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c# - DateTime.DayOfWeek 微优化

转载 作者:IT王子 更新时间:2023-10-29 04:36:00 26 4
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首先:

  1. 我问这个问题只是为了好玩和渴望学习。我不得不承认我喜欢搞乱微优化(尽管它们从未导致我的任何开发速度显着提高)。

  2. DateTime.DayOfWeek 方法并不代表我的任何应用程序中的瓶颈。

  3. 而且它极不可能成为任何其他方面的问题。如果有人认为此方法对其应用程序的性能有影响,他应该考虑When to optimize然后,他应该进行分析。

用ILSpy反编译DateTime类,我们发现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:

  1. 当天对应的分时数除以一天中现有的分时数。

  2. 我们对上述结果加1,使7除的余数在0和6之间。

这是计算星期几的唯一方法吗?

是否可以重新实现它以使其运行得更快?

最佳答案

让我们做一些调整。

  1. TimeSpan.TicksPerDay (864000000000) 的质因数分解:Prime factorization of 864000000000

DayOfWeek 现在可以表示为:

public DayOfWeek DayOfWeek
{
get
{
return (DayOfWeek)(((Ticks>>14) / 52734375 + 1L) % 7L);
}
}

我们对 7 求模,52734375 % 7 是 1。所以,上面的代码等于:

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

证明一下
public static void proof()
{
DateTime date = DateTime.MinValue;
DateTime max_date = DateTime.MaxValue.AddDays(-1);
while (date < max_date)
{
if (date.DayOfWeek != date.dayOfWeekTurbo())
{
Console.WriteLine("{0}\t{1}", date.DayOfWeek, date.dayOfWeekTurbo());
Console.ReadLine();
}
date = date.AddDays(1);
}
}

如果你愿意,你可以运行它,但我向你保证它工作正常。

好的,唯一剩下的就是一些基准测试。

这是一个辅助方法,为了让代码更清晰:

public static IEnumerable<DateTime> getAllDates()
{
DateTime d = DateTime.MinValue;
DateTime max = DateTime.MaxValue.AddDays(-1);
while (d < max)
{
yield return d;
d = d.AddDays(1);
}
}

我想这不需要解释。

public static void benchDayOfWeek()
{

DateTime[] dates = getAllDates().ToArray();
// for preventing the compiler doing things that we don't want to
DayOfWeek[] foo = new DayOfWeek[dates.Length];
for (int max_loop = 0; max_loop < 10000; max_loop+=100)
{


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].dayOfWeekTurbo();
st2.Stop();

Console.WriteLine("{0},{1}", st1.ElapsedTicks, st2.ElapsedTicks);

}
Console.ReadLine();
Console.WriteLine(foo[0]);

}

输出:

96,28
172923452,50884515
352004290,111919170
521851120,168153321
683972846,215554958
846791857,264187194
1042803747,328459950
Monday

如果我们用数据制作图表,它看起来像这样:

Chart

╔══════════════════════╦════════════════════╦═════════════════════╦═════════════╗
║ 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:代码是使用 Visual Studio 2013 Release模式编译的,并在除应用程序之外的所有内容都关闭的情况下运行。 (当然包括VS)。

我在 a toshiba Satellite C660-2JK 中运行了测试,英特尔® 酷睿™ i3-2350M 处理器和 Windows® 7 家庭高级版 64 位。

编辑:

正如 Jon Skeet 所注意到的,此方法在不在日期边界上时可能会失败。

由于 Jon Skeet 对这个答案的评论,

dayOfWeekTurbo can fail when it's not on a date boundary. For example, consider new DateTime(2014, 3, 11, 21, 39, 30) - your method thinks it's Friday when actually it's Tuesday. The "we are working in modulo 7" is the wrong way round, basically... by removing that extra division, the day-of-week changes during the day.

我决定编辑它。

如果我们改变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);
}
}

失败!

Jon Skeet 是对的。让我们听从 Jon Skeet 的建议并应用除法。

public static DayOfWeek dayOfWeekTurbo2(this DateTime date)
{
return (DayOfWeek)((((date.Ticks >> 14) / 52734375L )+ 1) % 7);
}

此外,我们更改方法 getAllDates()

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);
}
}

benchDayOfWeek()

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]);

}

还会更快吗?答案是

输出:

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

Chart

╔══════════════════════╦════════════════════╦════════════════════════╦═════════════╗
║ 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 ║
╚══════════════════════╩════════════════════╩════════════════════════╩═════════════╝

快 2 倍。

关于c# - DateTime.DayOfWeek 微优化,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/22258070/

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