带有(和不带)混合静态构造函数的C#静态初始化程序

Question

我已经阅读了C#语言规范(v5.0)的相关部分,但我找不到与我所看到的相关的部分.

如果您运行下面的代码,您将看到下面的输出,这是我所期望的:

using System;

class Test {
   static int count = 0;
   static void Main() {
      Console.WriteLine("In Main(), A.X=" + A.X);
   }

   public static int F(string message) {
      Console.WriteLine(message);
      A.X = ++count;

      Console.WriteLine("\tA.X has been set to " + A.X);
      B.Y = ++count;

      Console.WriteLine("\tB.Y has been set to " + B.Y);
      return 999;
   }
}
class A {
   static A() { }
   public static int U = Test.F("Init A.U");
   public static int X = Test.F("Init A.X");
}

class B {
   static B() { }
   public static int R = Test.F("Init B.R");
   public static int Y = Test.F("Init B.Y");
}

输出是:

Init A.U
    A.X has been set to 1
Init B.R
    A.X has been set to 3
    B.Y has been set to 4
Init B.Y
    A.X has been set to 5
    B.Y has been set to 6
    B.Y has been set to 2
Init A.X
    A.X has been set to 7
    B.Y has been set to 8
In Main(), A.X=999

这正是我所期望的输出.特别要注意的是,即使方法F()正在使用参数"Init AU"执行,一旦遇到对BY的引用,它就会被再次调用(如果您愿意,则会中断),从而导致B的静态初始化程序执行.一旦B的静态构造函数完成,我们再次返回到F()的AU调用,它将BY设置为6然后设置为2.因此,希望这个输出对每个人都有意义.

这是我不理解的:如果你注释掉B的静态构造函数,这就是你看到的输出:

Init B.R
        A.X has been set to 1
        B.Y has been set to 2
Init B.Y
        A.X has been set to 3
        B.Y has been set to 4
Init A.U
        A.X has been set to 5
        B.Y has been set to 6
Init A.X
        A.X has been set to 7
        B.Y has been set to 8
In Main(), A.X=999

C#Spec(v5.0)的第10.5.5.1节和第10.12节表示当引用"类的任何静态成员"时,将触发A的静态构造函数(及其静态初始化程序)执行.然而,我们在F()中引用了AX,并且没有触发A的静态构造函数(因为它的静态初始化器没有运行).

由于A有一个静态构造函数,我希望那些初始化程序运行(并中断)对F()的"Init BR"调用,就像B的静态构造函数在我展示的"Init AU"调用中中断A对F()的调用一样在开始.

谁能解释一下？Af面值看起来违反规范,除非规范的其他部分允许这样做.

谢谢

Answer 1

我想我明白这里发生了什么，尽管我没有很好的解释为什么会这样。

测试程序有点太粗糙，无法看出发生了什么。让我们做一个小调整：

class Test {
   static int count = 0;
   static void Main() {
      Console.WriteLine("In Main(), A.X=" + A.X);
   }

   public static int F(string message) {
       Console.WriteLine("Before " + message);
       return FInternal(message);
   }

   private static int FInternal(string message) {
      Console.WriteLine("Inside " + message);
      A.X = ++count;

      Console.WriteLine("\tA.X has been set to " + A.X);
      B.Y = ++count;

      Console.WriteLine("\tB.Y has been set to " + B.Y);
      return 999;
   }
}
class A {
   static A() { }
   public static int U = Test.F("Init A.U");
   public static int X = Test.F("Init A.X");
}

class B {
   static B() { }
   public static int R = Test.F("Init B.R");
   public static int Y = Test.F("Init B.Y");
}

输出与问题中的输出类似，但有更多细节：

Before Init A.U  
Inside Init A.U  
    A.X has been set to 1  
Before Init B.R  
Inside Init B.R  
    A.X has been set to 3  
    B.Y has been set to 4  
Before Init B.Y  
Inside Init B.Y  
    A.X has been set to 5  
    B.Y has been set to 6  
    B.Y has been set to 2  
Before Init A.X  
Inside Init A.X  
    A.X has been set to 7  
    B.Y has been set to 8  
In Main(), A.X=999

这里没什么奇怪的。删除 B 的静态构造函数，这就是您得到的结果：

Before Init A.U  
Before Init B.R  
Inside Init B.R  
    A.X has been set to 1  
    B.Y has been set to 2  
Before Init B.Y  
Inside Init B.Y  
    A.X has been set to 3  
    B.Y has been set to 4  
Inside Init A.U  
    A.X has been set to 5  
    B.Y has been set to 6  
Before Init A.X  
Inside Init A.X  
    A.X has been set to 7  
    B.Y has been set to 8  
In Main(), A.X=999

现在这很有趣。我们可以看到原始输出具有误导性。我们实际上是从尝试初始化开始的A.U。这并不奇怪，因为 A 应该首先初始化，因为A.X它是在 Main 中访问的。下一部分很有趣。看起来，当 B 没有静态构造函数时，CLR 会在进入方法之前FInternal中断将要访问 B 的字段 ( )的方法。将此与其他情况进行对比。B 的初始化被延迟，直到我们实际访问 B 的字段。

我不完全确定为什么事情会按照这个特定的顺序完成，但是您可以看到 B 的初始化没有被中断以初始化 A 的原因是 A 的初始化已经开始。