blocks|key|2904230|text|#3实际上是#1:静态初始化直到第一次使用它所属的类时才开始。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2904231|如果从静态构造函数或静态初始化块调用MyMethod，则可能会发生这种情况。如果您不直接或间接地从静态构造函数调用MyMethod，那么您应该没问题。|offset|length|style|CODE|2904232|entityMap^0|0|I|8|1L|8|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|K|8|@$D|L|E|M|F|G]|$D|N|E|O|F|G]]|9|@]|A|$]]|$1|H|3|-4|5|6|7|P|8|@]|9|@]|A|$]]]|I|$]]

The #3 is actually #1: static initialization does not start until the first use of the class to which it belongs.

It is possible if <code>MyMethod</code> is called from the static constructor or a static initialization block. If you do not invoke <code>MyMethod</code> directly or indirectly from your static constructor, you should be fine.

blocks|key|235127|text|根据MSDN，静态构造函数：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|235128|235129|在创建第一个实例或引用任何静态成员之前，会自动调用静态构造函数来初始化类。|blockquote|235130|235131|235132|因此，静态构造函数将在静态方法MyClass.MyMethod()被调用之前被调用(当然，假设在静态构造或静态字段初始化期间也没有被调用)。|style|CODE|235133|现在，如果您在该static+constructor中执行任何异步操作，那么您的工作就是同步它。|235134|entityMap|0|LINK|mutability|MUTABLE|url|http://msdn.microsoft.com/en-us/library/k9x6w0hc%2528v=vs.80%2529.aspx^0|2|4|0|0|0|0|0|0|F|I|0|8|I|0^^$0|@$1|2|3|4|5|6|7|X|8|@]|9|@$A|Y|B|Z|1|10]]|C|$]]|$1|D|3|-4|5|6|7|11|8|@]|9|@]|C|$]]|$1|E|3|F|5|G|7|12|8|@]|9|@]|C|$]]|$1|H|3|-4|5|6|7|13|8|@]|9|@]|C|$]]|$1|I|3|-4|5|6|7|14|8|@]|9|@]|C|$]]|$1|J|3|K|5|6|7|15|8|@$A|16|B|17|L|M]]|9|@]|C|$]]|$1|N|3|O|5|6|7|18|8|@$A|19|B|1A|L|M]]|9|@]|C|$]]|$1|P|3|-4|5|6|7|1B|8|@]|9|@]|C|$]]]|Q|$R|$5|S|T|U|C|$V|W]]]]

According to the <a href="http://msdn.microsoft.com/en-us/library/k9x6w0hc%28v=vs.80%29.aspx" rel="noreferrer">MSDN</a>, a static constructor:

<blockquote>
 A static constructor is called automatically to initialize the class
 before the first instance is created or any static members are
 referenced.
</blockquote>

So the static constructor will be called before the static method <code>MyClass.MyMethod()</code> is invoked (assuming not also invoked during static construction or static field initialization of course). 

Now, if you are doing anything asynchronous in that <code>static constructor</code>, then it's your job to synchronize that.

blocks|key|2904344|text|来自documentation+(重点是我的)：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|2904345|2904346|在创建第一个实例或引用任何静态成员之前，会自动调用静态构造函数来初始化类。|blockquote|style|BOLD|2904347|2904348|entityMap|0|LINK|mutability|MUTABLE|url|http://msdn.microsoft.com/en-us/library/k9x6w0hc%2528v=vs.80%2529.aspx^0|2|D|0|0|0|8|9|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@$A|T|B|U|1|V]]|C|$]]|$1|D|3|-4|5|6|7|W|8|@]|9|@]|C|$]]|$1|E|3|F|5|G|7|X|8|@$A|Y|B|Z|H|I]]|9|@]|C|$]]|$1|J|3|-4|5|6|7|10|8|@]|9|@]|C|$]]|$1|K|3|-4|5|6|7|11|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]]]

From the <a href="http://msdn.microsoft.com/en-us/library/k9x6w0hc%28v=vs.80%29.aspx">documentation</a> (emphasis mine):

<blockquote>
 A static constructor is called automatically to initialize the class
 before the first instance is created or any static members are
 referenced.
</blockquote>

blocks|key|235179|text|你可以保证4总是在2之后(如果你不从你的静态方法创建类的实例)，但是对于1和3不是这样的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|235180|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

You can guarantee 4 will always come after 2 (if you don't create a instance of your class from your static method), however the same is not true for 1 and 3.

blocks|key|3118670|text|在执行mymethod之前，将调用静态构造函数。然而，如果你搞砸了，如果4在2之前被调用，那么我建议你重新考虑你的设计。无论如何，不应该在静态构造函数中做复杂的事情。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3118671|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

The static constructor will be called before mymethod is executed. However if you are screwed if 4 is called before 2 then I suggest you re-think your design. Should not be doing complicated stuff in a static constructor anyway.

blocks|key|238816|text|CLR保证在访问任何静态成员之前运行静态构造函数。但是，你的设计有点难闻。这样做会更简单：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|238817|static+void+Main(string[]+args)+
{+
+++++bool+userIsAuthenticated+=+MyClass.AuthenticateUser();
+++++if+(userIsAuthenticated)
+++++++++MyClass.MyMethod();+
+}+|code-block|syntax|javascript|238818|在您的设计中，如果身份验证失败，阻止MyMethod运行的唯一方法就是抛出异常。|238819|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

The CLR guarantees that the static constructor runs before any static members are accessed. However, your design is a bit smelly. It would be more straightforward to do something like this:

<pre><code>static void Main(string[] args) 
{ 
 bool userIsAuthenticated = MyClass.AuthenticateUser();
 if (userIsAuthenticated)
 MyClass.MyMethod(); 
 } 
</code></pre>

With your design, if authentication fails, the only way to prevent MyMethod from running is by throwing an exception.

blocks|key|2904407|text|它确保了静态类的构造函数在其任何方法执行之前都已被调用。示例：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2904408|class+Program
{
++++static+void+Main(string[]+args)
++++{
++++++++Console.WriteLine("Press+enter");
++++++++Console.ReadLine();
++++++++Boop.SayHi();
++++++++Boop.SayHi();
++++++++Console.ReadLine();
++++}

}

static+class+Boop
{
++++static+Boop()
++++{
++++++++Console.WriteLine("Hi+incoming+...");
++++}

++++public+static+void+SayHi()
++++{
++++++++Console.WriteLine("Hi+there!");
++++}
}|code-block|syntax|javascript|2904409|输出：|2904410|2904411|按enter键|blockquote|2904412|2904413|//回车后|2904414|嗨，来电...|2904415|你好啊！|2904416|你好！|2904417|entityMap^0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|X|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Y|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Z|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|10|8|@]|9|@]|A|$]]|$1|J|3|K|5|L|7|11|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|12|8|@]|9|@]|A|$]]|$1|N|3|O|5|6|7|13|8|@]|9|@]|A|$]]|$1|P|3|Q|5|6|7|14|8|@]|9|@]|A|$]]|$1|R|3|S|5|6|7|15|8|@]|9|@]|A|$]]|$1|T|3|U|5|6|7|16|8|@]|9|@]|A|$]]|$1|V|3|-4|5|6|7|17|8|@]|9|@]|A|$]]]|W|$]]

It's ensured that a static class's constructor has been called before any of its methods get executed. Example:

<pre><code>class Program
{
 static void Main(string[] args)
 {
 Console.WriteLine("Press enter");
 Console.ReadLine();
 Boop.SayHi();
 Boop.SayHi();
 Console.ReadLine();
 }

}

static class Boop
{
 static Boop()
 {
 Console.WriteLine("Hi incoming ...");
 }

 public static void SayHi()
 {
 Console.WriteLine("Hi there!");
 }
}
</code></pre>

Output:

<blockquote>
 Press enter
 
 // after pressing enter
 
 Hi incoming ...
 
 Hi there!
 
 Hi there!
</blockquote>

blocks|key|3118718|text|你在这里只问了一个问题，但还有十几个问题是你应该问的，所以我将回答所有问题。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3118719|3118720|这里是我假设的序列|blockquote|3118721|3118722|3118723|类构造函数的开始(也称为cctor)|offset|length|style|CODE|3118724|End+of+|unordered-list-item|3118725|+|ordered-list-item|3118726|start+of+Main+|3118727|start+of+MyMethod+|3118728|3118729|3118730|3118731|这是正确的吗？|3118732|3118733|3118734|不是的。正确的顺序是：|3118735|程序的|3118736|启动器(如果有)。没有。程序的cctor的|3118737|End(如果有)。没有。|3118738|Start+of+Main|3118739|Start+of+cctor+for+MyClass|3118740|End+of+cctor+for+MyClass|3118741|Start+of+MyClass.MyMethod|3118742|3118743|3118744|如果有静态字段初始值设定项怎么办？|3118745|3118746|3118747|在某些情况下，允许CLR更改静态字段初始值设定项的运行顺序。有关详细信息，请参阅Jon的主题页面：|3118748|The+differences+between+static+constructors+and+type+initializers|3118749|3118750|像MyMethod这样的静态方法有没有可能在该类的cctor完成之前被调用？|3118751|3118752|3118753|是。如果cctor本身调用MyMethod，那么很明显在cctor完成之前就会调用MyMethod。|3118754|3118755|+cctor不调用MyMethod。有没有可能在MyClass的cctor完成之前调用像MyMethod这样的静态方法？|3118756|3118757|3118758|是。如果cctor使用其cctor调用MyMethod的另一种类型，则在MyClass+cctor完成之前将调用MyMethod。|3118759|3118760|没有直接或间接调用MyMethod的接收器！现在有没有可能在MyClass的cctor完成之前调用像MyMethod这样的静态方法呢？|3118761|3118762|3118763|不是的。|3118764|3118765|即使涉及到多个线程，这仍然是真的吗？|3118766|3118767|3118768|是。cctor将在一个线程上完成，然后才能在任何线程上调用静态方法。|3118769|3118770|可以多次调用cctor吗？假设两个线程都导致cctor运行。|3118771|3118772|3118773|无论涉及多少线程，cctor都保证最多被调用一次。如果两个线程“同时”调用MyMethod，那么它们就会竞争。其中一个线程输掉了比赛，并阻塞，直到MyClass+cctor在获胜的线程上完成。|3118774|3118775|失败的线程阻塞，直到cctor完成？真的？|3118776|3118777|3118778|真的。|3118779|3118780|那么，如果获胜线程上的cctor调用了阻塞先前由失败线程获取的锁的代码，该怎么办？|3118781|3118782|3118783|然后你就有了一个经典的锁顺序反转条件。你的程序死锁了。永远不变。|3118784|3118785|，这看起来很危险。如何避免死锁？|3118786|3118787|3118788|如果你这么做的时候很痛，那就别这么做了。永远不会做可能在cctor中阻塞的事情。|BOLD|3118789|3118790|依赖cctor初始化语义来执行复杂的安全需求是个好主意吗？拥有一个可以进行用户交互的cctor是个好主意吗？|3118791|3118792|3118793|这两个都不是好主意。我的建议是，您应该找到一种不同的方法来确保满足您的方法的影响安全的前提条件。|3118794|entityMap|0|LINK|mutability|MUTABLE|url|http://csharpindepth.com/Articles/General/Beforefieldinit.aspx^0|0|0|0|0|0|C|5|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|1T|0|0|0|1|8|0|0|0|0|0|18|8|0|0|0|0|0|1E|8|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|K|K|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|3U|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|3V|8|@]|9|@]|A|$]]|$1|C|3|D|5|E|7|3W|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|3X|8|@]|9|@]|A|$]]|$1|G|3|-4|5|6|7|3Y|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|3Z|8|@$J|40|K|41|L|M]]|9|@]|A|$]]|$1|N|3|O|5|P|7|42|8|@]|9|@]|A|$]]|$1|Q|3|R|5|S|7|43|8|@]|9|@]|A|$]]|$1|T|3|U|5|S|7|44|8|@]|9|@]|A|$]]|$1|V|3|W|5|S|7|45|8|@]|9|@]|A|$]]|$1|X|3|-4|5|6|7|46|8|@]|9|@]|A|$]]|$1|Y|3|-4|5|6|7|47|8|@]|9|@]|A|$]]|$1|Z|3|-4|5|6|7|48|8|@]|9|@]|A|$]]|$1|10|3|11|5|E|7|49|8|@]|9|@]|A|$]]|$1|12|3|-4|5|6|7|4A|8|@]|9|@]|A|$]]|$1|13|3|-4|5|6|7|4B|8|@]|9|@]|A|$]]|$1|14|3|15|5|6|7|4C|8|@]|9|@]|A|$]]|$1|16|3|17|5|6|7|4D|8|@]|9|@]|A|$]]|$1|18|3|19|5|S|7|4E|8|@]|9|@]|A|$]]|$1|1A|3|1B|5|S|7|4F|8|@]|9|@]|A|$]]|$1|1C|3|1D|5|S|7|4G|8|@]|9|@]|A|$]]|$1|1E|3|1F|5|S|7|4H|8|@]|9|@]|A|$]]|$1|1G|3|1H|5|S|7|4I|8|@]|9|@]|A|$]]|$1|1I|3|1J|5|S|7|4J|8|@]|9|@]|A|$]]|$1|1K|3|-4|5|6|7|4K|8|@]|9|@]|A|$]]|$1|1L|3|-4|5|6|7|4L|8|@]|9|@]|A|$]]|$1|1M|3|1N|5|E|7|4M|8|@]|9|@]|A|$]]|$1|1O|3|-4|5|6|7|4N|8|@]|9|@]|A|$]]|$1|1P|3|-4|5|6|7|4O|8|@]|9|@]|A|$]]|$1|1Q|3|1R|5|6|7|4P|8|@]|9|@]|A|$]]|$1|1S|3|1T|5|6|7|4Q|8|@]|9|@$J|4R|K|4S|1|4T]]|A|$]]|$1|1U|3|-4|5|6|7|4U|8|@]|9|@]|A|$]]|$1|1V|3|1W|5|E|7|4V|8|@$J|4W|K|4X|L|M]]|9|@]|A|$]]|$1|1X|3|-4|5|6|7|4Y|8|@]|9|@]|A|$]]|$1|1Y|3|-4|5|6|7|4Z|8|@]|9|@]|A|$]]|$1|1Z|3|20|5|6|7|50|8|@]|9|@]|A|$]]|$1|21|3|-4|5|6|7|51|8|@]|9|@]|A|$]]|$1|22|3|23|5|E|7|52|8|@$J|53|K|54|L|M]]|9|@]|A|$]]|$1|24|3|-4|5|6|7|55|8|@]|9|@]|A|$]]|$1|25|3|-4|5|6|7|56|8|@]|9|@]|A|$]]|$1|26|3|27|5|6|7|57|8|@]|9|@]|A|$]]|$1|28|3|-4|5|6|7|58|8|@]|9|@]|A|$]]|$1|29|3|2A|5|E|7|59|8|@$J|5A|K|5B|L|M]]|9|@]|A|$]]|$1|2B|3|-4|5|6|7|5C|8|@]|9|@]|A|$]]|$1|2C|3|-4|5|6|7|5D|8|@]|9|@]|A|$]]|$1|2D|3|2E|5|6|7|5E|8|@]|9|@]|A|$]]|$1|2F|3|-4|5|6|7|5F|8|@]|9|@]|A|$]]|$1|2G|3|2H|5|E|7|5G|8|@]|9|@]|A|$]]|$1|2I|3|-4|5|6|7|5H|8|@]|9|@]|A|$]]|$1|2J|3|-4|5|6|7|5I|8|@]|9|@]|A|$]]|$1|2K|3|2L|5|6|7|5J|8|@]|9|@]|A|$]]|$1|2M|3|-4|5|6|7|5K|8|@]|9|@]|A|$]]|$1|2N|3|2O|5|E|7|5L|8|@]|9|@]|A|$]]|$1|2P|3|-4|5|6|7|5M|8|@]|9|@]|A|$]]|$1|2Q|3|-4|5|6|7|5N|8|@]|9|@]|A|$]]|$1|2R|3|2S|5|6|7|5O|8|@]|9|@]|A|$]]|$1|2T|3|-4|5|6|7|5P|8|@]|9|@]|A|$]]|$1|2U|3|2V|5|E|7|5Q|8|@]|9|@]|A|$]]|$1|2W|3|-4|5|6|7|5R|8|@]|9|@]|A|$]]|$1|2X|3|-4|5|6|7|5S|8|@]|9|@]|A|$]]|$1|2Y|3|2Z|5|6|7|5T|8|@]|9|@]|A|$]]|$1|30|3|-4|5|6|7|5U|8|@]|9|@]|A|$]]|$1|31|3|32|5|E|7|5V|8|@]|9|@]|A|$]]|$1|33|3|-4|5|6|7|5W|8|@]|9|@]|A|$]]|$1|34|3|-4|5|6|7|5X|8|@]|9|@]|A|$]]|$1|35|3|36|5|6|7|5Y|8|@]|9|@]|A|$]]|$1|37|3|-4|5|6|7|5Z|8|@]|9|@]|A|$]]|$1|38|3|39|5|E|7|60|8|@]|9|@]|A|$]]|$1|3A|3|-4|5|6|7|61|8|@]|9|@]|A|$]]|$1|3B|3|-4|5|6|7|62|8|@]|9|@]|A|$]]|$1|3C|3|3D|5|6|7|63|8|@$J|64|K|65|L|3E]]|9|@]|A|$]]|$1|3F|3|-4|5|6|7|66|8|@]|9|@]|A|$]]|$1|3G|3|3H|5|E|7|67|8|@]|9|@]|A|$]]|$1|3I|3|-4|5|6|7|68|8|@]|9|@]|A|$]]|$1|3J|3|-4|5|6|7|69|8|@]|9|@]|A|$]]|$1|3K|3|3L|5|6|7|6A|8|@]|9|@]|A|$]]|$1|3M|3|-4|5|6|7|6B|8|@]|9|@]|A|$]]]|3N|$3O|$5|3P|3Q|3R|A|$3S|3T]]]]

You only asked one question here but there are a dozen or so questions that you should have asked, so I'll answer them all.

<blockquote>
 Here is the sequence which I assumed
</blockquote>

<ol>
<li>Start of class constructor (also known as <code>cctor</code>)</li>
<li>End of cctor </li>
<li>start of Main </li>
<li>start of MyMethod </li>
</ol>

<blockquote>
 Is this correct?
</blockquote>

No. The correct sequence is:

<ol>
<li>Start of cctor for Program, if there is one. There is not.</li>
<li>End of cctor for Program, if there is one. There is not.</li>
<li>Start of Main</li>
<li>Start of cctor for MyClass</li>
<li>End of cctor for MyClass</li>
<li>Start of MyClass.MyMethod</li>
</ol>

<blockquote>
 What if there is a static field initializer?
</blockquote>

The CLR is permitted to change the order in which static field initializers run in some cases. See Jon's page on the subject for details:

<a href="http://csharpindepth.com/Articles/General/Beforefieldinit.aspx">The differences between static constructors and type initializers</a>

<blockquote>
 Is it ever possible for a static method like <code>MyMethod</code> to be called before the cctor of that class completes?
</blockquote>

Yes. If the cctor itself calls MyMethod then obviously MyMethod will be called before the cctor completes.

<blockquote>
 The cctor does not call MyMethod. Is it ever possible for a static method like <code>MyMethod</code> to be called before the cctor of MyClass completes?
</blockquote>

Yes. If the cctor uses another type whose cctor calls MyMethod then MyMethod will be called before the MyClass cctor completes.

<blockquote>
 No cctors call MyMethod, directly or indirectly! Now is it ever possible for a static method like <code>MyMethod</code> to be called before the cctor of MyClass completes?
</blockquote>

No.

<blockquote>
 Is that still true even if there are multiple threads involved?
</blockquote>

Yes. The cctor will finish on one thread before the static method can be called on any thread.

<blockquote>
 Can the cctor be called more than once? Suppose two threads both cause the cctor to be run.
</blockquote>

The cctor is guaranteed to be called at most once, no matter how many threads are involved. If two threads call MyMethod "at the same time" then they race. One of them loses the race and blocks until the MyClass cctor completes on the winning thread. 

<blockquote>
 The losing thread blocks until the cctor is done? Really?
</blockquote>

Really.

<blockquote>
 So what if the cctor on the winning thread calls code that blocks on a lock previously taken by the losing thread?
</blockquote>

Then you have a classic lock order inversion condition. Your program deadlocks. Forever.

<blockquote>
 That seems dangerous. How can I avoid the deadlock?
</blockquote>

If it hurts when you do that then stop doing that. Never do something that can block in a cctor.

<blockquote>
 Is it a good idea to rely upon cctor initialization semantics to enforce complex security requirements? And is it a good idea to have a cctor that does user interactions?
</blockquote>

Neither are good ideas. My advice is that you should find a different way to ensure that the security-impacting preconditions of your methods are met.

blocks|key|2904486|text|或者，您可以在调试器中单步执行。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2904487|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

Or you could step through in the debugger.

blocks|key|238851|text|下面是事情的实际顺序：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|238852|the+static+Main|offset|length|style|CODE|238853|Start+constructor|unordered-list-item|238854|
|238855|End+of++static+Main|238856|Start+constructor|ordered-list-item|238857|Start+of+Main|238858|+of+Main|238859|
static|238860|Start+of+|238861|static+MyClass+constructor|238862|End+of+static|238863|constructor|238864|End+ofstaticMyClassconstructor|238865|Start+of+static+MyClass+constructor|238866|Start+of+the+staticMyClassconstructor|238867|End+ofstatic|238868|Start+of+the+staticMyClassconstructor|238869|End+of+the+staticMyClassconstructor|238870|Start+of+the+staticMyClassconstructor|238871|Start+of+the+staticMyClassconstructor|238872|Start+of+MyMethod|238873|End+ofMain|238874|238875|238876|238877|238878|entityMap^0|0|B|4|0|0|0|F|4|0|0|9|4|0|4|4|0|0|0|7|7|0|1|1|C|7|2|G|7|2|J|7|2|3|J|7|3|H|7|3|J|7|3|J|7|3|9|8|3|6|4|3|2|1|0|0^^$0|@$1|2|3|4|5|6|7|1V|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|1W|8|@$D|1X|E|1Y|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|1Z|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|20|8|@]|9|@]|A|$]]|$1|M|3|N|5|J|7|21|8|@$D|22|E|23|F|G]]|9|@]|A|$]]|$1|O|3|P|5|Q|7|24|8|@]|9|@]|A|$]]|$1|R|3|S|5|Q|7|25|8|@$D|26|E|27|F|G]]|9|@]|A|$]]|$1|T|3|U|5|6|7|28|8|@$D|29|E|2A|F|G]]|9|@]|A|$]]|$1|V|3|W|5|6|7|2B|8|@]|9|@]|A|$]]|$1|X|3|Y|5|Q|7|2C|8|@]|9|@]|A|$]]|$1|Z|3|10|5|Q|7|2D|8|@$D|2E|E|2F|F|G]]|9|@]|A|$]]|$1|11|3|12|5|Q|7|2G|8|@]|9|@]|A|$]]|$1|13|3|14|5|Q|7|2H|8|@]|9|@]|A|$]]|$1|15|3|16|5|Q|7|2I|8|@$D|2J|E|2K|F|G]]|9|@]|A|$]]|$1|17|3|18|5|Q|7|2L|8|@$D|2M|E|2N|F|G]]|9|@]|A|$]]|$1|19|3|1A|5|Q|7|2O|8|@$D|2P|E|2Q|F|G]]|9|@]|A|$]]|$1|1B|3|1C|5|Q|7|2R|8|@]|9|@]|A|$]]|$1|1D|3|1E|5|Q|7|2S|8|@$D|2T|E|2U|F|G]]|9|@]|A|$]]|$1|1F|3|1G|5|Q|7|2V|8|@$D|2W|E|2X|F|G]]|9|@]|A|$]]|$1|1H|3|1I|5|Q|7|2Y|8|@$D|2Z|E|30|F|G]]|9|@]|A|$]]|$1|1J|3|1K|5|Q|7|31|8|@$D|32|E|33|F|G]]|9|@]|A|$]]|$1|1L|3|1M|5|Q|7|34|8|@$D|35|E|36|F|G]]|9|@]|A|$]]|$1|1N|3|1O|5|Q|7|37|8|@$D|38|E|39|F|G]]|9|@]|A|$]]|$1|1P|3|-4|5|6|7|3A|8|@]|9|@]|A|$]]|$1|1Q|3|-4|5|6|7|3B|8|@]|9|@]|A|$]]|$1|1R|3|-4|5|6|7|3C|8|@]|9|@]|A|$]]|$1|1S|3|-4|5|6|7|3D|8|@]|9|@]|A|$]]|$1|1T|3|-4|5|6|7|3E|8|@]|9|@]|A|$]]]|1U|$]]

Here's the actual order in which things go down:

<ol>
<li>Start of <code>Main</code></li>
<li>Start of the static <code>MyClass</code> constructor</li>
<li>End of the static <code>MyClass</code> constructor</li>
<li>Start of <code>MyMethod</code></li>
<li>End of <code>Main</code></li>
</ol>

<pre><code>namespace MyNameSpace
{
 static class MyClass
 {
 static MyClass()
 {
 //Authentication process.. User needs to enter password
 }

 public static void MyMethod()
 {
 //Depends on successful completion of constructor
 }
 }

 class Program
 {
 static void Main(string[] args)
 {
 MyClass.MyMethod();
 }
 }
}
</code></pre>

Here is the sequence which I assumed

<ol>
<li>Start of static constructor</li>
<li>End of static constructor</li>
<li>Start of main</li>
<li>Start of MyMethod</li>
<li>End of main</li>
</ol>

Now in any scenario if 4 will start before 2 I am screwed. Is it possible?

How does a static constructor work?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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namespace MyNameSpace{    static class MyClass    {        static MyClass()        {            //Authentication process.. User needs to enter password        }...

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