如何计算精确的时差（C ++）？ [英] How can I calculate a precise time difference (C++)?

问题描述

我有两个UTC时间，分别存储为 day：hour：minute：second：millisecond，也存储为day，hour，min等的五个整数变量。我想找到两者之间的区别次，精确到毫秒。

I have two UTC times, stored as "day:hour:minute:second:millisecond", and also as five integer variables for day, hour, minute, etc. I'd like to find the difference between the two times, with millisecond precision. How can this be done in C++?

推荐答案

以毫秒精度计算时差：

1. 将两个时间戳都转换为 std :: chrono :: system_clock :: time_point 。

 struct std::tm thetime  { .tm_sec = sec, .tm_min = min, .tm_hour = hour,
                           .tm_mday = mday, .tm_mon = mon, .tm_year = year };

 auto mytime = (std::chrono::system_clock::from_time_t(std::mktime(&thetime)))
       + std::chrono::milliseconds(msec);

减去两个生成的时间戳。

Subtract the two resulting timestamps.

 auto diffms = std::chrono::duration_cast<std::chrono::milliseconds(mytime2 - mytime1).count();

编辑：这不考虑跳跃秒或夏令时。如果您对毫秒感兴趣，则可能应该使用既不包含毫秒的时间刻度（因此，本地时间和UTC都不是不错的选择）。如果您的日期和时间是从 UTC获得的，首先，POSIX时钟很不错，因为 UTC POSIX时钟不代表leap秒（当然也没有DST）。

this does not take into account leap seconds or daylight saving time. If you are interested in milliseconds, you should probably use a time scale which has neither of these (so neither local time nor UTC are good choices). If your date and time were derived from a "UTC" POSIX clock in the first place, you are probably good to go, since "UTC" POSIX clock doesn't account for leap seconds (and of course has no DST).

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