How to convert std::chrono::system_clock::duration into struct timeval

问题

The title says it all.

I have to implement a function that receives a std::chrono::system_clock::duration value and that needs to convert it into a timeval sruct so I can pass it to some system function.

回答1:

More general implementation.

template<typename Duration>
void to_timeval(Duration&& d, struct timeval & tv) {
    std::chrono::seconds const sec = std::chrono::duration_cast<std::chrono::seconds>(d);

    tv.tv_sec  = sec.count();
    tv.tv_usec = std::chrono::duration_cast<std::chrono::microseconds>(d - sec).count();
}

UPDATE:

Separate methods like to_timeval() are not too convenient. Where is overloading power? We have just hardcoded type timeval into name of function to_timeval(). It's not C++ way. I want to pass struct timeval into, for example, std::chrono::duration_cast() and get my chrono-result and vice versa.

So, we can extend std::chrono::duration_cast (of course, at your own risk). Enjoy.

namespace std {
namespace chrono {
namespace detail {

template<typename From, typename To>
struct posix_duration_cast;

// chrono -> timeval caster
template<typename Rep, typename Period>
struct posix_duration_cast< std::chrono::duration<Rep, Period>, struct timeval > {

    static struct timeval cast(std::chrono::duration<Rep, Period> const& d) {
        struct timeval tv;

        std::chrono::seconds const sec = std::chrono::duration_cast<std::chrono::seconds>(d);

        tv.tv_sec  = sec.count();
        tv.tv_usec = std::chrono::duration_cast<std::chrono::microseconds>(d - sec).count();

        return std::move(tv);
    }

};

// timeval -> chrono caster
template<typename Rep, typename Period>
struct posix_duration_cast< struct timeval, std::chrono::duration<Rep, Period> > {

    static std::chrono::duration<Rep, Period> cast(struct timeval const & tv) {
        return std::chrono::duration_cast< std::chrono::duration<Rep, Period> >(
                    std::chrono::seconds(tv.tv_sec) + std::chrono::microseconds(tv.tv_usec)
                    );
    }

};

}

// chrono -> timeval    
template<typename T, typename Rep, typename Period>
auto duration_cast(std::chrono::duration<Rep, Period> const& d)
-> std::enable_if_t< std::is_same<T, struct timeval>::value, struct timeval >
{
    return detail::posix_duration_cast< std::chrono::duration<Rep, Period>, timeval >::cast(d);
}

// timeval -> chrono
template<typename Duration>
Duration duration_cast(struct timeval const& tv) {
    return detail::posix_duration_cast< struct timeval, Duration >::cast(tv);
}

} // chrono
} // std

It's just example. As alternative we can implement own duration_cast() and in some cases forward it into std::chrono::duration_cast().

And we remember about struct timespec.

回答2:

Let d be the input duration value and tv the output timeval structure filled by the convert function. Note: the function sets the timeval to 0 if the duration is negative.

void convert( const std::chrono::system_clock::duration &d, timeval &tv )
{
  chrono::microseconds usec = chrono::duration_cast<chrono::microseconds>(d);
  if( usec <= chrono::microseconds(0) )
    tv.tv_sec = tv.tv_usec = 0;
  else
  {
    tv.tv_sec = usec.count()/1000000;
    tv.tv_usec = usec.count()%1000000;
  }
}

来源：https://stackoverflow.com/questions/17402657/how-to-convert-stdchronosystem-clockduration-into-struct-timeval

标签

chrono

timeval