C / C++ / MFC

5-Mar-09 3:47

* is backward compatible with C. Big Grin | :-D

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

Cedric Moonen5-Mar-09 3:50

Cedric Moonen

5-Mar-09 3:50

You can use pointers in C too (thus using the '*' is valid in C).

Cédric Moonen
Software developer

Charting control [v1.5]
OpenGL game tutorial in C++

PIEBALDconsult5-Mar-09 3:55

5-Mar-09 3:55

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.

bulg5-Mar-09 7:02

bulg

5-Mar-09 7:02

I can't believe there're people who know C and not C++ still! Smile | :)

PIEBALDconsult5-Mar-09 10:38

5-Mar-09 10:38

C++ is all hype; it's not necessary for most real work. Big Grin | :-D

jeansea5-Mar-09 4:14

jeansea

5-Mar-09 4:14

Is all the position the '*' use can replace by '&'? D'Oh! | :doh:

ky_rerun5-Mar-09 4:39

ky_rerun

5-Mar-09 4:39

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 />
 // CppTest.cpp : Defines the entry point for the console application.<br />
//<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

jeansea5-Mar-09 5:19

jeansea

5-Mar-09 5:19

<br />
void NoSideEffect(int *T)<br />
{<br />
<br />
int G = 5;<br />
T = &amp;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 Poke tongue | ;-P

ky_rerun5-Mar-09 5:56

ky_rerun

5-Mar-09 5:56

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

Maximilien5-Mar-09 4:42

Maximilien

5-Mar-09 4:42

using references is safer, you do not have to check for NULL pointers.

This signature was proudly tested on animals.

PIEBALDconsult5-Mar-09 10:38

5-Mar-09 10:38

No?

Can anybody let me know how to handle the move event in SDI

«_Superman_»5-Mar-09 17:34

«_Superman_»

5-Mar-09 17:34

That's right. You cannot pass a null to a reference like you can for a pointer.

«_Superman_»
I love work. It gives me something to do between weekends.

ashokbngr5-Mar-09 3:24

ashokbngr

5-Mar-09 3:24

Hi All,

I m developing the sdi application.I want to handle the System menu move option.
If I press ALT+SPACE on my application and select move ..then I need the event handler..

can anybody pls help me out in this..

Thanks in Advance,
Ashok.

Re: Can anybody let me know how to handle the move event in SDI

Code-o-mat5-Mar-09 4:02

Code-o-mat

5-Mar-09 4:02

I believe this[^] is what you need.

> 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. <

Unable to control mouse events

Apr^Ngp5-Mar-09 0:23

Apr^Ngp

5-Mar-09 0:23

I am attempting to control mouse events, like moving the mouse and pressing any mouse key ...
I am using the following code to set the mouse position as (0,0)

LPINPUT i;
   MOUSEINPUT mi;
   mi.dx=0;
   mi.dy=0;
   mi.dwFlags=MOUSEEVENTF_LEFTDOWN;
   mi.time=0;
   mi.dwExtraInfo=GetMessageExtraInfo();
   mi.mouseData=0;
   i[0].mi=mi;
   i[0].type=INPUT_MOUSE;
   SendInput(1,i,sizeof(i));

But, I am getting an exception at the line
i[0].mi=mi;

Actually I want to control the mouse pointer through an external hardware, which will be connected at the serial port.
I will read the data from the serial port like mouse movement and any mouse key being pressed, then SET this mouse status to the system mouse ....this is my ultimate aim ...

So I wrote the above code to set the mouse position first, but its not working ...

Please someone help me here ...

Apurv

A man is but the product of his thoughts. What he thinks, he becomes.
.......Mahatma Gandhi

Be the change you want to see in the world.
.......Mahatma Gandhi

Akt_4_U5-Mar-09 0:29

Akt_4_U

5-Mar-09 0:29

LPINPUT i;

LPINPUT means pointer to structure INPUT. Here i is of type LPINPUT. You have to allocate memory for this before you use this variable.
ie LPINPUT i = new INPUT;

akt

Apr^Ngp5-Mar-09 0:32

Apr^Ngp

5-Mar-09 0:32

oh ... thanks ...
its not giving anymore errors ...
but the mouse position is not set ....
here is my updated code ...

LPINPUT i = new INPUT;
    MOUSEINPUT mi;
    mi.dx=0;
    mi.dy=0;
    mi.dwFlags=MOUSEEVENTF_MOVE;
    mi.time=0;    
    mi.dwExtraInfo=GetMessageExtraInfo();
    mi.mouseData=0;
    i[0].mi=mi;
    i[0].type=INPUT_MOUSE;
    SendInput(1,i,sizeof(i));

Apurv

A man is but the product of his thoughts. What he thinks, he becomes.
.......Mahatma Gandhi

Be the change you want to see in the world.
.......Mahatma Gandhi

«_Superman_»5-Mar-09 0:42

«_Superman_»

5-Mar-09 0:42

Try this

INPUT i;
    MOUSEINPUT mi;
    mi.dx=0;
    mi.dy=0;
    mi.dwFlags=MOUSEEVENTF_MOVE;
    mi.time=0;    
    mi.dwExtraInfo=GetMessageExtraInfo();
    mi.mouseData=0;
    i.mi=mi;
    i.type=INPUT_MOUSE;
    SendInput(1, &i, sizeof(i));

«_Superman_»
I love work. It gives me something to do between weekends.

Apr^Ngp5-Mar-09 0:49

Apr^Ngp

5-Mar-09 0:49

its working ... superman .... thank you very much ....
will this SendInput() function work outside the application window ?
I mean, while i interface it with my hardware, so that the key inputs are coming from the hardware, know suppose the mouse is over the start button, and i send the Left-button pressed event to the SendInput() function ... will this result the Start button being pressed ???

Apurv

A man is but the product of his thoughts. What he thinks, he becomes.
.......Mahatma Gandhi

Be the change you want to see in the world.
.......Mahatma Gandhi

«_Superman_»5-Mar-09 0:57

«_Superman_»

5-Mar-09 0:57

I believe that it will work where ever the mouse pointer is located.

«_Superman_»
I love work. It gives me something to do between weekends.