Why does cout print char arrays differently from other arrays?

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;

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.