Why is address of char data not displayed?

When you are taking the address of b, you get char *. operator<< interprets that as a C string, and tries to print a character sequence instead of its address.

try cout << "address of char :" << (void *) &b << endl instead.

[EDIT] Like Tomek commented, a more proper cast to use in this case is static_cast, which is a safer alternative. Here is a version that uses it instead of the C-style cast:

cout << "address of char   :" << static_cast<void *>(&b) << endl;

There are 2 questions:

• Why it does not print the address for the char:

Printing pointers will print the address for the int*and the string* but will not print the contents for char* as there is a special overload in operator<<. If you want the address then use: static_cast<const void *>(&c);

• Why the address difference between the int and the string is 8

On your platform sizeof(int) is 4 and sizeof(char) is 1 so you really should ask why 8 not 5. The reason is that string is aligned on a 4-byte boundary. Machines work with words rather than bytes, and work faster if words are not therefore "split" a few bytes here and a few bytes there. This is called alignment

Your system probably aligns to 4-byte boundaries. If you had a 64-bit system with 64-bit integers the difference would be 16.

(Note: 64-bit system generally refers to the size of a pointer, not an int. So a 64-bit system with a 4-byte int would still have a difference of 8 as 4+1 = 5 but rounds up to 8. If sizeof(int) is 8 then 8+1 = 9 but this rounds up to 16)