将 int64_t 转换为 time_duration

我想转让一个boost::posix_time::ptime通过网络作为boost::int64_t。根据将 boost::posix_time::ptime 转换为 __int64 的方法 https://stackoverflow.com/questions/1574647/a-way-to-turn-boostposix-timeptime-into-an-int64，我可以轻松定义自己的epoch并且只传输time_duration从该参考纪元开始作为 64 位整数。但如何转换回ptime?

#include <iostream>
#include <cassert>
#include <boost/date_time/posix_time/posix_time.hpp>
#include <boost/date_time/gregorian/greg_month.hpp>

using namespace std;

using boost::posix_time::ptime;
using boost::posix_time::time_duration;
using boost::gregorian::date;

int main(int argc, char ** argv){
    ptime t = boost::posix_time::microsec_clock::local_time();

    // convert to int64_t
    ptime myEpoch(date(1970,boost::gregorian::Jan,1));
    time_duration myTimeFromEpoch = t - myEpoch;
    boost::int64_t myTimeAsInt = myTimeFromEpoch.ticks();

    // convert back to ptime
    ptime test = myEpoch + time_duration(myTimeAsInt);

    assert(test == t);
    return 0;
}

这不起作用，因为time_duration以刻度计数作为参数的构造函数是私有的。我也对任何其他方式简单地转移它感兴趣ptime超过简单的数据类型。

毫秒分辨率的工作解决方案：

int main(int argc, char ** argv){
    ptime t = boost::posix_time::microsec_clock::local_time();

    // convert to int64_t
    ptime myEpoch(date(1970,boost::gregorian::Jan,1));
    time_duration myTimeFromEpoch = t - myEpoch;
    boost::int64_t myTimeAsInt = myTimeFromEpoch.total_milliseconds();

    // convert back to ptime
    ptime test = myEpoch + boost::posix_time::milliseconds(myTimeAsInt);

    cout << test << endl;
    cout << t << endl;

    time_duration diff = test - t;

    assert(diff.total_milliseconds()==0);
    return 0;
}

谢谢12a6。