UINT64 UTC时间 [英] uint64 UTC time

问题描述

我有一个UTC日期时间而不存储在UINT64的格式，例如： 20090520145024798
我需要得到的小时，分​​钟，秒和毫秒出这段时间。我可以通过将其转换为字符串，并使用子做到这一点很容易。然而，这code需要非常快，所以我想避免字符串操作。有没有更快的方法，可能使用位操作做到这一点？哦，这个需要用C做++在Linux上的方式。

解决方案

  UINT64 U = 20090520145024798;
无符号长W = U％10亿;
无符号毫秒= W％1000;
W / = 1000;
无符号秒= W％100;
W / = 100;
无符号分钟= W％100;
无符号小时= W / 100;
无符号长V = W / 10亿;
签名天= V％100;
体积/ = 100;
无符号月= v％的100;
无符号年= V / 100;
 

为什么这个解决方案从 UINT64 U 切换到 unsigned long类型是W （原因和 v ），在中间的是，YYYYMMDD和HHMMSSIII适合32位和32位除法比在一些系统上的64位除法快。

I have a UTC date time without the formatting stored in a uint64, ie: 20090520145024798 I need to get the hours, minutes, seconds and milliseconds out of this time. I can do this very easily by converting it to a string and using substring. However, this code needs to be very fast so I would like to avoid string manipulations. Is there faster way, perhaps using bit manipulation to do this? Oh by the way this needs to be done in C++ on Linux.

解决方案

uint64 u = 20090520145024798;
unsigned long w = u % 1000000000;
unsigned millisec = w % 1000;
w /= 1000;
unsigned sec = w % 100;
w /= 100;
unsigned min = w % 100;
unsigned hour = w / 100;
unsigned long v = w / 1000000000;
unsigned day = v % 100;
v /= 100;
unsigned month = v % 100;
unsigned year = v / 100;

The reason why this solution switches from uint64 u to unsigned long w (and v) in the middle is that the YYYYMMDD and HHMMSSIII fit to 32 bits, and 32-bit division is faster than 64-bit division on some systems.

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