C / C++ / MFC

#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";

char array [] = "hello world";
char *p;
char **pp

p = array;   // happy
ASSERT(*p == 'h';
p = &array;  // not happy.

pp = &array;
ASSERT(**p == 'h');

Iain Clarke wrote:

char array [] = "hello world";
char *p;
char **pp

p = array; // happy
ASSERT(*p == 'h';
p = &array; // not happy.

Iain Clarke wrote:

char **pp;
pp = &array;
ASSERT(**p == 'h');

Iain Clarke wrote:

So, &array is a pointer to a pointer to chars.

int values[] = { 10, 20, 30, 40, 55, 60, 70, 80, 90, 100 };
int (*pval)[10] = &values;
wcout << *pval[0] << endl;
// or
wcout << **pval << endl;

  wcout << *(*pval + 1) << endl;
  wcout << *(pval[0] + 1) << endl;