c++ ampersand operator with char arrays

Question

I was just playing with pointers and arrays when I got confused with this piece of code that I was testing with.

#include <iostream>
using namespace std;

int main(void) {
    char a[] = "hello";
    cout << &a[0] << endl;
    char b[] = {'h', 'e', 'l', 'l', 'o', '\0'};
    cout << &b[0] << endl;
    int c[] = {1, 2, 3};
    cout << &c[0] << endl;
    return 0;
}

I expected that this would print three addresses(that of a[0], b[0] and c[0]). But the result was:

hello
hello
0x7fff1f4ce780

Why is that for the first two cases with char, '&' gives the whole string or am i missing something here?

Answer 1

Because cout's operator << prints a string if you pass it a char* as parameter, which is what &a[0] is. If you want to print the address, you'll have to explicitly cast it to void*:

cout << static_cast<void*>(&a[0]) << endl;

or just

cout << static_cast<void*>(a) << endl;