std::chrono...rono::year)(3)

💖 std::chrono::operator+, std::chrono::operator-(std::chrono::year) 💖

okay bestie, this is std::chrono...rono::year)(3), from the <pink>C++ Standard Libary</pink> (their spelling, not mine 💀), and it's the year-math operators page 🎀 lowkey iconic, since C++20.

🌸 NAME 🌸

std::chrono::operator+,std::chrono::operator-(std::chrono::year) — literally std::chrono::operator+ and std::chrono::operator- for std::chrono::year, fr fr ✨

🦋 Synopsis 🦋

Here's the whole overload lineup, no cap:

constexpr std::chrono::year operator+( const std::chrono::year& y,
                                       const std::chrono::years& ys ) noexcept;
(1) (since C++20)

constexpr std::chrono::year operator+( const std::chrono::years& ys,
                                       const std::chrono::year& y ) noexcept;
(2) (since C++20)

constexpr std::chrono::year operator-( const std::chrono::year& y,
                                       const std::chrono::years& ys ) noexcept;
(3) (since C++20)

constexpr std::chrono::years operator-( const std::chrono::year& y1,
                                        const std::chrono::year& y2 ) noexcept;
(4) (since C++20)

What each one is giving:

• 1,2) — adds ys.count() years to y. same energy, just swapped argument order 💅
• 3) — subtracts ys.count() years from y.
• 4) — returns the difference in years between y1 and y2.

✨ Return value ✨

• 1,2) → std::chrono::year(int(y) + ys.count())
• 3) → std::chrono::year(int(y) - ys.count())
• 4) → std::chrono::years(int(y1) - int(y2))

🧚 Notes 🧚

Important vibe check: if the resulting year value for (1-3) is outside the range [-32767, 32767], the actual value stored is unspecified 😭 so don't act surprised.

The result of subtracting two year values is a duration of type std::chrono::years. This duration unit represents the length of the average Gregorian year, and the resulting duration bears no relationship to the number of days in the particular years represented by the operands. For example, the result of 2018y - 2017y is std::chrono::years(1), which represents 365.2425 days, not 365 days 🍓

🎀 Example 🎀

// Run this code

#include <cassert>
#include <chrono>

int main()
{
    std::chrono::year y{2020};

    y = std::chrono::years(12) + y; // overload (2): duration + time point
    assert(y == std::chrono::year(2032));

    y = y - std::chrono::years(33); // overload (3): time point - duration
    assert(y == std::chrono::year(1999));

    // y = std::chrono::years(33) - y; // not supported: duration - time point

    using namespace std::chrono;
    constexpr std::chrono::years ys = 2025y - 2020y; // overload (4)
    static_assert(ys == std::chrono::years(5));
}

see how overload (2) does duration + time point, (3) does time point - duration, and (4) is year minus year giving std::chrono::years? slay. also that commented-out line is not supported because duration - time point is a no-go 🙈

🔥 See also 🔥

• operator++, operator++(int), operator--, operator--(int) — increments or decrements the month; public member function of std::chrono::month
• operator+=, operator-= — adds or subtracts a number of months; public member function of std::chrono::month

k that's the whole page bestie 🫶 source/version: http://cppreference.com; date: 2024.06.10; footer title: std::chrono...rono::year)(3). periodt 💅