The address of character type variable

Question

Here is just a test prototype :

#include <iostream>
#include <string>
using namespace std;

int main()
{
   int a=10;
   char b='H';
   string c="Hamza";

   cout<<"The value of a is : "<<a<<endl;
   cout<<"The value of b is : "<<b<<endl;
   cout<<"The value of c is : "<<c<<endl<<endl;

   cout<<"address of a : "<<&a<<endl;
   cout<<"address of b : "<<&b<<endl;
   cout<<"address of c : "<<&c<<endl;

   return 0;
}

Why the address of variable 'b', which is of character type, not printing?

Answer 1

The << operator in your code is overloaded in C++ 11. It doesn't conflict with any of other types like int or string, but it takes pointer to char which if used can produce undesired results.

You can do it like:-

cout << static_cast<void*>(&b)

Answer 2

There is an overload for << which takes a pointer to char and interprets it as a terminated C-style string. Using this for the address of any other char will go horribly wrong.

Instead, convert to a typeless pointer so that << doesn't get too clever:

cout << static_cast<void*>(&b)

Answer 3

Expression &b has type char *. When operator << used fo an object of type char * it considers it as a string and outputs it as a string. To output the address you should write

( void * ) &b

or

reinterpret_cast<void *>( &b )