为什么在 Threadlocal 委托中初始化线程静态属性不会为第一个线程启动它？

Question

我已经创建了3个线程，并且所有线程都在线程列表属性上具有增量操作，除了thread1。我还以图11的值初始化ThreadLocal委托中的线程属性。在这里，我总是在第一个线程中获得num = 0的值。为什么呢？

class Program
    {
        //static int numDuplicate = 0;
        [ThreadStatic]
        static int num = 5;
        public static ThreadLocal<int> _field = new ThreadLocal<int>(() =>
        {
            num = 11;
            //numDuplicate = num;
            Console.WriteLine("Threadstatic variable value in Threadlocal's delegate = " + num.ToString());
            return Thread.CurrentThread.ManagedThreadId;
        });

        public static void Main(string[] args)
        {
            Thread t1 = new Thread(new ThreadStart(() =>
            {
                Console.WriteLine("Threadlocal attribute value for thread 1: " + _field + ". Threadstatic variable value = " + num.ToString());
            }));
            Thread t2 = new Thread(new ThreadStart(() =>
            {
                _field.Value++;
            Console.WriteLine("Threadlocal attribute value for thread 2: " + _field + ". Threadstatic variable value = " + num.ToString());
            }));
            Thread t3 = new Thread(new ThreadStart(() =>
            {
                _field.Value++;
                Console.WriteLine("Threadlocal attribute value for thread 3: " + _field + ". Threadstatic variable value = " + num.ToString());
            }));
            t1.Start();
            t2.Start();
            t3.Start();
            Console.ReadLine();
        }
    }
//Output:
Threadstatic variable value in Threadlocal's delegate = 11
Threadstatic variable value in Threadlocal's delegate = 11
Threadstatic variable value in Threadlocal's delegate = 11
Threadlocal attribute value for thread 1: 10. Threadstatic variable value = 0
Threadlocal attribute value for thread 3: 13. Threadstatic variable value = 11
Threadlocal attribute value for thread 2: 12. Threadstatic variable value = 11

Answer 1

，因为'不能为线程构图分配一个初始值。

请勿指定用线程构图标记的字段的初始值，因为当类构造函数执行时，这种初始化仅发生一次，因此仅影响一个线程。如果您不指定初始值，则可以依靠该字段（如果是值类型），或者如果是参考类型，则可以依靠其默认值。

。

来源：MSDN

---

您可以看一下ThreadLocal<>

static ThreadLocal<int> num  = new ThreadLocal<int>(() => 5);

Func<>将作为for foreach新线程执行。在这种情况下，它只是返回5。