如何创建用于std :: chrono函数的自定义时钟？ [英] How to create a custom clock for use in std::chrono functions?

问题描述

我有一些随意的纪元，例如1988年7月13日。本质上，我想测量相对于此的时间。我当时正在考虑编写一个自定义时钟类，以便可以这样编写代码：

I have some arbitrary epoch, like July 13, 1988. Essentially I want to measure the time relative to this. I was thinking of writing a custom clock class, so that I could write code like this:

using std::chrono;
time_point<My_Clock> tp;
std::cout << duration_cast<seconds>(tp.time_since_epoch()).count() << std::endl;

这可能吗？如果不是，最干净的方法是什么？

Is this possible? If not, what's the cleanest way to accomplish this?

推荐答案

编写此自定义时钟的困难之处在于弄清楚如何编写其 now（）函数。在下面的示例中，我将 now（）基于 system_clock 的 now（） 。首先，我做了一些侦探性工作，发现我的 system_clock 出现了1970年元旦，忽略了 le秒。这就是 unix时间。事实证明，我知道的每个实现（并且我都已经检查了所有实现）都具有相同的纪元（但这在C ++ 11标准中未指定）。

The hard part of writing this custom clock is figuring out how to write its now() function. In the example below I base the now() off of system_clock's now(). First I do some detective work to discover that my system_clock has an epoch of New Years 1970, neglecting leap seconds. This is known as unix time. As it turns out, every implementation I'm aware of (and I think I've checked them all) have this very same epoch (but this is unspecified by the C++11 standard).

接下来，我计算1988-07-13是1970-01-01之后的6768天。使用这两个事实，其余的操作很容易：

Next I compute that 1988-07-13 is 6768 days after 1970-01-01. Using these two facts, the rest is easy:

#include <chrono>

struct My_Clock
{
    typedef std::chrono::seconds              duration;
    typedef duration::rep                     rep;
    typedef duration::period                  period;
    typedef std::chrono::time_point<My_Clock> time_point;
    static const bool is_steady =             false;

    static time_point now() noexcept
    {
        using namespace std::chrono;
        return time_point
          (
            duration_cast<duration>(system_clock::now().time_since_epoch()) -
            hours(6768*24)
          );
    }
};

MyClock 需要嵌套的typedef来描述其持续时间， rep ，期间和 time_point 。根据您的问题，我选择了 seconds 作为持续时间 ，但是您可以选择任何内容。

MyClock needs nested typedefs to describe its duration, rep, period, and time_point. Based on your question, I've chosen seconds as the duration, but you can choose anything you want.

对于 now（）函数，我只调用 system_clock :: now（）并减去以秒为单位的纪元。通过使用 MyClock :: duration 编写所有内容，我对计算有了一点点聪明，以便我可以更轻松地更改 duration 。请注意，我能够从小时减去时代，这隐式转换为 duration （即 seconds ）。或者，我可以为自己建立自定义的持续时间天：

For the now() function I just call the system_clock::now() and subtract off the epoch in units of seconds. I got just a little clever with this computation by writing everything in terms of MyClock::duration so that I can more easily change duration. Note that I was able to subtract off the epoch in terms of hours, which implicitly converts to duration (which is seconds). Alternatively I could have built myself a custom duration of days:

typedef std::chrono::duration<int, std::ratio_multiply<std::chrono::hours::period,
                                                       std::ratio<24>>> days;

然后返回 now（）可能是这样写的：

And then the return of now() could have been written:

        return time_point
          (
            duration_cast<duration>(system_clock::now().time_since_epoch()) -
            days(6768)
          );

无论如何，现在您都可以这样使用：

At any rate, now you can use this like:

#include <iostream>

int
main()
{
    using namespace std::chrono;
    time_point<My_Clock> tp = My_Clock::now();
    std::cout << tp.time_since_epoch().count() << '\n';
}

对我来说这只是打印出来：

Which for me just printed out:

786664963

今天证明了这一点（ 2013-06-16）大约是1988-07-13年之后的24.9年。

Which demonstrates that today (2013-06-16) is approximately 24.9 years after 1988-07-13.

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