Question

boost :: date_time和std :: chrono的互操作性如何？

例如，有没有办法在boost :: posix_time :: ptime和std :: chrono :: time_point之间进行转换？

我尝试搜索有关此类转换的文档，但找不到任何文档。

Answer 1

我在boost commits邮件列表中找到了这个：http://lists.boost.org/boost-commit/2009/04/15209.php

以下是相关功能：

template < class Clock, class Duration> 
struct convert_to<posix_time::ptime, chrono::time_point<Clock, Duration> > { 
    inline static posix_time::ptime apply(const chrono::time_point<Clock, Duration>& from) 
    { 
        typedef chrono::time_point<Clock, Duration> time_point_t; 
        typedef chrono::nanoseconds duration_t; 
        typedef duration_t::rep rep_t; 
        rep_t d = chrono::duration_cast<duration_t>(from.time_since_epoch()).count(); 
        rep_t sec = d/1000000000; 
        rep_t nsec = d%1000000000; 
        return boost::posix_time::from_time_t(0)+ 
        boost::posix_time::seconds(static_cast<long>(sec))+ 
        #ifdef BOOST_DATE_TIME_HAS_NANOSECONDS 
        boost::posix_time::nanoseconds(nsec); 
        #else 
        boost::posix_time::microseconds((nsec+500)/1000); 
        #endif 
    } 
}; 

template < class Clock, class Duration> 
struct convert_to<chrono::time_point<Clock, Duration>, posix_time::ptime> { 
    inline static chrono::time_point<Clock, Duration> apply(const posix_time::ptime& from) 
    { 
        boost::posix_time::time_duration const time_since_epoch=from-boost::posix_time::from_time_t(0); 
        chrono::time_point<Clock, Duration> t=chrono::system_clock::from_time_t(time_since_epoch.total_seconds()); 
        long nsec=time_since_epoch.fractional_seconds()*(1000000000/time_since_epoch.ticks_per_second()); 
        return t+chrono::nanoseconds(nsec); 

    } 
};

我不确定他们什么时候会成为助推器的一部分。他们现在似乎没有加速行李......

Answer 2

您可以将time_t转换为std :: chrono :: system_clock :: time_point：

class system_clock
{
public:
    ...
    static time_t     to_time_t  (const time_point& __t);
    static time_point from_time_t(time_t __t);
};

您可以将time_t转换为ptime：

ptime from_time_t(time_t t);

但是我没有看到将ptime转换为time_t的方法。

boost :: date_time和std :: chrono之间的互操作性

2 个答案: