Why does cout print char arrays differently from o

Question

I'm using C++ to understand how exactly pointers work. I have this piece of code using arrays, which I'm using just to understand how the equivalent works with pointers.

int main() {    
    int arr[10] = {1,2,3};    
    char arr2[10] = {'c','i','a','o','\0'};
    cout << arr << endl;
    cout << arr2 << endl;
}

However when I run this, arr outputs the address of the first element of the array of ints (as expected) but arr2 doesn't output the address of the first element of the array of chars; it actually prints "ciao".

What is it that I'm missing or that I haven't learned yet about this?

Answer 1

It's the operator<< that is overloaded for const void* and for const char*. Your char array is converted to const char* and passed to that overload, because it fits better than to const void*. The int array, however, is converted to const void* and passed to that version. The version of operator<< taking const void* just outputs the address. The version taking the const char* actually treats it like a C-string and outputs every character until the terminating null character. If you don't want that, convert your char array to const void* explicitly when passing it to operator<<:

cout << static_cast<const void*>(arr2) << endl;

Answer 2

While casting is probably a more meaningful approach, you could also use the addressof operator:

cout << &arr2 << endl;

Answer 3

Because cout's operator << is overloaded for char* to output strings, and arr2 matches that.

If you want the address, try casting the character array as a void pointer.

Answer 4

There is a standard overload for char* that outputs a NUL terminated string.