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(&b)