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yeah, in my application also there is one main message loop and few other make shift message loop.
MSN Messenger.
prakashnadar@msn.com
Tip of the day of visual C++ IDE.
"We use it before you do! Visual C++ was developed using Visual C++"
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You can have as many message loops in your thread as you wish. I commonly have three or four. Another question is how many message queues can a thread have. The answer to that is only one.
When you display a dialog (using standard Windows), it uses it's own internal message loop to process the message while your main application has the primary message loop used for normal processing.
Often, when an application needs to process messages during an operation that will take a lot of time, another message loop is used to pump the messages from the message queue.
So as you can see, you can have many message loops. But you usually only worry about the primary message loop.
Tim Smith
I'm going to patent thought. I have yet to see any prior art.
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When you display a dialog (using standard Windows), it uses it's own internal message loop to process the message while your main application has the primary message loop used for normal processing.
Hi Tim,
Is this the case with both Modal and Modeless Dialog Boxes ?? How it is actually implemented?? What are the classes involved in doing this?? How we can create multiple message loops and link them with the application.
Regards
Satya
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Hi,
I want to copy the shortcut of an executable to a different location, from my program. Is there any function to do that? I have used CopyFile() to copy the exe, but how to create the shortcut to the exe and send/paste it to a particular location?
Kindly help.
Thanks..
Vini
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http://www.codeproject.com/shell/create_shortcut.asp
greatest thing is to do wot others think you cant
suhredayan@omniquad.com
messenger :suhredayan@hotmail.com
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hi,
you can use the following code to create a shortcut instead of copying an existing one. below code will create a shortcut for notepad in c:\ . Change the path accordingly.
CString strFile("notepad.exe");<br />
IShellLink* pLink;<br />
IPersistFile* pPersistFile;<br />
<br />
<br />
if(FAILED(CoInitialize(NULL))) throw CString("Failed to initialize COM Library.");<br />
<br />
if(FAILED(CoCreateInstance(CLSID_ShellLink, NULL, CLSCTX_INPROC_SERVER, IID_IShellLink, (void **) &pLink)))<br />
throw CString("Failed to initialize COM Library.");<br />
<br />
pLink->SetPath((LPCTSTR)strFile);<br />
pLink->SetDescription("Notepad fooooooooooo");<br />
pLink->SetShowCmd(SW_SHOW);<br />
if(FAILED(pLink->QueryInterface(IID_IPersistFile, (void **)&pPersistFile))) <br />
throw CString("Failed to query IID_IPersistFile.");<br />
CString str("c:\\notepad.lnk");<br />
CComBSTR path;<br />
path = str.AllocSysString(); <br />
pPersistFile->Save(path, true);<br />
pPersistFile->Release();<br />
pLink->Release();<br />
<br />
CoUninitialize();
rgds... mil10.
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Got it... Thanks so much.  ![Rose | [Rose]](https://codeproject.freetls.fastly.net/script/Forums/Images/rose.gif)
Vini
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What are the #include's needed for this code?
I'm compiling for Win32 and I haven't been able to figure out the right combination of files
thanks
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Hey everyone,
I'm busy working on a mini POS system, and they have an Epson LX400 printer that they use to print till slips. They feed a 7.5cm wide roll of paper through for the till slip. The problem arises when I use GDI printing and call EndPage(), the printer feeds the equivalent of an A4 page upwards. I've tried modifying the DEVMODE structure before I call CreateDC() but nothing works. How do I stop the printer from form feeding?
The software needs to run on Win98, so I can't use AddForm(), GetForm() etc.
Any ideas?
Regards,
Peter
PS If I use standard fopen and fprintf functionality, everything is fine, but then I can't get the client's logo onto the slip, so thats not an option.
controlSHIFT
[Glossary Manager] [AfterThought Backup Lite]
99 little bugs in the code, 99 little bugs,
Fix 1 bug, recompile....
101 little bugs in the code...
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Peter
You say you've tried modifying DEVMODE, presumably this is to set a custom form size, by setting dmPaperSize to zero, and setting dmPaperLength and dmPaperWidth ? If so, are you setting DM_PAPERSIZE|DM_PAPERWIDTH|DM_PAPERLENGTH in dmFields ?
Your other alternative is to get the logo onto the slip; I'm assuming the logo is monochrome (not sure if it's safe to use the B and W words these days), and you have a bitmap of it. AFAICR, you can do limited graphics printing with an LX series by sending escape sequences. You might be able to do a generalised PrintBitMap method for yourself instead. It sounds painful, but might be easier than trying to figure out what's wrong with the GDI version.
Steve S
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Hi,
Can somebody tell me how to draw a filled triangle. Our requirement is to draw a filled triangle as an arrow head above a circle. and depending on the direction, i need to change the position of the triangle.
The inputs given are direction of the arrow, centre and radius of the circle.
I will be thankful if somebody can suggest any algorithm for that.
Regards,
P. Vamsy Krishna
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You clearly need to use trig to calculate the points of the triangle, then you can use ExtFloodFill to fill it, if the background is a uniform color. Actually, it may work either way, I seem to recall some funky parameter that allowed that.
Christian
I have drunk the cool-aid and found it wan and bitter. - Chris Maunder
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Hi Christian,
Thanks for your immediate reply..
I am new to the graphics programming..so, I am not aware of ExtFloodFill.Can you please explain about that.
Tell me where can i get the algo for that.
Regards,
P Vamsy Krishna
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*sigh*
CDC::ExtFloodFill See Also
CDC Overview | Class Members | Hierarchy Chart | CDC::FloodFill | CDC::GetDeviceCaps | ExtFloodFill
Fills an area of the display surface with the current brush.
BOOL ExtFloodFill(
int x,
int y,
COLORREF crColor,
UINT nFillType
);
Parameters
x
Specifies the logical x-coordinate of the point where filling begins.
y
Specifies the logical y-coordinate of the point where filling begins.
crColor
Specifies the color of the boundary or of the area to be filled. The interpretation of crColor depends on the value of nFillType.
nFillType
Specifies the type of flood fill to be performed. It must be either of the following values:
FLOODFILLBORDER The fill area is bounded by the color specified by crColor. This style is identical to the filling performed by FloodFill.
FLOODFILLSURFACE The fill area is defined by the color specified by crColor. Filling continues outward in all directions as long as the color is encountered. This style is useful for filling areas with multicolored boundaries.
Return Value
Nonzero if the function is successful; otherwise 0 if the filling could not be completed, if the given point has the boundary color specified by crColor (if FLOODFILLBORDER was requested), if the given point does not have the color specified by crColor (if FLOODFILLSURFACE was requested), or if the point is outside the clipping region.
Remarks
This member function offers more flexibility than FloodFill because you can specify a fill type in nFillType.
If nFillType is set to FLOODFILLBORDER, the area is assumed to be completely bounded by the color specified by crColor. The function begins at the point specified by x and y and fills in all directions to the color boundary.
If nFillType is set to FLOODFILLSURFACE, the function begins at the point specified by x and y and continues in all directions, filling all adjacent areas containing the color specified by crColor.
Only memory-device contexts and devices that support raster-display technology support ExtFloodFill. For more information, see the GetDeviceCaps member function.
As I said, the nFillType variable allows you to fill an area with differing colours inside it, so long as none of them are the colour your shape is bound by.
msdn.microsoft.com[^]
Christian
I have drunk the cool-aid and found it wan and bitter. - Chris Maunder
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You are far too kind Christian. In my opinion, using the resources available is one of the most important things one needs to learn. He can look it up on MSDN as easy as you can.
a two cent stamp short of going postal.
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Just out of interest, isn't a triangle a polygon? So (ignoring the trig to calculate the triangle points at the right places) isn't it just a simple matter of selecting a solid brush (or whatever you want) and rendering the polygon?
Steve S
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Nice solution
Ant.
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Good point I have to admit that I've not done anything with GDI for a good two years now. 
Christian
I have drunk the cool-aid and found it wan and bitter. - Chris Maunder
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I've got this code, which gets words from a file into a char matrix([20][50]).
But as it's a 50 long string, what i get for example is this:
"hello (lots of unwanted spaces) "
so i've written some code to put 0 (\0) when the first space starts (' '), but it doesn't do the job. I'd want the string to be like this:
"hello"
so i can compare it, but it stays like this:
"hello (lots of unwanted spaces) "
Here's the code? what could i be doing wrong?
while ((palabras[j][0]!='\n') && (!(feof(archini))))
{
j++;
fgets(palabras[j],50,archini);
}
for (k=0;k<j;k++)
{
i=0;
while (palabras[k][i]!=' ')
{
i++;
}
palabras[k][i]=0;
}
Thanks!!!
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This is terribly ugly. Are you using C ? If not, start to use std::string to hold your strings, and I'd also use a vector for your array.
Christian
I have drunk the cool-aid and found it wan and bitter. - Chris Maunder
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Hello,
It's been a while since I learned about recursion in Java. Can somebody explain exactly how recursion works in C++? (ie. A recursive function call)
Thanks.
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Recursion
A function can call itself. This is called recursion, and recursion can be direct or indirect. It is direct when a function calls itself; it is indirect recursion when a function calls another function that then calls the first function.
Some problems are most easily solved by recursion, usually those in which you act on data and then act in the same way on the result. Both types of recursion, direct and indirect, come in two varieties: those that eventually end and produce an answer, and those that never end and produce a runtime failure. Programmers think that the latter is quite funny (when it happens to someone else).
It is important to note that when a function calls itself, a new copy of that function is run. The local variables in the second version are independent of the local variables in the first, and they cannot affect one another directly, any more than the local variables in main() can affect the local variables in any function it calls, as was illustrated in Listing 5.4.
To illustrate solving a problem using recursion, consider the Fibonacci series:
1,1,2,3,5,8,13,21,34...
Each number, after the second, is the sum of the two numbers before it. A Fibonacci problem might be to determine what the 12th number in the series is.
One way to solve this problem is to examine the series carefully. The first two numbers are 1. Each subsequent number is the sum of the previous two numbers. Thus, the seventh number is the sum of the sixth and fifth numbers. More generally, the nth number is the sum of n - 2 and n - 1, as long as n > 2.
Recursive functions need a stop condition. Something must happen to cause the program to stop recursing, or it will never end. In the Fibonacci series, n < 3 is a stop condition.
The algorithm to use is this:
1. Ask the user for a position in the series.
2. Call the fib() function with that position, passing in the value the user entered.
3. The fib() function examines the argument (n). If n < 3 it returns 1; otherwise, fib() calls itself (recursively) passing in n-2, calls itself again passing in n-1, and returns the sum.
If you call fib(1), it returns 1. If you call fib(2), it returns 1. If you call fib(3), it returns the sum of calling fib(2) and fib(1). Because fib(2) returns 1 and fib(1) returns 1, fib(3) will return 2.
If you call fib(4), it returns the sum of calling fib(3) and fib(2). We've established that fib(3) returns 2 (by calling fib(2) and fib(1)) and that fib(2) returns 1, so fib(4) will sum these numbers and return 3, which is the fourth number in the series.
Taking this one more step, if you call fib(5), it will return the sum of fib(4) and fib(3). We've established that fib(4) returns 3 and fib(3) returns 2, so the sum returned will be 5.
This method is not the most efficient way to solve this problem (in fib(20) the fib() function is called 13,529 times!), but it does work. Be careful: if you feed in too large a number, you'll run out of memory. Every time fib() is called, memory is set aside. When it returns, memory is freed. With recursion, memory continues to be set aside before it is freed, and this system can eat memory very quickly. Listing 5.10 implements the fib() function.
--------------------------------------------------------------------------------
WARNING: When you run Listing 5.10, use a small number (less than 15). Because this uses recursion, it can consume a lot of memory.
--------------------------------------------------------------------------------
Listing 5.10. Demonstrates recursion using the Fibonacci series.
1: // Listing 5.10 - demonstrates recursion
2: // Fibonacci find.
3: // Finds the nth Fibonacci number
4: // Uses this algorithm: Fib(n) = fib(n-1) + fib(n-2)
5: // Stop conditions: n = 2 || n = 1
6:
7: #include <iostream.h>
8:
9: int fib(int n);
10:
11: int main()
12: {
13:
14: int n, answer;
15: cout << "Enter number to find: ";
16: cin >> n;
17:
18: cout << "\n\n";
19:
20: answer = fib(n);
21:
22: cout << answer << " is the " << n << "th Fibonacci number\n";
23: return 0;
24: }
25:
26: int fib (int n)
27: {
28: cout << "Processing fib(" << n << ")... ";
29:
30: if (n < 3 )
31: {
32: cout << "Return 1!\n";
33: return (1);
34: }
35: else
36: {
37: cout << "Call fib(" << n-2 << ") and fib(" << n-1 << ").\n";
38: return( fib(n-2) + fib(n-1));
39: }
40: }
Output: Enter number to find: 5
Processing fib(5)... Call fib(3) and fib(4).
Processing fib(3)... Call fib(1) and fib(2).
Processing fib(1)... Return 1!
Processing fib(2)... Return 1!
Processing fib(4)... Call fib(2) and fib(3).
Processing fib(2)... Return 1!
Processing fib(3)... Call fib(1) and fib(2).
Processing fib(1)... Return 1!
Processing fib(2)... Return 1!
5 is the 5th Fibonacci number
Analysis: The program asks for a number to find on line 15 and assigns that number to target. It then calls fib() with the target. Execution branches to the fib() function, where, on line 28, it prints its argument.
The argument n is tested to see whether it equals 1 or 2 on line 30; if so, fib() returns. Otherwise, it returns the sums of the values returned by calling fib() on n-2 and n-1.
In the example, n is 5 so fib(5) is called from main(). Execution jumps to the fib() function, and n is tested for a value less than 3 on line 30. The test fails, so fib(5) returns the sum of the values returned by fib(3) and fib(4). That is, fib() is called on n-2 (5 - 2 = 3) and n-1 (5 - 1 = 4). fib(4) will return 3 and fib(3) will return 2, so the final answer will be 5.
Because fib(4) passes in an argument that is not less than 3, fib() will be called again, this time with 3 and 2. fib(3) will in turn call fib(2) and fib(1). Finally, the calls to fib(2) and fib(1) will both return 1, because these are the stop conditions.
The output traces these calls and the return values. Compile, link, and run this program, entering first 1, then 2, then 3, building up to 6, and watch the output carefully. Then, just for fun, try the number 20. If you don't run out of memory, it makes quite a show!
Recursion is not used often in C++ programming, but it can be a powerful and elegant tool for certain needs.
--------------------------------------------------------------------------------
NOTE: Recursion is a very tricky part of advanced programming. It is presented here because it can be very useful to understand the fundamentals of how it works, but don't worry too much if you don't fully understand all the details.
--------------------------------------------------------------------------------
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AAMOI, where did you cut that from?
Steve S
(I am not a copyright lawyer)
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it's from C++ in 21 days
is it allowed to copy some parts of the book??
if not, how cares? 
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Why don't you just post a link next time?
http://newdata.box.sk/bx/c/htm/ch05.htm#Heading43
"The pointy end goes in the other man." - Antonio Banderas (Zorro, 1998)
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