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Did you use the DC you got with wM_ERASEBKGND?
> The problem with computers is that they do what you tell them to do and not what you want them to do. <
> Life: great graphics, but the gameplay sux. <
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right now i'm using GetDC() on the ctreectrl but i could easily change that to use the one that came from WM_ERASEBKGND
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Then do so (there's a HDC in either wParam or lParam, look it up), that DC probably has a clipping region selected in it which will permit any paintsings onto areas which should not be overpainted.
> The problem with computers is that they do what you tell them to do and not what you want them to do. <
> Life: great graphics, but the gameplay sux. <
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cool
works perfect, thanks
one last question though, is there a way to get the rect for that region that updated?
cause i'm drawing bitmaps on the background after erasing and i need to calculate like sizes and positions and stuff for this updated region
thanks
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You could use CDC::GetClipBox[^] or CDC::GetBoundsRect[^] for that.
> The problem with computers is that they do what you tell them to do and not what you want them to do. <
> Life: great graphics, but the gameplay sux. <
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I have a SDI app with one single view, derived from CView at the execute time. now I want to create a splitter window at run time and make a new view of different class (see CRichEdit). But as I look inside the MSDN:
"In "dynamic" splitters, the views are of the same class, In "static" splitters, the views can be of different classes."
I dislike to have a splitter window initially (that's static). What do you suggest to do?
Thank you masters!
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Hello,
I would like to move something on a graph based on Left click mouse. This works fine, but How do I make it keep moving if the left Mmouse is still down, I want it to stop moving only when Left Mouse is up...as if I was dragging it... Do I need a timer to do this, or could this be done in other ways?
Thanks
sft
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You need to move it when the mouse is moving I guess ? If yes, when you click the button, set a flag to true, and when the button is released, reset the flag to false.
Handle every move event and check if the flag is set: if that's the case, then move your object.
If that doesn't answer your question, then please provide more information.
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Not a bad idea. I will give it a shot.
Thanks
sft
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WM_MOUSEMOVE
WM_LBUTTONUP
WM_LBUTTONDOWN
Thank you masters!
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Software2007 wrote: Do I need a timer to do this, or could this be done in other ways?
void CSam1Dlg::OnMouseMove(UINT nFlags, CPoint point)
{
if( MK_LBUTTON == (MK_LBUTTON&nFlags))
{
}
CDialog::OnMouseMove(nFlags, point);
}
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Its what I wanted - works fine - thanks.
sft
modified on Thursday, March 5, 2009 12:50 PM
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the two Different Code create the same result i want to know the
different between the operator '*' and '&'
the code is below:
first:
<br />
#include<iostream><br />
<br />
using namespace std;<br />
<br />
int Add(int &P1,int &P2)<br />
{<br />
P1=5;<br />
P2=10;<br />
return P1+P2;<br />
}<br />
<br />
int main()<br />
{<br />
int P1=2,P2=3;<br />
<br />
cout<<"As Parameter Before The P1 is:"<<P1<<" P2 is:"<<P2<<endl;<br />
<br />
int Sum=Add(P1,P2);<br />
cout<<"P1 Parameter After The P1 is: "<<P1<<" P2 is:"<<P2<<endl;<br />
}<br />
Second:
<br />
#include<iostream><br />
<br />
using namespace std;<br />
<br />
int Add(int *P1,int *P2)<br />
{<br />
*P1=5;<br />
*P2=10;<br />
return *P1+*P2;<br />
}<br />
<br />
int main()<br />
{<br />
<br />
int *P1,*P2;<br />
<br />
int PA=2,PB=3;<br />
<br />
P1=&PA;<br />
<br />
P2=&PB;<br />
<br />
cout<<"As Parameter Before The P1 is:"<<*P1<<" P2 is:"<<*P2<<endl;<br />
<br />
int Sum=Add(P1,P2);<br />
cout<<"P1 Parameter After The P1 is: "<<*P1<<" P2 is:"<<*P2<<endl;<br />
}<br />
end the variable P1 and P2 's value are all change .
could you tell me the difference between them
thank you 
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The difference is in how you will access the variable in your function: if you pass the variables by reference (using &), you can access the variables directly. If you pass the pointer to the variables (using *), then in your function you will receive the pointer.
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It is more neat sometimes to use &.
Take this simple example of a multiply function.
Using *
double multiply(double *x, double *y)
{
return *x * *y;
}
double d = multiply(&a, &b);
Using &
double multiply(double &x, double &y)
{
return x * y;
}
double d = multiply(a, b);
«_Superman_»
I love work. It gives me something to do between weekends.
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* is backward compatible with C. 
Edit: I was off by one character, I meant to indicate that C doesn't support references.
modified on Thursday, March 5, 2009 9:56 AM
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You can use pointers in C too (thus using the '*' is valid in C).
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Ah, I'm not awake yet, coffee just finished dripping.
I must have meant it the other way around. I only dabbled in C++ a little, and not recently.
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I can't believe there're people who know C and not C++ still! 
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C++ is all hype; it's not necessary for most real work.
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Is all the position the '*' use can replace by '&'? 
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in a declaration ie int *X means a a pointer to x named *;
You cannot define a reference with initializing it
so
int &X //doesn't work however
---
int Y;
int &x = Y; does
Y and X now refer to the same place in memory changing the value of one will change the value of the other. When you use a & in a function prototype we are saying I want my parameter to access memory at the location of the variable the caller specified; When you declare a pointer in the prototype you are going to explicitly pass in a pointer that pointer can however be reassigned but once reassinged the caller will no longer be able to read changes to what was passed in
I wrote a little example.
<br />
<br />
#include "stdafx.h"<br />
<br />
<br />
void NoSideEffect(int *T)<br />
{<br />
<br />
int G = 5;<br />
T = &G;<br />
<br />
<br />
}<br />
<br />
void SideEffect(int &T)<br />
{<br />
int G = 5;<br />
T = G;<br />
}<br />
<br />
void PointerSideEffect(int *T)<br />
{<br />
int G = 10;<br />
*T = G;<br />
}<br />
<br />
<br />
<br />
int _tmain(int argc, _TCHAR* argv[])<br />
{<br />
char buffer[2];<br />
int X = 2;<br />
NoSideEffect(&X);<br />
printf("X is Now %d \n",X);<br />
SideEffect(X);<br />
printf("X is Now %d \n",X);<br />
PointerSideEffect(&X);<br />
printf("X is Now %d \n",X);<br />
gets(buffer);<br />
}<br />
<br />
this will print out
X is Now 2
X is Now 5
X is Now 10
a programmer traped in a thugs body
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<br />
void NoSideEffect(int *T)<br />
{<br />
<br />
int G = 5;<br />
T = &G;<br />
<br />
<br />
}<br />
int X = 2;<br />
NoSideEffect(&X);<br />
printf("X is Now %d \n",X);<br />
X is Now 2
don't change why?could you explain more detail 
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This is because c++ accurately always passes by value When you pass a pointer you are passing the value of the pointer which is a memory address. when the compiler creates a stack frame the value of the memory address is put on the stack as a local variable when you change the value of that variable you point it to a new memory location and no longer maps to memory address of the variable in the caller. So any change will not affect the caller since you are changing the location associated with another variable in this case another local variable which is allso on the stack.
a programmer traped in a thugs body
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