Question

For a school prject we have to work with pointers, for this i wanted to see the adresses of the memory but …
When i use the next piece of code :

int _tmain(int argc, _TCHAR* argv[]){
  char a;
  char b;
  char * pa;
  char * pb;
  pa = &a;
  pb = &b;
   
  cout << "adress pa "<< pa <<endl;
  cout << "adress pb "<< pb <<endl;

  cout << "a is  " << a << endl;
  cout << "b is  " << b << endl;

i get this as output :

Is this a characterset problem and most importantly can i correct this ?

I have tested another piece of code :

#include "stdafx.h"
#include <iostream>
using namespace std;

int _tmain(int argc, _TCHAR* argv[])
{int x = 25;
int * p;
p = &x;
cout << "adres" << p << endl;
*p = 10;
cout << "x"<< x << endl;
cin.get();

    return 0;
}

and the output is readable :

What is different ?

Answer 1

It is because you are dereferencing uninitialized pointers. This is undefined behavior.

One of the overloads of << operator in the C++ standard library interprets char* as a C string, not as a pointer. Sine your C string is not initialized, the << operator prints junk. There is no similar overload for int* that would interpret it as anything other than a pointer, hence you see the correct behavior in the second case.

If you do not want your char* pointer to be interpreted as a C string, cast your pointer to void*.

cout << hex << static_cast<void*>(pa) << endl;

(hex lets you show the pointer using a more conventional base-16 notation).