我是C ++的新手，在使用指向char数组（C样式字符串）的指针时，我对它与ostream对象的行为感到困惑。
const char* items {"sox"};
cout << items << endl;
cout << items[0] << endl;
cout << *items << endl;
cout << &items << endl;
cout << &items[1] << endl;
运行此命令将导致：
sox
s
s
0x7fff2e832870
ox
In contrary to pointer of other data types, printing the variable doesn't output the address, but the string as a whole. By what I understand, this is due to the << operator being overloaded for char arrays to treat them as strings.
What I don't understand is, that cout << &items[1] prints the string from index 1 onward (ox), instead of the address of the char at index 1. Is this also due to << operator being overloaded or what is the reason for this behavior?
items[1]is the second character of the array and its address, i.e.&items[1], is a pointer to the second character (with index 1) as well. So, with the same rule that you have mentioned foroperator <<, the second character of the string till the end is printed.The type of
&items[1]ischar *. Therefore theconst char *overload ofoperator <<is used, which prints the string from index 1 onwards.OTOH, the type of
&itemsisstd::string *, for which no specific overload exists, so the address of the string variable itself is printed.