Age calculation

[seeplus]

With C++20/23 std::chrono and a birthdate as either type sys_days or year_month_day, is there any 'easy' way to obtain their age as years, months and days. It's quite easy to get the age as a number of days (get the birthdate and today's date as sys_days and subtract) - but as years, months, days without doing calculations involving 365 and an array of days in months?

[seeplus]

We've come up with this:

struct Age {
	unsigned years {};
	unsigned months {};
	unsigned days {};
	bool good {};
};

...

// Obtain today's date
[[nodiscard]]
inline std::chrono::sys_days get_today() noexcept {
	return std::chrono::floor<std::chrono::days>(std::chrono::system_clock::now());
}

// Obtain age as years/months/days
[[nodiscard]]
inline Age get_age(std::chrono::sys_days from, std::chrono::sys_days to = get_today()) noexcept {
	// Check if been born!
	if (from > to)
		return {};

	Age age { .days = unsigned((to - from).count()), .good = true };
	std::chrono::sys_days sd1 {};

	for (std::chrono::year_month_day ymd { from }, ymd1 { ymd + std::chrono::months(1) }; (sd1 = std::chrono::sys_days(ymd1)) <= to; ++age.months, ymd = ymd1, ymd1 = ymd + std::chrono::months(1))
		age.days -= (sd1 - std::chrono::sys_days(ymd)).count();

	age.years = age.months / 12;
	age.months %= 12;
	return age;
}

We were hoping that with std::chrono to remove the loop, but this doesn't seem possible??