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  • C# 委托和方法

    委托是一种特殊的引用类型,它将方法也作为特殊的对象封装起来,从而将方法作为变量或参数进行传递

     1 using System;
     2 using System.Collections.Generic;
     3 using System.Linq;
     4 using System.Text;
     5 
     6 namespace ConsoleApplication1
     7 {
     8     delegate void DualFunction(double x,double y);
     9     class Program
    10     {
    11         static void Main()
    12         {
    13             DualFunction fun1;
    14             double a = 2.5, b = 2;
    15             Console.WriteLine("请选择函数(加0减1乘2除3):");
    16             int i = int.Parse(Console.ReadLine());
    17             if (i == 1)
    18                 fun1 = new DualFunction(Sub);
    19             else if (i == 2)
    20                 fun1 = new DualFunction(Mul);
    21             else if (i == 3)
    22                 fun1 = new DualFunction(Div);
    23             else
    24                 fun1 = new DualFunction(Add);
    25             fun1(a, b);
    26             Console.ReadLine();
    27         }
    28         static void Add(double x,double y)
    29         {
    30             Console.WriteLine("{0}+{1}={2}", x, y, x + y);
    31         }
    32         static void Sub(double x,double y)
    33         {
    34             Console.WriteLine("{0}-{1}={2}", x, y, x - y);
    35         }
    36         static void Mul(double x,double y)
    37         {
    38             Console.WriteLine("{0}*{1}={2}", x, y, x * y);
    39         }
    40         static void Div(double x,double y)
    41         {
    42             Console.WriteLine("{0}/{1}={2}", x, y, x / y);
    43         }
    44     }
    45 }
    View Code

    DaulFunction fun1=new DualFunction(Sub);

    DualFunction fun1=Sub;//这两种方式也行

    还可以将委托对象放在数组中  比如

    static void Main()
    {
    DualFunction[] funs = new DualFunction[] { Add, Sub, Mul, Div };
    double a=2.5, b=2.0;
    Console.WriteLine("请选择函数(加0减1乘2除3):");
    int i = int.Parse(Console.ReadLine());
    funs[i](a, b);
    Console.ReadLine();
    }

    一个委托对象还可以封装多个方法   委托对象的合并和删除  比如DualFunction fun1 = funs[1] + funs[2];

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  • 原文地址:https://www.cnblogs.com/wangkaipeng/p/4516021.html
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