确定时间戳是否是确切的小时(Determining if a time stamp is an exact hour)

确定时间戳是否是确切的小时(Determining if a time stamp is an exact hour)

我需要确定时间戳是否是精确的小时（即它表示没有分钟，秒或毫秒组件的时间） - 仅使用基元。 代码每秒调用几百次，我不想要Calendar或其他对象的开销。

我试过这个但是舍入错误导致比较失败。

float hours = (time / 3600000f); boolean isExactHour = hours == (int)hours;

例如，如果time = 1373763600470，则小时= 381601.03125。 该时间戳代表今天格林尼治标准时间01：00：00：000，小时数应为381601。

有什么想法快速简单地做到这一点？ 谢谢！

[编辑]

看起来这比第一眼看上去更复杂（什么时候不是？:)

为清楚起见，我不关心时区，也不关心闰秒。 时间戳由一种方法生成，该方法返回前一个午夜 - 即2013年7月13日00：00：00：000。 我正在计算，对于长数组中的任何时间戳，总是这个初始时间戳加上/减去5分钟的精确倍数。 我的目的是确定一个给定的邮票是否是“最重要的时刻”。 我可能会遇到边缘情况，其中5分钟的倍数与年末重叠，但我可以忍受这些。

I need to determine if a time stamp is an exact hour (i.e. it represents a time with no minutes, seconds or milliseconds components) - using primitives only. The code is called several hundred times per second and I don't want the overhead of a Calendar or other object.

I've tried this but rounding errors cause the comparison to fail.

float hours = (time / 3600000f); boolean isExactHour = hours == (int)hours;

For example, if time = 1373763600470, hours = 381601.03125. That time stamp represents 01:00:00:000 GMT today and hours should be 381601.

Any ideas for a quick and simple way to do this? Thanks!

[EDIT]

It seems that this is more complex than at first sight (when is it not? :)

For clarity, I don't care about time zones, nor leap seconds. The time stamp is generated by a method which returns the previous midnight - i.e. 13 July 2013 00:00:00:000 for today. I am then calculating, for any time stamp in an array of longs which is always this initial time stamp plus/minus an exact multiple of 5 minutes. My aim to to determine if a given stamp is "top of the hour". I might have edge cases where the multiple of 5 minutes overlaps a year end but I can live with those.

最满意答案

(time % 36000000) == 0

当然这很明显？

编辑

为了获得闰秒的准确性，假设一个自1970年以来以年为索引的闰秒查找表，类似于：

((time-leapSeconds[(int)(time/(1000*60*60*24*365.2425))-1970]*1000) % 3600000) == 0

需要这种精确度的程序员也应该看到以下所有Einar的评论。

(time % 36000000) == 0

Surely this is obvious?

EDIT

To get accuracy w.r.t. leap seconds, assuming a lookup table of leap seconds indexed by year since (say) 1970, something like:

((time-leapSeconds[(int)(time/(1000*60*60*24*365.2425))-1970]*1000) % 3600000) == 0

Programmers needing this level of accuracy should also see all of Einar's comments below.