C / C++ / MFC

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;

George_George wrote:

Sometimes, I noticed that we either use,

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

DavidCrow wrote:

1. array (buf)

array

buf

CPallini wrote:

If it stands then the syntax (apart of case (2)) it is used quite a lot.

char array[] = "hello world";

printf("%s\n", array);

printf("%s\n", &array);

printf("%s\n", &array[0]);

DavidCrow wrote:

But none of array(buf), &array(&buf), or &array[0](&buf[0]) are correct.

array(buf)

a number is even (odd) if, divided by two, gives 0 (1) as reminder.

wchar_t me[] = L"George Jackson\n";

wprintf(me);
wprintf(&me[0]);
wprintf(&me); // Won't compile: expects const wchar_t* not wchar_t (*)[16]

wchar_t you[] = L"George Jackson\n";
wprintf(L"%s\n", you);
wprintf(L"%s\n", &you[0]);
wprintf(L"%s\n", &you);

#include <iostream>
#include <string>
 
using namespace std;
 
int _tmain(int argc, _TCHAR* argv[])
{
	wchar_t me[] = L"George Jackson\n";
 
	wcout << me << endl;
	wcout << &me[0] << endl;
	wcout << &me << endl; // You get the address of "me";
 
	wchar_t *me_ptr = me;
	wcout << me_ptr << endl;
 	
	me_ptr = &me[0];
	wcout << me_ptr << endl;
 	
	//me_ptr = &me; // Does not compile
	wchar_t (*me2_ptr)[16] = &me;
	wcout << *me2_ptr << endl;
 
	return 0;
}