Let's have using duration = std::chrono::steady_clock::duration
. I would like to convert duration to double in seconds with maximal precition elegantly.
I have found the reverse way (convert seconds as double to std::chrono::duration? ), but it didn't help me finding it out.
Alternatively expressed, I want optimally some std function double F(duration)
, which returns seconds.
Thank you.
Simply do:
std::chrono::duration<double>(d).count()
Or, as a function:
template <class Rep, class Period>
constexpr auto F(const std::chrono::duration<Rep,Period>& d)
{
return std::chrono::duration<double>(d).count();
}
If you need more complex casts that cannot be fulfilled by the std::chrono::duration
constructors, use std::chrono::duration_cast
.