c++ - 将 time_duration 转换为 DATE

Question

我想转换一个 time_duration到 DATE格式，即自1899年以来的天数、12、30。

DATE date_from_duration(time_duration td)
{
   double days = td.hours()/24.+td.minutes()/(24.*60.)+td.seconds()/(24.*60.*60.);
   return days;
}

此代码几乎可以工作，但有时会出现舍入错误，f.i time_duration(1007645, 15, 0) 应该导致 2014-12-12 00:15:00，但实际上是 2014-12 -12 00:14:59。

DATE 的检查是用这个方法完成的，偷自here :

ptime pTime_from_DATE(double date)
{
    using boost::math::modf;

    static const ptime::date_type base_date(1899, Dec, 30);
    static const ptime base_time(base_date, ptime::time_duration_type(0,0,0));

    int dayOffset, hourOffset, minuteOffset, secondOffset;
    double fraction = fabs(modf(date, &dayOffset)) * 24; // fraction = hours
    fraction = modf(fraction, &hourOffset) * 60; // fraction = minutes
    fraction = modf(fraction, &minuteOffset) * 60; // fraction = seconds
    modf(fraction, &secondOffset);
    ptime t(base_time);
    t += ptime::time_duration_type(hourOffset, minuteOffset, secondOffset);
    t += ptime::date_duration_type(dayOffset);
    return t;
}

有什么想法可以有效地纠正这个舍入问题吗？

Answer 1

我可能忽略了一些复杂性，但对我来说它看起来很简单:

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#include <boost/date_time/posix_time/posix_time.hpp>
#include <iostream>
using DATE = double;

boost::posix_time::ptime pTime_from_DATE(double date)
{
    static const boost::posix_time::ptime::date_type base_date(1899, boost::gregorian::Dec, 30);
    return boost::posix_time::ptime(
            base_date, 
            boost::posix_time::milliseconds(date * 1000 * 60 * 60 * 24));
}

int main() {
    boost::posix_time::time_duration duration(1007645, 15, 0);

    DATE date = duration.total_milliseconds() / 1000.0 / 60 / 60 / 24;

    std::cout << date << ": " << pTime_from_DATE(date);
}

打印

41985.2: 2014-Dec-12 05:15:00

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