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c++ - chrono::year 对象是如何构造的？

问题描述

我刚刚注意到将具有chrono::year. 它的构造函数int在以下范围内：[-32767, 32767]，但是我不清楚这个参数代表什么。

编辑：
这是理解is_leap功能chrono::year提供的含义的关键。没有来源，不清楚这里代表的是哪一年。

标签： c++datetimeepochchrono

解决方案

In 25.8.1 [time.cal.general]:

The types in 25.8 describe the civil (Gregorian) calendar and its relationship to sys_days and local_days.

The wording on this was (is) challenging as the intent is to model the Gregorian calendar (as does C++ currently via the C API) without offending those who follow other calendars.

I also am just now noting that the word "proleptic" is missing from the spec, and should probably be added in a strategic spot.

To directly answer the question, the integral associated with std::chrono::year is the Anno Domini reference, as defined by Pope Gregory in 1582, but running both backwards and forwards in time. As I write this, the year is 2018y.

And (answering Jonathan Mee's comment below), this program:

#include <chrono>
#include <iostream>

int
main()
{
    using namespace std;
    using namespace std::chrono;
    const auto foo = 2018y;
    cout << int{foo} << '\n';
}

Outputs:

2018

Live demo that you can experiment with with the proviso that the "date.h" example implementation puts things in namespace date instead of namespace std::chrono.

I should also note that this software allows for user-written calendars to interoperate with the std::chrono system. Here is an example of the Julian calendar. There are a couple more examples here.

Finally, a brief note on the rationale as to why the current year is represented as year{2018} (Anno Domini), as opposed to year{48} (time_t's 1970 origin), or year{118} (tm_year's 1900 origin):

This philosophy is hysterical when used in movies. But gets tiresome when used in software design. This library tries to do the expected.

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