简单的异步编程入门例子

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading;

namespace 异步编程例子
{
    class Program
    {
        public delegate int CaculateDelegate(int min, int max);
        static void Main(string[] args)
        {
            CaculateDelegate d=Caculate;
            Console.WriteLine("请输入:");
            int min = int.Parse(Console.ReadLine());
            Console.WriteLine("请输入");
            int max = int.Parse(Console.ReadLine());
            IAsyncResult ret = d.BeginInvoke(min,max,null,null);
            Console.WriteLine("正在计算中,请耐心等待.....");  
            while (!ret.IsCompleted)
            {
                Console.Write(".");
                Thread.Sleep(1000);
            }
            int count = (int)d.EndInvoke(ret);
            Console.WriteLine();
            Console.WriteLine("{0}+{1}={2}",min,max,count);
            Console.ReadKey();
        }

        public static int Caculate(int min, int max)
        {
            int count;
            Thread.Sleep(5000);
            count = min + max;
            return count;
        }
    }
}

上例使用轮询方法不断询问异步调用是否完成,会浪费不少CPU时间在循环等待上,下例可以让异步调用方法在结束时自动调用一个函数,并在这个函数中显示处理结果。

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading;

namespace 异步编程例子
{
    class Program
    {
        public delegate int CaculateDelegate(int min, int max);
        public static CaculateDelegate d = Caculate;
        static void Main(string[] args)
        {
            Console.WriteLine("请输入:");
            int min = int.Parse(Console.ReadLine());
            Console.WriteLine("请输入");
            int max = int.Parse(Console.ReadLine());
            IAsyncResult ret = d.BeginInvoke(min,max,ShowResult,(min+","+max));
            Console.WriteLine("正在计算中,请耐心等待.....");
            Console.ReadKey();
        }

        public static int Caculate(int min, int max)
        {
            int count;
            Thread.Sleep(5000);
            count = min + max;
            return count;
        }

        public static void ShowResult(IAsyncResult ret)
        {
            int count = (int)d.EndInvoke(ret);  
            string[] n = ((string)ret.AsyncState).Split(',');
            Console.WriteLine();
            Console.WriteLine("{0}+{1}={2}", n[0],n[1], count);
        }
    }
}
原文地址:https://www.cnblogs.com/jiekk/p/1977505.html