c – 如何将std :: chrono :: time_point转换为std :: tm而不使用time_t？

我想打印或提取年/月/日值.

我不想使用time_t因为2038年的问题,但是我在互联网上发现的所有例子都使用它来将time_point转换为tm.

有没有一个简单的方法来从time_point转换为tm(最好不用提升)？

像libc这样的timesub的实现将是我最后的手段：
http://www.opensource.apple.com/source/Libc/Libc-262/stdtime/localtime.c

编辑：阅读建议的链接并进行更多的研究后,我得出以下结论.

>使用time_t,它是64位长的是(大多数目的).
>使用Boost.Date_Time进行便携式代码.

值得注意的是,Boost.Date_Time可以是一个只有标题的库.资源：
http://www.boost.org/doc/libs/1_53_0/more/getting_started/unix-variants.html#header-only-libraries

解决方法

答案更新更好的算法,链接到算法的详细描述,并完成转换到std :: tm.

I would like to print or extract year/month/day values.
Is there a simple way to convert from time_point to tm (preferably
without boost)?

首先要注意的是,std :: chrono :: time_point不仅在持续时间,而且在时钟上进行模板化.时钟意味着一个时代.不同的时钟可以有不同的时代.

例如,在我的系统上,std :: chrono :: high_resolution_clock和std :: chrono :: stable_clock有一个时代：每当计算机启动.如果您不知道计算机何时启动,则无法将该time_point转换为任何日历系统.

话虽如此,你可能正在谈论std :: chrono :: system_clock :: time_point,因为这个time_point,只有这个time_point需要与民用(gregorian)日历确定性的关系.

事实证明,我所知道的std :: chrono :: system_clock的每个实现都是使用unix time.这有一个1970年的新纪元忽略了闰秒.

标准不能保证.但是,如果要使用以下公式,可以利用这一事实：

07001

首先,警告,我使用最新的C 1y草案,其中包括伟大的新的constexpr工具.如果您需要退出编译器的某些constexpr属性,那么这样做.

给定上述链接中的算法,您可以转换std :: chrono :: time_point< std :: chrono :: system_clock,Duration>到std :: tm,而不使用带有以下功能的time_t：

template <class Duration>
std::tm
make_utc_tm(std::chrono::time_point<std::chrono::system_clock,Duration> tp)
{
    using namespace std;
    using namespace std::chrono;
    typedef duration<int,ratio_multiply<hours::period,ratio<24>>> days;
    // t is time duration since 1970-01-01
    Duration t = tp.time_since_epoch();
    // d is days since 1970-01-01
    days d = round_down<days>(t);
    // t is Now time duration since midnight of day d
    t -= d;
    // break d down into year/month/day
    int year;
    unsigned month;
    unsigned day;
    std::tie(year,month,day) = civil_from_days(d.count());
    // start filling in the tm with calendar info
    std::tm tm = {0};
    tm.tm_year = year - 1900;
    tm.tm_mon = month - 1;
    tm.tm_mday = day;
    tm.tm_wday = weekday_from_days(d.count());
    tm.tm_yday = d.count() - days_from_civil(year,1,1);
    // Fill in the time
    tm.tm_hour = duration_cast<hours>(t).count();
    t -= hours(tm.tm_hour);
    tm.tm_min = duration_cast<minutes>(t).count();
    t -= minutes(tm.tm_min);
    tm.tm_sec = duration_cast<seconds>(t).count();
    return tm;
}

另请注意,所有现有实现的std :: chrono :: system_clock :: time_point是UTC(忽略闰秒)时区的持续时间.如果要使用另一个时区转换time_point,则需要将时区偏移量添加/减去std :: chrono :: system_clock :: time_point,然后将其转换为天数精度.而且如果您进一步想要考虑到闰秒,则在截断之前使用this table调整适当的秒数,并了解unix时间与UTC的对应关系.

此功能可以通过以下方式验证：

#include <iostream>
#include <iomanip>

void
print_tm(const std::tm& tm)
{
    using namespace std;
    cout << tm.tm_year+1900;
    char fill = cout.fill();
    cout << setfill('0');
    cout << '-' << setw(2) << tm.tm_mon+1;
    cout << '-' << setw(2) << tm.tm_mday;
    cout << ' ';
    switch (tm.tm_wday)
    {
    case 0:
        cout << "Sun";
        break;
    case 1:
        cout << "Mon";
        break;
    case 2:
        cout << "Tue";
        break;
    case 3:
        cout << "Wed";
        break;
    case 4:
        cout << "Thu";
        break;
    case 5:
        cout << "Fri";
        break;
    case 6:
        cout << "Sat";
        break;
    }
    cout << ' ';
    cout << ' ' << setw(2) << tm.tm_hour;
    cout << ':' << setw(2) << tm.tm_min;
    cout << ':' << setw(2) << tm.tm_sec << " UTC.";
    cout << setfill(fill);
    cout << "  This is " << tm.tm_yday << " days since Jan 1\n";
}

int
main()
{
    print_tm(make_utc_tm(std::chrono::system_clock::Now()));
}

我目前打印出来的

2013-09-15 Sun 18:16:50 UTC. This is 257 days since Jan 1

如果chrono-Compatible Low-Level Date Algorithms脱机或被移动,这里是make_utc_tm中使用的算法.在上述链接中对这些算法进行了深入的解释.它们经过了很好的测试,具有非常大的有效性.

// Returns number of days since civil 1970-01-01.  Negative values indicate
//    days prior to 1970-01-01.
// Preconditions:  y-m-d represents a date in the civil (Gregorian) calendar
//                 m is in [1,12]
//                 d is in [1,last_day_of_month(y,m)]
//                 y is "approximately" in
//                   [numeric_limits<Int>::min()/366,numeric_limits<Int>::max()/366]
//                 Exact range of validity is:
//                 [civil_from_days(numeric_limits<Int>::min()),//                  civil_from_days(numeric_limits<Int>::max()-719468)]
template <class Int>
constexpr
Int
days_from_civil(Int y,unsigned m,unsigned d) noexcept
{
    static_assert(std::numeric_limits<unsigned>::digits >= 18,"This algorithm has not been ported to a 16 bit unsigned integer");
    static_assert(std::numeric_limits<Int>::digits >= 20,"This algorithm has not been ported to a 16 bit signed integer");
    y -= m <= 2;
    const Int era = (y >= 0 ? y : y-399) / 400;
    const unsigned yoe = static_cast<unsigned>(y - era * 400);      // [0,399]
    const unsigned doy = (153*(m + (m > 2 ? -3 : 9)) + 2)/5 + d-1;  // [0,365]
    const unsigned doe = yoe * 365 + yoe/4 - yoe/100 + doy;         // [0,146096]
    return era * 146097 + static_cast<Int>(doe) - 719468;
}

// Returns year/month/day triple in civil calendar
// Preconditions:  z is number of days since 1970-01-01 and is in the range:
//                   [numeric_limits<Int>::min(),numeric_limits<Int>::max()-719468].
template <class Int>
constexpr
std::tuple<Int,unsigned,unsigned>
civil_from_days(Int z) noexcept
{
    static_assert(std::numeric_limits<unsigned>::digits >= 18,"This algorithm has not been ported to a 16 bit signed integer");
    z += 719468;
    const Int era = (z >= 0 ? z : z - 146096) / 146097;
    const unsigned doe = static_cast<unsigned>(z - era * 146097);          // [0,146096]
    const unsigned yoe = (doe - doe/1460 + doe/36524 - doe/146096) / 365;  // [0,399]
    const Int y = static_cast<Int>(yoe) + era * 400;
    const unsigned doy = doe - (365*yoe + yoe/4 - yoe/100);                // [0,365]
    const unsigned mp = (5*doy + 2)/153;                                   // [0,11]
    const unsigned d = doy - (153*mp+2)/5 + 1;                             // [1,31]
    const unsigned m = mp + (mp < 10 ? 3 : -9);                            // [1,12]
    return std::tuple<Int,unsigned>(y + (m <= 2),m,d);
}

template <class Int>
constexpr
unsigned
weekday_from_days(Int z) noexcept
{
    return static_cast<unsigned>(z >= -4 ? (z+4) % 7 : (z+5) % 7 + 6);
}

template <class To,class Rep,class Period>
To
round_down(const std::chrono::duration<Rep,Period>& d)
{
    To t = std::chrono::duration_cast<To>(d);
    if (t > d)
        --t;
    return t;
}

更新

最近我把上面的算法包装成一个可以免费使用的日期/时间库documented and available here.这个库可以很容易地从std :: system_clock :: time_point提取一个年/月/日,甚至是几个小时：分钟：秒：分数秒.所有没有经过time_t.

这是一个使用上述标题库的简单程序,以打印UTC时区中的当前日期和时间,以及system_clock :: time_point提供的精度(在这种情况下为微秒)：

#include "date.h"
#include <iostream>


int
main()
{
    using namespace date;
    using namespace std;
    using namespace std::chrono;
    auto const Now = system_clock::Now();
    auto const dp = time_point_cast<days>(Now);
    auto const date = year_month_day(dp);
    auto const time = make_time(Now-dp);
    cout << date << ' ' << time << " UTC\n";
}

哪个只是为我输出：

2015-05-19 15:03:47.754002 UTC

该库有效地将std :: chrono :: system_clock :: time_point转换成易于使用的日期时间类型.

原文地址：https://www.jb51.cc/c/115205.html

c – 如何将std :: chrono :: time_point转换为std :: tm而不使用time_t？

解决方法

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