1

想知道如果扩展方法仍然有效，我的静态构造函数是否失败并抛出异常？记录器旨在帮助检测扩展方法中的问题。如果它们仍然无法工作，我将不得不尝试 catch 并确保构造函数成功。我希望能够让它抛出异常，因为希望调用代码可以记录错误。（这只是类似于我正在考虑的示例代码）

     public static class Extensions 
     {

        private static readonly log4net.ILog log;
        private const TIME_TO_CHECK;

        static Extensions() 
        {
            log = log4net.LogManager.GetLogger           (System.Reflection.MethodBase.GetCurrentMethod().DeclaringType);  //could maybe throw exception

            TIME_TO_CHECK = readInFromFile();  //could maybe           throw                           exception            }           

        public static DateTime StartOfWeek(this DateTime dt, DayOfWeek startOfWeek) {
        int diff = dt.DayOfWeek - startOfWeek;
        if (diff < 0) {
        diff += 7;
      }
      return dt.AddDays(-1 * diff).Date;
    }
  }

我确实四处搜索（希望这不是重复），发现从静态构造函数中抛出异常通常不是很好。在大多数情况下，我认为这些类是可以实例化的，而不仅仅是扩展方法。

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1 回答 1

6

想知道如果扩展方法仍然有效，我的静态构造函数是否失败并抛出异常？

不。如果任何类型的类型初始化程序（无论是否使用静态构造函数）失败，则该类型此后基本上无法使用。

很容易证明这一点......

using System;

static class Extensions
{
    static Extensions()
    {
        Console.WriteLine("Throwing exception");
        throw new Exception("Bang");
    }

    public static void Woot(this int x)
    {
        Console.WriteLine("Woot!");
    }
}

class Test
{

    static void Main()
    {
        for (int i = 0; i < 5; i++)
        {
            try
            {
                i.Woot();
            }
            catch (Exception e)
            {
                Console.WriteLine("Caught exception: {0}", e.Message);
            }
        }
    }
}

输出：

Throwing exception
Caught exception: The type initializer for 'Extensions' threw an exception.
Caught exception: The type initializer for 'Extensions' threw an exception.
Caught exception: The type initializer for 'Extensions' threw an exception.
Caught exception: The type initializer for 'Extensions' threw an exception.
Caught exception: The type initializer for 'Extensions' threw an exception.
于 2013-07-05T15:56:52.900 回答