C# -- 等待异步操作执行完成的方式

1. 等待异步操作的完成，代码实现

 1 class Program
 2     {
 3         static void Main(string[] args)
 4         {
 5             Func<int, int> mySum1 = SumNumbers;
 6             Func<int, int> mySum2 = SumNumbers;
 7             Func<int, int> mySum3 = SumNumbers;
 8 
 9             AsyncCallback callback = c => Console.WriteLine("线程ID:{0},回调函数执行:{1}",Thread.CurrentThread.ManagedThreadId,c.AsyncState);
10 
11 
12             IAsyncResult result1= mySum1.BeginInvoke(10000, callback, "第一个异步");
13             IAsyncResult result2= mySum2.BeginInvoke(10000, callback, "第二个异步");
14             IAsyncResult result3= mySum3.BeginInvoke(10000, callback, "第三个异步");
15 
16 
17             //异步等待方式1：使用EndInvoke
18             int asyc1 = mySum1.EndInvoke(result1);
19             Console.WriteLine("第一个异步已经执行完成，结果:{0}",asyc1);
20 
21             //异步等待方式2：使用AsyncWaitHandle.WaitOne，参数-1表示一直等待到执行完成
22             if (result2.AsyncWaitHandle.WaitOne(-1))
23             {
24                 Console.WriteLine("第二个异步执行完成");
25             }
26 
27             //异步等待方式3：使用IsCompleted判断
28             while (!result3.IsCompleted)
29             {
30                 Console.WriteLine("第三个异步还在执行中....");
31                 Thread.Sleep(300);
32             }
33             Console.WriteLine("第三个异步执行完成");
34 
35             Console.ReadKey();
36         }
37 
38         public static int SumNumbers(int count)
39         {
40             int sum = 0;
41             for (int i = 0; i < count; i++)
42             {
43                 sum += i;
44             }
45 
46             Thread.Sleep(3000);
47             return sum;
48         }
49     }