pointer to char

Question

Hi everybody ,

Why a pointer to a data type 'char' doesn't give an address ?

char letter = 'G';
cout << letter; // outputs G
cout << &letter // outputs G (why is it G and a squarre instead of an hexadecimal address?)

Thanks

Answer 1

Because the ostream operator << is specially overloaded for char pointers. This is used to display C-style strings.

const char* cstr = "Hello World";
char cstr2[] = "Hello World";

std::cout << cstr;

The pointer points to the first character in the string. Printing the pointer/array using std::cout would cause the program to display all characters until a terminating character '\0' located at the end of the string, is encountered.

Answer 2

It is defaulting to a string (e.g. char*x="test";). To get an address, you need to cast it.
cout << static_cast<void*>(&letter);

Answer 3

Rabilu Muhammad Ibrahim since a character address is the default for strings, it must be. Otherwise, both of these would output the address.
char str[] = "test";
char *strg = "test";
cout << str << strg;

Answer 4

Really suprising that you have to cast to get the address of char .

Answer 5

John Wells , Hatsy Rei, thank you very much for you answer.
unfortunatly , i'm beginner with c++ and pointer so i need more progress to understand correctly your answer.
but , i understood that there is a logical reason for this output, i'm happy with that !

Answer 6

Donna it'll print the address of letr, not the address of letter.