,20,5);
r = calc2(10,20);
}
I have compiled this with 'cl' command. Then, with "dumpbin /disasm test.obj command I created following file: (I have removed the headers here)
?calc1@@YAHHHH@Z (int __cdecl calc1(int,int,int)):
00000000: 55 push ebp
00000001: 8B EC mov ebp,esp
00000003: 8B 45 08 mov eax,dword ptr [ebp+8]
00000006: 03 45 0C add eax,dword ptr [ebp+0Ch]
00000009: 2B 45 10 sub eax,dword ptr [ebp+10h]
0000000C: 5D pop ebp
0000000D: C3 ret
?calc2@@YAHHH@Z (int __cdecl calc2(int,int)):
0000000E: 55 push ebp
0000000F: 8B EC mov ebp,esp
00000011: 8B 45 08 mov eax,dword ptr [ebp+8]
00000014: 03 45 0C add eax,dword ptr [ebp+0Ch]
00000017: 5D pop ebp
00000018: C3 ret
_main:
00000019: 55 push ebp
0000001A: 8B EC mov ebp,esp
0000001C: 51 push ecx
0000001D: 6A 05 push 5
0000001F: 6A 14 push 14h
00000021: 6A 0A push 0Ah
00000023: E8 00 00 00 00 call 00000028
00000028: 83 C4 0C add esp,0Ch
0000002B: 89 45 FC mov dword ptr [ebp-4],eax
0000002E: 6A 14 push 14h
00000030: 6A 0A push 0Ah
00000032: E8 00 00 00 00 call 00000037
00000037: 83 C4 08 add esp,8
0000003A: 89 45 FC mov dword ptr [ebp-4],eax
0000003D: 8B E5 mov esp,ebp
0000003F: 5D pop ebp
00000040: C3 ret
On line 00000023 in the main function it calls calc1 function,
But no where address of calc1 (00000000) is mentioned.
The same case happens when call to calc1 (0000000E) is made.
So, how the control will be transferred to respective functions?
Could anyone help me in understanding the working of call instruction and explain the calling of proper function in the above case?
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The first function called
r = calc1(10,20,5); becomes
0000001D: 6A 05 push 5 - Parameter 5
0000001F: 6A 14 push 14h - Parameter 20
00000021: 6A 0A push 0Ah - Parameter 10
00000023: E8 00 00 00 00 call 00000028
00000028: 83 C4 0C add esp,0Ch - Alter the next execution
0000002B: 89 45 FC mov dword ptr [ebp-4],eax - Get the result
The second function called
r = calc2(10,20); becomes
0000002E: 6A 14 push 14h - Parameter 20
00000030: 6A 0A push 0Ah - Parameter 10
00000032: E8 00 00 00 00 call 00000037
00000037: 83 C4 08 add esp,8 - Alter the next execution
0000003A: 89 45 FC mov dword ptr [ebp-4],eax - Get the result
When the function is called epb holds the return address. You will notice that this needs to be preserved pushed and popped at start and end of function respectively.
Ant.
I'm hard, yet soft.
I'm coloured, yet clear.
I'm fruity and sweet.
I'm jelly, what am I? Muse on it further, I shall return! - David Walliams (Little Britain)
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Rajas P wrote:
Could anyone help me in understanding the working of call instruction and explain the calling of proper function in the above case?
The problem is that you've dumped the object file rather than the executable. When the compiler compiles the file, it doesn't know where in memory the function is going to be located, so it generates a call instruction with a dummy address in it (in this case, 0x0000000), and generates a separate FIXUP record in the COFF object file that tells the linker to fill in the correct address when it links the object files and determines where in memory everything is going to be located. This is why the two calls are actually referencing the same address in the disassembly above - neither of them have had the actual code location inserted by the linker yet.
Ryan "Punctuality is only a virtue for those who aren't smart enough to think of good excuses for being late" John Nichol "Point Of Impact"
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I've to develop a SNTP server. I know the SNTP packet structure, and the SNTP port.
Is the SNTP server a simple server socket that replies to clients with an updated SNTP packet or there is something
more ???
Thanks a lot.
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Good morning everyone,
I am developing a small application at my work that reads data from a SQL Server database using CDatabase.
The problem that I am having is that it works fine on 9 out of 10 computers (some are Windows XP w/out SP2, another has SP2, another has W2K). There are three Windows 2000 machine that I have tested this out on, and one of them does not work. All of the machines are joined to the domain, and can physically see and ping the server that the DB is on. The one machine that is not working out of the whole bunch is getting an error message. The code is -1 and the error string is "SQL Server Not Found". I have tried fully qualifying the domain name in the connection string (ie, SSV001.mydomain.com) but that one machine cannot connect to it. (The machine can access other servers on the network, so I know the network connection is good). Here's what I am using to connect
strConnection.Format("Driver={SQL Server};Server=SSV001.mydomain.com;Database=TESTDB;Uid=blah;Pwd=blahblah;"
CDatabase myDB;
TRY
{
myDB.OpenEx(strConnection,CDatabase::noOdbcDialog);
}
CATCH (CDBException,e)
{
CString errMsg.
errMsg.Format("Error code: %i. Error string: %s",e->m_nRetCode,e->m_strError);
}
9 of the machines open it fine, one of the machines triggers the catch block and shows the SQL Server not found error message.
Any help to resolve this would be appreciated a whole lot.
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What if you use the IP address instead of the host name (SSV001.mydomain.com)?
"Opinions are neither right nor wrong. I cannot change your opinion. I can, however, change what influences your opinion." - David Crow
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Thanks so much for the quick response.
I just tried replacing the host name w/ the IP address. It still does not work on that one PC  (It does work on the other ones though, so I know I have the IP Address correct).
One correction: The error message is actually "Specified SQL Server Not Found"
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If you change noOdbcDialog to forceOdbcDialog, you should get an ODBC dialog. Is the information in it correct?
"Opinions are neither right nor wrong. I cannot change your opinion. I can, however, change what influences your opinion." - David Crow
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Hello to all.
I use functions FindText() and FindText2() for class CHtmlView to organize searching for text on current page. But I need to search on different pages (I have their URL's).
Had anybody faced with this problem before?
I'll be very appreciate for every post below.
Aleksey.
Sorry for my poor English
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the type of Report for ListCtrl
LVCFMT_CENTER have been setted ,but the first colum shows like LVCFMT_LEFT
How can I make the text to the center of the head?
Thanks a lot!
E-mail:
wushlimail@163.com
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How are you inserting that column?
"Opinions are neither right nor wrong. I cannot change your opinion. I can, however, change what influences your opinion." - David Crow
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Remember that the CListCtrl methods are zero based. So the first column starts at 0.
Maybe thats the problem.
I Dream of Absolute Zero
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This is a well-known bug with the list control. The first column is always left-justified. Solution? Create an extra column as the first, and then delete it after you've added the second. Now the second is the first, actually uses the alignment you set for it, and everyone is happy.
(For reference: Microsoft's wonderfully-helpful writeup  )
Shog9
I'm not the Jack of Diamonds... I'm not the six of spades.
I don't know what you thought; I'm not your astronaut...
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I have a problem making a program, that takes one fraction and simplifies it....
for example the fraction (2/4) is simplified to (1/2)....
that´s my small problem..
my larger problem is then adding/substracting two fractions....
With previous thanks...
JohnnyIce
C++ Beginner
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you need to define a "fraction" structure to hold the numerator and the denominator, and do basic operations as defined there link at mathworld[^]
and to reduce the fraction you need to find the greatest common divisor of the numerator and denominator.
( I will let you have a look at it ... since my connection to mathworld is really slow this morning)
Maximilien Lincourt
Your Head A Splode - Strong Bad
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thanks..
but i know the mathmatical definitions of fractions.....
my largest problem is to put it on C++ form...
So that i can simplify fractions.....
My weakest link in this chain, is the programming it self...
Having the program figuring out the smallest fraction of f.example (2/4).
with previous thanks..
JohnnyIce
C++ Beginner
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To simplify, or reduce, 2/4, you must divide both the numerator and denominator by their greatest common divisor, which is 2. To simplify 12/16, you must divide both the numerator and denominator by their GCD, which is 4, yielding 3/4.
Here is a code snippet:
CString strFraction = "12/16";
int nSolidus = strFraction.Find('/');
int nNumerator = atoi(strFraction.Left(nSolidus));
int nDenominator = atoi(strFraction.Mid(nSolidus + 1));
"Opinions are neither right nor wrong. I cannot change your opinion. I can, however, change what influences your opinion." - David Crow
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 well.....actually tis the programming part that comez to be easy.....
letz split that up ur problem into small functionalitiez.....
like said above u r to find the GCD.....
so from thiz u can conclude that there are three thingz u ought to do .....
1)
(a)write a function to find all the possible factorz of the numerator(place the factorz in an array say a[])
(b)write a function to find all the possible factorz of the denominator(place the factorz in an array say b[])
2)write a function which findz the greatest comman factor from the arrayz a & b
3)divide the numerator and denominator with this GCD......and voila....
u did it.....
Happy Programmin....
cheerz.....
"faith, hope, love remain, these three.....; but the greatest of these is love" -1 Corinthians 13:13
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