Defining two versions of << overload

Question

Can someone tell me what is wrong when I try to overload << in a slightly different way by using another symbol like <<=

#include <iostream>

struct Date { 
    int day, month, year, hour, minute, second;
    Date (int d, int m, int y, int h, int min, int s) { day = d;  month = m;  year = y;  hour = h;  minute = min;  second = s; }
    friend std::ostream& operator << (std::ostream&, const Date&);
    friend std::ostream& operator <<= (std::ostream&, const Date&);  // Carries out << but without the hour, minute, and second.
};

std::ostream& operator << (std::ostream& os, const Date& d) {
    os << d.day << ' ' << d.month << ' ' << d.year << ' ' << d.hour << ' ' << d.minute << ' ' << d.second;
    return os;
}

std::ostream& operator <<= (std::ostream& os, const Date& d) {
    os << d.day << ' ' << d.month << ' ' << d.year;
    return os;
}

int main () {     
    Date date(25, 12, 2021, 8, 30, 45);
    std::cout << "Today is " << date << '\n';  // Works fine
    std::cout << "Today is " <<= date << '\n';  // Does not work
}

If I use

std::cout << "Today is " <<= date;

It works fine, so what is the problem with adding in << '\n' when std::ostream& is returned by <<= ?

Answer 1

Due to the operator precedence this statement

std::cout << "Today is " <<= date << '\n';

is equivalent to

( std::cout << "Today is " ) <<= ( date << '\n' );

and the right most expression

( date << '\n' )

produces an error because such an operator is not defined for objects of the type struct Date.

Answer 2

Thanks to Vlad from Moscow's answer,

(std::cout << "Today is " <<= date) << '\n';

is the simple fix. Not as pretty as was desired, but works correctly. Perhaps defining a derived class DateAndTime of Date that has the hours, minutes, and seconds and then overloading << for DateAndTime may be a more elegant, but longer, solution.