C++ character array not showing address

Question

I was experimenting with char array and then I tried to run this program:

#include <iostream>

using namespace std ;

int main ( )
{
    char *str = "Hello!" ;

    cout << &str[0] << endl ;
    cout << &str[1] << endl ;
    cout << &str[2] << endl ;
    cout << &str[3] << endl ;
    cout << &str[4] << endl ;

    return 0 ;
}

And I keep getting these outputs:

Hello!
ello!
llo!
lo!
o!

What exactly is happening here? I was expecting hex values.

`char *str = "Hello";` is illegal in C++11 and deprecated before that. — chris, Jul 09 '14 at 22:08

score 5 · Accepted Answer · answered Jul 09 '14 at 21:41

When you take the address of an array element, you get a pointer into the array.

In c++, like c, a character array (or pointer to character) is interpreted as a string, so the characters are printed out as a string.

If you want addresses, just add a cast to (void *).

#include <iostream>

using namespace std ;

int main ( )
{
    const char *str = "Hello!" ;

    cout << (void*) &str[0] << endl ;
    cout << (void*) &str[1] << endl ;
    cout << (void*) &str[2] << endl ;
    cout << (void*) &str[3] << endl ;
    cout << (void*) &str[4] << endl ;

    return 0 ;
}

Yeah, `&str[0]` is of type `char*`, so that's a C-string as far as `cout` is concerned. — tadman, Jul 09 '14 at 21:42

score 3 · Answer 2 · answered Jul 09 '14 at 21:43

3

A char * is assumed to be a C-style string. You need to cast to void * to get the pointer value.

(And you're missing a const - string literals are immutable.)

answered Jul 09 '14 at 21:43

Alan Stokes

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C++ character array not showing address

2 Answers2