C / C++ / MFC

10-Dec-07 4:02

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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

10-Dec-07 3:15

I made a simple test

printf(buf, "array = %x, &array=%x", array, &array);

And (surprisingly to me) they indeed hold the same address.
It seems that the address-of operator behaviour, applied on array names, returns always the pointed address (i.e. on practical grounds has no effect on it).

You can find some info here
http://www.fredosaurus.com/notes-cpp/arrayptr/arraysaspointers2.html[^]

and here
http://personal.denison.edu/~bressoud/cs372-f04/pointersAndArrays.htm#seq334[^]
(Look at the bottom of paragraph 3.3.4)
Smile | :)

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.

jhwurmbach10-Dec-07 3:51

10-Dec-07 3:51

CPallini wrote:
And (surprisingly to me) they indeed hold the same address.

I am glad that I am not the only one who is surprised.
Roll eyes | :rolleyes:

CPallini10-Dec-07 4:04

10-Dec-07 4:04

Often a little test is worthy to do.
Big Grin | :-D

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.

jhwurmbach10-Dec-07 4:06

10-Dec-07 4:06

I did[^], but it was you who found the bit of explanation.
Thanks!

George_George10-Dec-07 15:27

10-Dec-07 15:27

Thanks ,

I think the exact answer is? Right?

http://www.fredosaurus.com/notes-cpp/arrayptr/arraysaspointers2.html[^]

--------------------
The address of operator (&) does nothing when applied to array names (see discussion below), but does with pointers..
--------------------

regards,
George

CPallini10-Dec-07 21:46