|
How and what you erase is entirely up to your program to decide.
Basically you need to write the erasing code in the WM_MOUSEMOVE[^] handler.
You get if the right mouse button is down by checking wParam for MK_RBUTTON.
And you will get the mouse coordinates from lParam.
«_Superman_»
I love work. It gives me something to do between weekends.
|
|
|
|
|
I know everything except how this erasing code should look like. So I beg for any help with this code.
And this is exactly what I want to do:
Using left button I'm drawing a sketch, but in case of a mistake I want to use right button to remove part of this sketch.
Thanks for any help.
|
|
|
|
|
kudlaty79 wrote: I know everything except how this erasing code should look like.
That's exactly what Superman was explaining to you: there's nothing like a standard erasing tool. You have to implement it yourself and how you're going to implement it is fully dependent on how you draw your data. So, if you want some suggestions here, you will need to explain in details how you draw your data.
|
|
|
|
|
Here is drawing code:
void COrder::OnLButtonDown(UINT nFlags, CPoint point)
{
if(drawing_status==1)
{
last_point = point;
}
CDialog::OnLButtonDown(nFlags, point);
}
void COrder::OnLButtonUp(UINT nFlags, CPoint point)
{
if(drawing_status==1)
{
UpdateWindow();
}
CDialog::OnLButtonUp(nFlags, point);
}
void COrder::OnMouseMove(UINT nFlags, CPoint point)
{
if(drawing_status==1)
{
CClientDC draw(this);
CPen pen;
if((nFlags & MK_LBUTTON)==MK_LBUTTON)
{
pen.CreatePen(PS_SOLID, 1, RGB(0,0,255));
draw.SelectObject(pen);
if(last_point.x<322 || last_point.x>626 || last_point.y<372 || last_point.y>596)
{
}
else
{
draw.MoveTo(last_point);
draw.LineTo(point);
}
last_point = point;
}
CDialog::OnMouseMove(nFlags, point);
}
|
|
|
|
|
Please use the pre tags to format your code properly as specified in the posting guidelines.
Now to your problem: you approach on drawing your data is wrong. You shouldn't draw directly on the screen in response to a mouse event. Doing so will causes many problems: try to put another window in front of your window (or minimizing your window) and then making your window visible again. You'll see that your drawings have disappeared.
What you should do instead is store all the points where you clicked with the mouse as pure data and refresh your view. The OnPaint handler will be called and there you need to redraw everything. Search for the scribble example on MSDN, this will show you exactly how to do so.
Once this is done, erasing becomes "easier" because you only need to remove points of data which are stored in your document. This depends a bit how you want your erase tool to work: remove a complete line, or only sections, ...
|
|
|
|
|
Hi,
I want to write some code which will double click a desktop shortcut. or the alternate way is how can i run a desktop shurtcut using Createprocess() api.
Which exe is responsible to lunch Start->Run window?
Thanks in advance..
Birajendu
SonicWALL
Bangalore
India
|
|
|
|
|
I see three separate (and unrelated) questions here? Which one are you interested in?
"Old age is like a bank account. You withdraw later in life what you have deposited along the way." - Unknown
"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
|
|
|
|
|
birajendu wrote: I want to write some code which will double click a desktop shortcut. or the alternate way is how can i run a desktop shurtcut using Createprocess() api.
If you know the path to the desktop shortcut, then you should be able to use ShellExecute to open it. If you need a process handle (the only reason I can think that you might want to use CreateProcess for this), then ShellExecuteEx will let you do that.
birajendu wrote: Which exe is responsible to lunch Start->Run window?
Windows Explorer.
Java, Basic, who cares - it's all a bunch of tree-hugging hippy cr*p
|
|
|
|
|
I tried with ShellExecute to open this, But for network connection shortcuts it seems not working. i found another thing about network shortcuts is, the target type is a GUID rather than the actual application.
The reason why i asked about the start->run is, if you drag the network shortcut in to start->run it functions perfactly. So can i implement some code to simulate that action?
Birajendu
SonicWALL
Bangalore
India
|
|
|
|
|
|
Do i have to use advance loops? I thought about it for quite a while but could not figure it out. Example
int numbs[4][4] =
[1,2,3,4
5,6,7,8
9,10,11,12
13,14,15,16];
when i print it, it should print like this.
1 2 3 4, then 8, 12,16, 15,14,13,9,5,67,11,10, (ie clockwise direction).
The function should be generic for any size matrix.Any help will be appreciated.
|
|
|
|
|
without actually writing the code, i would think something like this would work:
xMin = 0, xMax = arrayWidth
yMin = 0, yMax = arrayHeight
printed_something = false
do
{
loop across top edge from xMin to xMax, print the values
increase yMin
loop across right edge from yMin to yMax, print the values
decrease xMax
loop across bottom edge from xMax to xMin, print the values
decrease yMax
loop across left edge from yMax to yMin, print the values
increase xMin
} while (printed_something)
|
|
|
|
|
Consider this as a rough idea:
#include <stdio.h>
void PrintArrayClockwise(int *array, int rows, int columns)
{
int r = 0;
int rh = rows / 2;
int i;
while (r < rh)
{
for (i=r; i<(columns - r); i++) printf("%d ",array[(r*columns)+i]);
printf("\n");
for (i=r+2; i<=(rows-r); i++) printf("%d ",array[(i*columns)-1-r]);
printf("\n");
for (i=(columns-r-2); i>=r; i--) printf("%d ",array[((rows-r-1)*columns)+i]);
printf("\n");
for (i=rows-r-2; i>r; i--) printf("%d ",array[(i*columns)+r]);
printf("\n");
r++;
}
}
int main()
{
int numbs[16];
int i;
for (i=0; i<16; i++) numbs[i]=i+1;
PrintArrayClockwise(numbs, 4, 4);
return 0;
}
It should work with arrays of different sizes, both square and not. I didn't check it thoroughfully but it should be a good starting point. 
2+2=5 for very large amounts of 2
(always loved that one hehe!)
|
|
|
|
|
This works perfectly! I have not yet understood how it works though but will think on it.
I was thinking on the idea to print the top row of the matrix then rotate the matric anticlockwise while removing the top row and printing the next top row and so on. But then I needed a function to rotate the matrix anticlockwise, which was not easy to implement when you consider you have to eliminate the top row as well. Your code is much simpler and very neat. I still have to understand it fully though. What do you think about my idea. Can you explain a little bit of your code.
Thanks a lot for sharing
|
|
|
|
|
Happy to be of help.
About rotating the matrix, it would take more time and more space, so I think it's not a good solution.
Yeah I know that my code wasn't that readable, C pointer math can be hellish...
It works by using a single iterator on rows r. This iterator goes from the first row (0) to the middle row ((rows/2)-1).
First I print the row pointed by r:
for (i=r; i<(columns - r); i++) printf("%d ",array[(r*columns)+i]);
then I proceed printing the rightmost column, starting from the last item I printed for row r:
for (i=r+2; i<=(rows-r); i++) printf("%d ",array[(i*columns)-1-r]);
I continue with the lowest row, which is printed right to left:
for (i=(columns-r-2); i>=r; i--) printf("%d ",array[((rows-r-1)*columns)+i]);
and finally with the first column, bottom to top:
for (i=rows-r-2; i>r; i--) printf("%d ",array[(i*columns)+r]);
With each iteration r grows by one. You will see that it's used in all the for loops, in order to move them towards the centre of the matrix one row and one column each.
Hope it's more clear now, if you think it would help I can rewrite this using distinct row and column indexes on the array, in order to make it clearer.
2+2=5 for very large amounts of 2
(always loved that one hehe!)
|
|
|
|
|
Thanks for explaining. I got the idea. This is a beautiful logic but it also requires some arithmetic and pointer skills If you could implement it in general arrow notation, that would be great. But of course not required
You are absolutely right, the rotate array matrix approach would be too much waste of time and computationally expensive.
I think this is great logic. I would not have seen it unless it was implemented.
Thanks again. I greatly appreciate your help
BTW Brave for implementing it in such a short amount of time!
|
|
|
|
|
No problem, here is the version with separated row and column indexes:
#include <stdio.h>
void PrintArrayClockwise(int array[][4], int rows, int columns)
{
int r = 0;
int rh = rows / 2;
int i;
while (r < rh)
{
for (i=r; i<(columns - r); i++) printf("%d ",array[r][i]);
printf("\n");
for (i=r+1; i<(rows-r); i++) printf("%d ",array[i][columns-r-1]);
printf("\n");
for (i=(columns-r-2); i>=r; i--) printf("%d ",array[rows-r-1][i]);
printf("\n");
for (i=rows-r-2; i>r; i--) printf("%d ",array[i][r]);
printf("\n");
r++;
}
}
int main()
{
int numbs[4][4] = {{1,2,3,4},{5,6,7,8},{9,10,11,12},{13,14,15,16}};
PrintArrayClockwise(numbs, 4, 4);
return 0;
}
Unluckily there's no way of passing a bydimensional array to a function without indicating the second dimension, at least none that I know of (but my ansi C is VERY rusty hehe). It sure can be done with int** anyway.
2+2=5 for very large amounts of 2
(always loved that one hehe!)
|
|
|
|
|
Thanks again! I appreciate it. I am definitely a fan of your programming skills. How much of experience do you have in programming btw. You have personnel contact?
|
|
|
|
|
Huh well thanks, you are definitely overvauling me hehe.
I'm into programming ever since I was 11 (started with a Philips game console which could be programmed in a kind of pseudo-assembler, then a year later I passed to a C64 and so on - typical for my generation), and have been working as a software engineer ever since 1989. Quite some time I'd say. :P
Don't get me wrong, but I won't be giving my personal contact in a forum. I've nothing against you in particular, but it's simply not good practice!
If you ever need any more help post here. If I'll be able to help I'll gladly do it, or someone else will. Besides, forums are good also for other people, who can learn by reading the answers to your questions.
2+2=5 for very large amounts of 2
(always loved that one hehe!)
|
|
|
|
|
Good to know a little bit about you. I actually looked at your webpage on codeproject but there was not much there except that you are from Italy  Well thanks for the little background. Codeproject is my resource for technical questions so I will definitely post here if i need any help.
Thanks again.
|
|
|
|
|
Hi
I want to copy needed dlls into an autorun CD that runs a VC++ 2008 exe
I don't want to install the Redistributable Package.
I tested it with dependency walker and copied all needed dlls for MFC app, but it did not worked!
(but I do it easy for VC++2003 exe!)
What other thing I must do?
Thanks
www.logicsims.ir
|
|
|
|
|
build the program statically linked.
Project Properties / Config Properties / general / Use Of MFC = Use MFC in a static library
|
|
|
|
|
I know it, Thanks
but I don't want to link it statically.
www.logicsims.ir
|
|
|
|
|
This is all due to the horribly designed side-by-side crap Microsoft started pushing with Windows 2000 (though it didn't really work until XP.) With SxS, you have to put your DLLs in the correct install spots (meaning let Microsoft's installer do it.) There are other solutions, but I've forgotten the rules. Here's one article on the subject (I neither endorse or condemn this article--I breezed through it):
http://69.10.233.10/KB/cpp/PrivateAssemblyProjects.aspx[^]
Do searches on "side-by-side" and/or SxS with MFC and DLL and you will start getting hits [on the nightmare Microsoft created.]
|
|
|
|
|
Thank you so much 
www.logicsims.ir
|
|
|
|
|
General 
News 
Suggestion 
Question 
Bug 
Answer 
Joke 
Praise 
Rant 
Admin 
Submit an article or tip
