自1970年1月1日00:00以来，我有一个秒，以int64表示(以纳秒为单位)，我正尝试将其转换为月/日/年/日。

迭代地执行此操作很容易，我可以正常工作，但我想按惯例进行。我正在寻找实际的数学。

最佳答案

旧问题的新答案:

这个新答案的原理:现有答案要么不显示纳秒到年/月/日转换的算法(例如，它们使用隐藏了源的库)，要么在显示的算法中使用迭代。

这个答案没有任何迭代。

The algorithms are here，并进行了详细说明。还对它们进行了+/-一百万年的跨度(远远超出您的需要)的单元测试。

该算法不计算leap秒。如果需要，可以完成，但需要查找表，并且该表会随着时间增长。

日期算法仅以天为单位，而不是纳秒。要将天转换为纳秒，请乘以86400*1000000000(注意确保您使用的是64位算术)。要将纳秒转换为天，请除以相同的数量。或者更好的方法是使用C++ 11 <chrono>库。

回答这个问题需要三种日期算法。
1. days_from_civil:

// 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 unsigned integer");
    static_assert(std::numeric_limits<Int>::digits >= 20,
             "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, unsigned>(y + (m <= 2), m, d);
}

// Returns day of week in civil calendar [0, 6] -> [Sun, Sat]
// Preconditions:  z is number of days since 1970-01-01 and is in the range:
//                   [numeric_limits<Int>::min(), numeric_limits<Int>::max()-4].
template <class Int>
constexpr
unsigned
weekday_from_days(Int z) noexcept
{
    return static_cast<unsigned>(z >= -4 ? (z+4) % 7 : (z+5) % 7 + 6);
}

#include <iostream>

int
main()
{
    int64_t z = days_from_civil(2015LL, 8, 22);
    int64_t ns = z*86400*1000000000;
    std::cout << ns << '\n';
    const char* weekdays[] = {"Sun", "Mon", "Tue", "Wed", "Thu", "Fri", "Sat"};
    unsigned wd = weekday_from_days(z);
    int64_t y;
    unsigned m, d;
    std::tie(y, m, d) = civil_from_days(ns/86400/1000000000);
    std::cout << y << '-' << m << '-' << d << ' ' << weekdays[wd] << '\n';
}

1440201600000000000
2015-8-22 Sat

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

int
main()
{
    using namespace date;
    auto now  = clock_cast<utc_clock>(sys_days{2016_y/September/26});
    auto then = clock_cast<utc_clock>(sys_days{1970_y/January/1});
    std::cout << now - then << '\n';
}

1474848026s

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

int
main()
{
    using namespace date;
    using namespace std::chrono_literals;
    auto now  = sys_days{2016_y/September/26} + 0s;
    auto then = sys_days{1970_y/January/1};
    std::cout << now - then << '\n';
}

1474848000s

关于c++ - 数学将1970年以来的秒转换为日期，反之亦然，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/7960318/