C / C++ / MFC

Matthew Faithfull10-Dec-07 2:53

Matthew Faithfull

10-Dec-07 2:53

You could try creating one Window but make it invisible. Not pretty but even Microsoft themselves have been known to resort to this at times. I'm sure WMI could also be used to get you the information you want but I don't knw any more about it than that I'm afraid.

Nothing is exactly what it seems but everything with seems can be unpicked.

Don Box10-Dec-07 3:02

Don Box

10-Dec-07 3:02

U could handle WM_ENDSESSION.

Come online at:-
jubinc@skype

David Crow10-Dec-07 3:40

David Crow

10-Dec-07 3:40

Without a window, how are you going to respond to the WM_QUERYENDSESSION or WM_ENDSESSION message?

Stephen Hewitt10-Dec-07 12:51

Stephen Hewitt

10-Dec-07 12:51

You have a hidden window. It's a common technique. If you use Spy you'll find a number of such windows.

Steve

David Crow11-Dec-07 4:21

David Crow

11-Dec-07 4:21

I know that, but Daniel's requirement was, "without any windows."

jhwurmbach10-Dec-07 4:02

10-Dec-07 4:02

Have your application be a window - hidden and with no size.
This way, you can catch the relevant messages.

Let's think the unthinkable, let's do the undoable, let's prepare to grapple with the ineffable itself, and see if we may not eff it after all.
Douglas Adams, "Dirk Gently's Holistic Detective Agency"

Log

john563210-Dec-07 2:20

john5632

10-Dec-07 2:20

Hi Experts,

How to get Log of a value with respect base of 2.

Thanx

Re: Log

CPallini10-Dec-07 2:32

10-Dec-07 2:32

log<sub>2</sub>(x)=log<sub>e</sub>(x)/log<sub>e</sub>(2)

If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler.
-- Alfonso the Wise, 13th Century King of Castile.

Re: Log

David Crow10-Dec-07 3:42

David Crow

10-Dec-07 3:42

Binary logarithm.

unsigned value

john563210-Dec-07 1:53

john5632

10-Dec-07 1:53

Hi Experts,

How can I read a byte in unsigned variable. Because this value may be nagetive.

I tried like this

<br />
short signed int m_nOneByte=0;<br />
memcpy(&m_nOneByte,btBuffer+0x40,1);<br />

Hex value is 0xf6 which is -10.But it is dispalying 246.

Re: unsigned value

jhwurmbach10-Dec-07 2:02

10-Dec-07 2:02

pther wrote:
How can I read a byte in unsigned variable. Because this value may be nagetive.

Sorry, you seem confused: Bytes *ARE* unsigned types. Range 0-255.
0xf6 is 246. It is NOT negative.

int (the signed is the default) can hold all positive values a byte can have.

Are you perchance missing the char-type?

Re: unsigned value

Cedric Moonen10-Dec-07 2:03

Cedric Moonen

10-Dec-07 2:03

If your array is a char array, you can simply assign it:
short newValue = *(btBuffer+0x40);

BTW, you should use the prefix "m_" on class members only (m stands for member).

Cédric Moonen
Software developer

Charting control [v1.2]

Re: unsigned value

CPallini10-Dec-07 2:10

10-Dec-07 2:10

pther wrote:
How can I read a byte in unsigned variable. Because this value may be nagetive.

Simple, for instance:

unsigned char buff[]={0xff, 0xf6, 0x01};
int i;
i = *((char *) &buf[1]); // i = -10
<pre>

 :)

If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler.
-- Alfonso the Wise, 13th Century King of Castile.

string buffer

George_George10-Dec-07 1:01

10-Dec-07 1:01

Hello everyone,

Suppose we defined a string buffer (array), like this,

char array[] = "hello world";
char buf[256]

Sometimes, I noticed that we either use,

1. array (buf)
or use,
2. &array (&buf)
or use
3. &array[0] (&buf[0])

as the beginning address of the array,

example like,

strcpy (buf, array);
strcpy (&buf, array);
...

I am wondering the differences between the 3 approaches, and which approach is the most correct?

thanks in advance,
George

jhwurmbach10-Dec-07 1:58

10-Dec-07 1:58

array and buf both denote pointers to chars (char*).
In fact, array is a const char*, as you can not alter the content of the string (well, even if you could, it would be wrong to try).

By convention, strings in C++ are null terminated, and it is your responsibility to count the \0 in. I the string literal "Hello World!", the \0 is appended automagically by the runtime. Go try and check the string lenght!

So, &buf is denoting the address of a pointer to a string of chars. When that is what you want, you want exactly that.

George_George10-Dec-07 2:12

10-Dec-07 2:12

Thanks jhwurmbach,

So you mean buf is not the same as $buf? But in my experience buf and &buf always have the same effect, examples,

1. strcpy (buf, "Hello World");
2. strcpy (&buf, "Hello World");

regards,
George

jhwurmbach10-Dec-07 2:19

10-Dec-07 2:19

George_George wrote:
n my experience buf and &buf always have the same effect, examples,

1. strcpy (buf, "Hello World");
2. strcpy (&buf, "Hello World");

And this *DOES* work? Unsure | :~

I would bet that given
char buf[256];

buf is of type char*
&buf is of type char** - that is a pointer to a char pointer.
OTOH, &buf[0] would be the same as buf, as you are dereferencing on pointer by using the operator[]

George_George10-Dec-07 2:50

10-Dec-07 2:50

Hi jhwurmbach,

I have tested that buf and &buf are of the same value -- means same effect. Here is my program to test. Any comments?

int main()
{
char buf[] = "Hello World";

int p1 = buf;

int p2 = &buf;

return 0;
}

regards,
George

jhwurmbach10-Dec-07 2:59

10-Dec-07 2:59

This:

#include "stdafx.h"
#include "iostream"

int _tmain(int argc, _TCHAR* argv[])
{
	char array[] = "Hello World!";

	if( array == &array)	{
		std::cout << "Equal" << std::endl;
	}
	else	{
		std::cout << "Inequal" << std::endl;
	}

	char* p1 = array;
	char* p2 = (char*)&array;
	char* p3 = &array[0];

	int pI1 = (int)array;
	int pI2 = (int)&array;
	int pI3 = (int)&array[0];

	return system("pause");
}

won't compile:
test.cpp(21): error C2446: '==' : no conversion from 'char (*__w64 )[13]' to 'char *'
test.cpp(21): error C2040: '==' : 'char [13]' differs in levels of indirection from 'char (*__w64 )[13]'
So array is of type char*, wheras &array is of type char[13].

But you are right, both point to the same chunk of memory.
It seems like the runtime is doing some magic with the adress-of-operator.

George_George10-Dec-07 15:18

10-Dec-07 15:18

Thanks jhwurmbach,

I have tested that you are correct. My further finding is when you rename the extension from .cpp to .c, then C compiler will be used and there will not be any compile error.

What magics do you think runtime or compiler is making? I am very confused why the value of address of a variable is the same as that the variable itself.

jhwurmbach wrote:
But you are right, both point to the same chunk of memory.
It seems like the runtime is doing some magic with the adress-of-operator.

regards,
George

CPallini10-Dec-07 2:19

10-Dec-07 2:19

jhwurmbach wrote:
array and buf both denote pointers to chars (char*).
In fact, array is a const char*, as you can not alter the content of the string (well, even if you could, it would be wrong to try).

Actually both are const pointers to char (i.e. const char *). That means you cannot change their value (i.e. pointed address) while it is perfectly legal to change their content.

hence:

array++; // MISTAKE

while:

array[0] = 'H'; // LEGAL and, maybe, useful.

If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler.
-- Alfonso the Wise, 13th Century King of Castile.

George_George10-Dec-07 2:53

10-Dec-07 2:53

Cool CPallini,

You are correct and I have tested it.

What about your comments about buf and &buf?

I have tested that buf and &buf are of the same value -- means same effect. Here is my program to test. Any comments?

int main()
{
char buf[] = "Hello World";

int p1 = buf;

int p2 = &amp;amp;buf;

return 0;
}

regards,
George

CPallini10-Dec-07 3:15