为什么减去两个本地 DateTime 值似乎不能说明夏令时？

Question

我正在使用一些 C# 代码来尝试了解 C# 中的减法 DateTime 对象在夏令时方面的工作原理。

根据 Google 和其他来源，2017 年东部标准时区的夏令时“提前”事件发生在 3 月 12 日凌晨 2:00。因此，当天的前几个小时是：

   12:00am - 1:00am
    1:00am - 2:00am
   (There was no 2:00am - 3:00am hour due to the "spring ahead")
    3:00am - 4:00am

因此，如果我要计算该日期在该时区的凌晨 1:00 和凌晨 4:00 之间的时差，我希望结果为 2 小时。

但是，我为尝试模拟此问题而编写的代码返回了 3 小时的 TimeSpan。

代码：

TimeZoneInfo easternStandardTime = TimeZoneInfo.FindSystemTimeZoneById("Eastern Standard Time");

DateTime oneAm = TimeZoneInfo.ConvertTime(new DateTime(2017, 03, 12, 01, 00, 00), easternStandardTime);
DateTime fourAm = TimeZoneInfo.ConvertTime(new DateTime(2017, 03, 12, 04, 00, 00), easternStandardTime);

TimeSpan difference = (fourAm - oneAm);

Console.WriteLine(oneAm);
Console.WriteLine(fourAm);
Console.WriteLine(TimeZoneInfo.Local.IsDaylightSavingTime(oneAm));
Console.WriteLine(TimeZoneInfo.Local.IsDaylightSavingTime(fourAm));
Console.WriteLine(difference);

在我的电脑上，这会生成：

2017-03-12 01:00:00.000 -5
2017-03-12 04:00:00.000 -4
False
True
03:00:00

所有输出都符合预期——除了 3 小时的最终值，正如我上面提到的，我希望改为 2 小时。

很明显，我的代码没有正确模拟我想到的情况。有什么缺陷？

Answer 1

观察：

// These are just plain unspecified DateTimes
DateTime dtOneAm = new DateTime(2017, 03, 12, 01, 00, 00);
DateTime dtFourAm = new DateTime(2017, 03, 12, 04, 00, 00);

// The difference is not going to do anything other than 4-1=3
TimeSpan difference1 = dtFourAm - dtOneAm;

// ... but we have a time zone to consider!
TimeZoneInfo eastern = TimeZoneInfo.FindSystemTimeZoneById("Eastern Standard Time");

// Use that time zone to get DateTimeOffset values.
// The GetUtcOffset method has what we need.
DateTimeOffset dtoOneAmEastern = new DateTimeOffset(dtOneAm, eastern.GetUtcOffset(dtOneAm));
DateTimeOffset dtoFourAmEastern = new DateTimeOffset(dtFourAm, eastern.GetUtcOffset(dtFourAm));

// Subtracting these will take the offset into account!
// It essentially does this: [4-(-4)]-[1-(-5)] = 8-6 = 2
TimeSpan difference2 = dtoFourAmEastern - dtoOneAmEastern;

// Let's see the results
Console.WriteLine("dtOneAm: {0:o} (Kind: {1})", dtOneAm, dtOneAm.Kind);
Console.WriteLine("dtFourAm: {0:o} (Kind: {1})", dtFourAm, dtOneAm.Kind);
Console.WriteLine("difference1: {0}", difference1);

Console.WriteLine("dtoOneAmEastern: {0:o})", dtoOneAmEastern);
Console.WriteLine("dtoFourAmEastern: {0:o})", dtoFourAmEastern);
Console.WriteLine("difference2: {0}", difference2);

结果：

dtOneAm: 2017-03-12T01:00:00.0000000 (Kind: Unspecified)
dtFourAm: 2017-03-12T04:00:00.0000000 (Kind: Unspecified)
difference1: 03:00:00

dtoOneAmEastern: 2017-03-12T01:00:00.0000000-05:00)
dtoFourAmEastern: 2017-03-12T04:00:00.0000000-04:00)
difference2: 02:00:00

请注意，DateTime 在其Kind 属性中带有DateTimeKind，默认为Unspecified。它不属于任何特定的时区。 DateTimeOffset 没有有一个种类，它有一个 Offset，它告诉你本地时间与 UTC 的偏移量。 这些都没有给你时区。这就是TimeZoneInfo 对象正在做的事情。请参阅the timezone tag wiki 中的“时区！= 偏移量”。

我认为您可能感到沮丧的部分是，由于几个历史原因，DateTime 对象在进行数学运算时永远不了解时区，即使您可能有 DateTimeKind.Local . 本可以观察当地时区的变化，但不是那样做的。

您可能还对Noda Time 感兴趣，它以更明智和更有目的性的方式为 .NET 中的日期和时间提供了一个非常不同的 API。

using NodaTime;

...

// Start with just the local values.
// They are local to *somewhere*, who knows where?  We didn't say.
LocalDateTime ldtOneAm = new LocalDateTime(2017, 3, 12, 1, 0, 0);
LocalDateTime ldtFourAm = new LocalDateTime(2017, 3, 12, 4, 0, 0);

// The following won't compile, because LocalDateTime does not reference
// a linear time scale!
// Duration difference = ldtFourAm - ldtOneAm;

// We can get the 3 hour period, but what does that really tell us?
Period period = Period.Between(ldtOneAm, ldtFourAm, PeriodUnits.Hours);

// But now lets introduce a time zone
DateTimeZone eastern = DateTimeZoneProviders.Tzdb["America/New_York"];

// And apply the zone to our local values.
// We'll choose to be lenient about DST gaps & overlaps.
ZonedDateTime zdtOneAmEastern = ldtOneAm.InZoneLeniently(eastern);
ZonedDateTime zdtFourAmEastern = ldtFourAm.InZoneLeniently(eastern);

// Now we can get the difference as an exact elapsed amount of time
Duration difference = zdtFourAmEastern - zdtOneAmEastern;


// Dump the output
Console.WriteLine("ldtOneAm: {0}", ldtOneAm);
Console.WriteLine("ldtFourAm: {0}", ldtFourAm);
Console.WriteLine("period: {0}", period);

Console.WriteLine("zdtOneAmEastern: {0}", zdtOneAmEastern);
Console.WriteLine("zdtFourAmEastern: {0}", zdtFourAmEastern);
Console.WriteLine("difference: {0}", difference);

ldtOneAm: 3/12/2017 1:00:00 AM
ldtFourAm: 3/12/2017 4:00:00 AM
period: PT3H

zdtOneAmEastern: 2017-03-12T01:00:00 America/New_York (-05)
zdtFourAmEastern: 2017-03-12T04:00:00 America/New_York (-04)
difference: 0:02:00:00

我们可以看到三个小时的时间段，但这并不意味着与经过的时间相同。这只是意味着两个本地值在时钟上的位置相隔三个小时。 NodaTime 了解这些概念之间的区别，而 .Net 的内置类型则不了解。

给你一些后续阅读：

• What's wrong with DateTime anyway?
• More Fun with DateTime
• The case against DateTime.Now
• Five Common Daylight Saving Time Antipatterns of .NET Developers

哦，还有一件事。你的代码有这个...

DateTime oneAm = TimeZoneInfo.ConvertTime(new DateTime(2017, 03, 12, 01, 00, 00), easternStandardTime);

由于您创建的DateTime 具有未指定的种类，您要求将从您计算机的本地时区转换为东部时间。如果您恰好不是在东部时间，那么您的 oneAm 变量可能根本就不是 1 AM！

Answer 2

好的，所以我对您的代码做了一些小改动。不确定这是否是您想要实现的目标，但这会给您想要的...

static void Main() {
        TimeZoneInfo easternStandardTime = TimeZoneInfo.FindSystemTimeZoneById("Eastern Standard Time");
        TimeZone timeZone = TimeZone.CurrentTimeZone;

        DateTime oneAm = TimeZoneInfo.ConvertTime(new DateTime(2017, 03, 12, 01, 00, 00), easternStandardTime);
        DateTime fourAm = TimeZoneInfo.ConvertTime(new DateTime(2017, 03, 12, 04, 00, 00), easternStandardTime);

        DaylightTime time = timeZone.GetDaylightChanges(fourAm.Year);

        TimeSpan difference = ((fourAm - time.Delta) - oneAm);

        Console.WriteLine(oneAm);
        Console.WriteLine(fourAm);
        Console.WriteLine(TimeZoneInfo.Local.IsDaylightSavingTime(oneAm));
        Console.WriteLine(TimeZoneInfo.Local.IsDaylightSavingTime(fourAm));
        Console.WriteLine(difference);
        Console.ReadLine();
    }

Answer 3

所以这在MSDN 文档中得到了解决。

基本上，当从另一个日期减去一个日期时，您应该使用DateTimeOffset.Subtract()，而不是像这里的算术减法。

TimeSpan difference = fourAm.Subtract(oneAm);

产生预期的 2 小时时差。