C# DateTime.DayOfWeek微观优化
首先:C# DateTime.DayOfWeek微观优化,c#,performance,datetime,micro-optimization,dayofweek,C#,Performance,Datetime,Micro Optimization,Dayofweek,首先: 96,28 172923452,50884515 352004290,111919170 521851120,168153321 683972846,215554958 846791857,264187194 1042803747,328459950 Monday 90,26 43772675,17902739 84299562,37339935 119418847,47236771 166955278,72444714 207441663,89852249 223981096,106
96,28
172923452,50884515
352004290,111919170
521851120,168153321
683972846,215554958
846791857,264187194
1042803747,328459950
Monday
90,26
43772675,17902739
84299562,37339935
119418847,47236771
166955278,72444714
207441663,89852249
223981096,106062643
275440586,125110111
327353547,145689642
363908633,163442675
407152133,181642026
445141584,197571786
495590201,217373350
520907684,236609850
511052601,217571474
610024381,260208969
637676317,275558318
DateTime.DayOfWeek
方法并不代表瓶颈
DateTime
类,我们将了解DateTime.DayOfWeek
是如何实现的:
public DayOfWeek DayOfWeek
{
[__DynamicallyInvokable, TargetedPatchingOptOut("Performance critical to inline across NGen image boundaries")]
get
{
return (DayOfWeek)((this.InternalTicks / 864000000000L + 1L) % 7L);
}
}
public long Ticks
{
[__DynamicallyInvokable, TargetedPatchingOptOut("Performance critical to inline this type of method across NGen image boundaries")]
get
{
return this.InternalTicks;
}
}
此方法执行以下操作:
(864000000000)
的素因子分解:DayOfWeek
现在可以表示为:
public DayOfWeek DayOfWeek
{
get
{
return (DayOfWeek)(((Ticks>>14) / 52734375 + 1L) % 7L);
}
}
我们在模7中工作,52734375%7
1。因此,上述代码等于:
public static DayOfWeek dayOfWeekTurbo(this DateTime date)
{
return (DayOfWeek)(((date.Ticks >> 14) + 1) % 7);
}
直觉上,它是有效的。但让我们用代码来证明它
如果我们用这些数据制作一个图表,它看起来是这样的:
╔══════════════════════╦════════════════════╦═════════════════════╦═════════════╗
║ Number of iterations ║ Standard DayOfWeek ║ Optimized DayOfWeek ║ Speedup ║
╠══════════════════════╬════════════════════╬═════════════════════╬═════════════╣
║ 0 ║ 96 ║ 28 ║ 3.428571429 ║
║ 100 ║ 172923452 ║ 50884515 ║ 3.398351188 ║
║ 200 ║ 352004290 ║ 111919170 ║ 3.145165301 ║
║ 300 ║ 521851120 ║ 168153321 ║ 3.103424404 ║
║ 400 ║ 683972846 ║ 215554958 ║ 3.1730787 ║
║ 500 ║ 846791857 ║ 264187194 ║ 3.205272156 ║
║ 600 ║ 1042803747 ║ 328459950 ║ 3.174827698 ║
╚══════════════════════╩════════════════════╩═════════════════════╩═════════════╝
快3倍
注意:该代码是使用Visual Studio 2013的发布模式编译的,运行时除应用程序外,所有内容都已关闭。(当然包括VS)
我在实验室进行了测试,
英特尔®核心™ i3-2350M处理器和Windows®7 Home Premium 64位
编辑:
如前所述,如果不在日期边界上,此方法可能会失败
根据Jon Skeet对这个答案的评论
dayOfWeekTurbo
不在日期边界时可能会失败。例如
考虑<代码>新日期时间(2014, 3, 11,21, 39, 30)< /代码> -您的方法认为
今天是星期五,实际上是星期二。“我们在模中工作”
7“是错误的方向,基本上。。。通过删除额外的
除此之外,一周中的某一天在一天中发生变化
我决定编辑它
如果我们更改proof()
方法
public static void proof()
{
DateTime date = DateTime.MinValue;
DateTime max_date = DateTime.MaxValue.AddSeconds(-1);
while (date < max_date)
{
if (date.DayOfWeek != date.dayOfWeekTurbo2())
{
Console.WriteLine("{0}\t{1}", date.DayOfWeek, date.dayOfWeekTurbo2());
Console.ReadLine();
}
date = date.AddSeconds(1);
}
}
此外,我们还更改了方法getAllDates()
还会更快吗?答案是肯定的
输出:
96,28
172923452,50884515
352004290,111919170
521851120,168153321
683972846,215554958
846791857,264187194
1042803747,328459950
Monday
90,26
43772675,17902739
84299562,37339935
119418847,47236771
166955278,72444714
207441663,89852249
223981096,106062643
275440586,125110111
327353547,145689642
363908633,163442675
407152133,181642026
445141584,197571786
495590201,217373350
520907684,236609850
511052601,217571474
610024381,260208969
637676317,275558318
快2倍 看看这个问题:。不同之处在于我已经回答了自己的问题question@AdamSears我找到了。它在我的书签栏上:)
dayOfWeekTurbo
如果不在日期边界上,它可能会失败。例如,考虑<代码>新的DATETIME(2014, 3, 11,21, 39, 30)< /C> >您的方法认为实际是星期二的星期五。“我们在模7中工作”是错误的,基本上。。。通过去掉那个额外的除法,一周中的一天在白天发生变化。双向飞碟是绝对正确的。原始的long
divisionInternalTicks/TimeSpan.TicksPerDay
具有一个重要特性,即它截断了值中的时间部分。您的子表达式date.Ticks>>14
每隔2**14==16384
tick“递增”,因此您将在每过1.6384毫秒时获得一周中的新一天?
public static DayOfWeek dayOfWeekTurbo2(this DateTime date)
{
return (DayOfWeek)((((date.Ticks >> 14) / 52734375L )+ 1) % 7);
}
public static IEnumerable<DateTime> getAllDates()
{
DateTime d = DateTime.MinValue;
DateTime max = DateTime.MaxValue.AddHours(-1);
while (d < max)
{
yield return d;
d = d.AddHours(1);
}
}
public static void benchDayOfWeek()
{
DateTime[] dates = getAllDates().ToArray();
DayOfWeek[] foo = new DayOfWeek[dates.Length];
for (int max_loop = 0; max_loop < 10000; max_loop ++)
{
Stopwatch st1, st2;
st1 = Stopwatch.StartNew();
for (int i = 0; i < max_loop; i++)
for (int j = 0; j < dates.Length; j++)
foo[j] = dates[j].DayOfWeek;
st1.Stop();
st2 = Stopwatch.StartNew();
for (int i = 0; i < max_loop; i++)
for (int j = 0; j < dates.Length; j++)
foo[j] = dates[j].dayOfWeekTurbo2();
st2.Stop();
Console.WriteLine("{0},{1}", st1.ElapsedTicks, st2.ElapsedTicks);
}
Console.ReadLine();
Console.WriteLine(foo[0]);
}
90,26
43772675,17902739
84299562,37339935
119418847,47236771
166955278,72444714
207441663,89852249
223981096,106062643
275440586,125110111
327353547,145689642
363908633,163442675
407152133,181642026
445141584,197571786
495590201,217373350
520907684,236609850
511052601,217571474
610024381,260208969
637676317,275558318
╔══════════════════════╦════════════════════╦════════════════════════╦═════════════╗
║ Number of iterations ║ Standard DayOfWeek ║ Optimized DayOfWeek(2) ║ Speedup ║
╠══════════════════════╬════════════════════╬════════════════════════╬═════════════╣
║ 1 ║ 43772675 ║ 17902739 ║ 2.445026708 ║
║ 2 ║ 84299562 ║ 37339935 ║ 2.257624766 ║
║ 3 ║ 119418847 ║ 47236771 ║ 2.528090817 ║
║ 4 ║ 166955278 ║ 72444714 ║ 2.304588821 ║
║ 5 ║ 207441663 ║ 89852249 ║ 2.308697504 ║
║ 6 ║ 223981096 ║ 106062643 ║ 2.111781205 ║
║ 7 ║ 275440586 ║ 125110111 ║ 2.201585338 ║
║ 8 ║ 327353547 ║ 145689642 ║ 2.246923958 ║
║ 9 ║ 363908633 ║ 163442675 ║ 2.226521519 ║
║ 10 ║ 407152133 ║ 181642026 ║ 2.241508433 ║
║ 11 ║ 445141584 ║ 197571786 ║ 2.25306251 ║
║ 12 ║ 495590201 ║ 217373350 ║ 2.279903222 ║
║ 13 ║ 520907684 ║ 236609850 ║ 2.201546909 ║
║ 14 ║ 511052601 ║ 217571474 ║ 2.348895246 ║
║ 15 ║ 610024381 ║ 260208969 ║ 2.344363391 ║
║ 16 ║ 637676317 ║ 275558318 ║ 2.314124725 ║
╚══════════════════════╩════════════════════╩════════════════════════╩═════════════╝