C / C++ / MFC

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

pther wrote:

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

pther wrote:

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

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

 :)

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");

#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");
}

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.

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).

array++; // MISTAKE

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

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

CPallini wrote:

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

George_George wrote:

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

char array [] = "hello world";