如何将 std::chrono::duration 转换为双倍（秒）？

Question

Let's have using duration = std::chrono::steady_clock::duration . I would like to convert duration to double in seconds with maximal precision elegantly.

I have found the reverse way , but it didn't help me finding it out.

Alternatively expressed, I want optimally some std function double F(duration) , which returns seconds.

Answer 1

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 , usestd::chrono::duration_cast .

Answer 2

Converting a std::chrono::duration to any numeric "count" is deceptively simple!

You can divide any two std::chrono::duration s to efficiently get a count.

Given a duration d of any period and any integer representation:

using std::chrono_literals;

// Want to know how many seconds?  Divide by one second!
std::int64_t int_seconds = d / 1s;
double       fp_seconds  = d / 1.0s;

As shown, it does not matter what the period is of the two durations. And the resulting type will be the common type of the two durations.

Answer 3

This is the most straightforward way for me:

auto start = std::chrono::steady_clock::now();

/*code*/

auto end = std::chrono::steady_clock::now();
std::chrono::duration<double> diff = end - start;
std::cout << "Duration [seconds]: " << diff.count() << std::endl;

however I do not know about precision... Although this method is pretty precise.