c# - 将局部变量传递给 .net 中的另一个线程

Question

在下面的代码中，为什么我们没有得到 NullReference 异常并且 var2 值为 56 尽管 TestMethod 肯定在 'Messagebox' 行之前完成？我读了this Eric Lippert 和 this 的出色回答博文，但我还是不明白。

void TestMethod()
{
    int var1 = 10;
    List<long> list1 = new List<long>();
    for (int i = 0; i < 5; i++)
        list1.Add(i);

    ThreadPool.QueueUserWorkItem(delegate
    {
        int var2 = var1;
        Thread.Sleep(1000);
        list1.Clear();
        MessageBox.Show(var2.ToString());
    });
    var1 = 56;
}

Answer 1

我认为这是因为委托(delegate)围绕变量 var1 形成了闭包。可能看看闭包在内部是如何工作的会对你有所帮助。可以引用说明here

The compiler (as opposed to the runtime) creates another class/type. The function with your closure and any variables you closed over/hoisted/captured are re-written throughout your code as members of that class. A closure in .Net is implemented as one instance of this hidden class.

有了那个，我相信编译器生成的代码大致如下所示:

void TestMethod()
{
    UnspeackableClosureClass closure = new UnspeackableClosureClass(10);
    List<long> list1 = new List<long>();
    for (int i = 0; i < 5; i++)
        list1.Add(i);

    ThreadPool.QueueUserWorkItem(closure.AutoGeneratedMethod);
    closure.closureVar = 56;
}

public class UnspeackableClosureClass
{
   public int closureVar;
   public UnspeackableClosureClass(int val){closureVar=val}

   public void AutoGeneratedMethod(){
     int var2 = closureVar;
     Thread.Sleep(1000);
     list1.Clear();
     MessageBox.Show(var2.ToString());
  }
}