Ampersand in String

[AmitH]

Hello everybody,

Until now, I thought that ampersand (&) is used to display the memory address.
For example:

int a[10] = { 1, 2, 3 };
cout << &a << endl; // Getting the address of array 'a' in pos [0] (a[0])
cout << &a[2] << endl; // The address of a[2] 

Until now, as expected, the output will be addresses only.

But when the array is a string, it changes everything:

char b[10] = "Good Luck";
cout << &b << endl; // Until now, it's OK, address of b[0]
cout << &b[0] << " " <<endl;
/*OR*/ cout << &b[3] << endl; // NOW WHAT?! 

The output is:

002DFA18
Good Luck
d Luck

As you can see, instead of getting the address of b[0] or b[3], the program outputs the string itself starting from the the position I wrote, and ends in '\0'.

Why is this happening?!

[coder777]

The reason is that the operator<< is overloaded for const char * and other pointer differently. So if you provide a pointer to char the string is printed instead of the address.

[Peter87]

If you want to print the address of a char* you can cast it to a void* before printing it.

cout << static_cast<void*>(&b[3]) << endl;

[AmitH]

The reason is that the operator<< is overloaded for const char * and other pointer differently.

Sorry, I did not understand you.
If I use strcpy(c, &b[3]) instead of "<<"- the action is similar: copy "d Luck" to c.

How can I "predict" when the program will output an address and when a value?

[coder777]

Let's phrase it so: The stream (cout) 'knows' what a string is and prints it accordingly. If a pointer to another type is provided it prints the value of the pointer.

See

http://www.cplusplus.com/reference/ostream/ostream/operator%3C%3C/

what types are 'known' by default

you can overload the operator<< anytime to extend the functionality.

How can I "predict" when the program will output an address and when a value?

read the documentation