The address of character type variable

Question

Here is just a test prototype :

#include 
#include 
using namespace std;

int main()
{
   int a=10;
   char b=\'H\';
   string          


        

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 )