73

我浏览了很多文章,但我仍然不清楚我们通常创建的普通委托和多播委托之间的区别。

public delegate void MyMethodHandler(object sender);
MyMethodHandler handler = new MyMethodHandler(Method1);
handler += Method2;
handler(someObject);

上面的委托 MyMethodHandler 将调用这两个方法。现在多播委托从何而来。我读过他们可以调用多种方法,但我担心我对委托的基本理解是不正确的。

4

6 回答 6

80

这篇文章很好地解释了它:

delegate void Del(string s);

class TestClass
{
    static void Hello(string s)
    {
        System.Console.WriteLine("  Hello, {0}!", s);
    }

    static void Goodbye(string s)
    {
        System.Console.WriteLine("  Goodbye, {0}!", s);
    }

    static void Main()
    {
        Del a, b, c, d;

        // Create the delegate object a that references 
        // the method Hello:
        a = Hello;

        // Create the delegate object b that references 
        // the method Goodbye:
        b = Goodbye;

        // The two delegates, a and b, are composed to form c: 
        c = a + b;

        // Remove a from the composed delegate, leaving d, 
        // which calls only the method Goodbye:
        d = c - a;

        System.Console.WriteLine("Invoking delegate a:");
        a("A");
        System.Console.WriteLine("Invoking delegate b:");
        b("B");
        System.Console.WriteLine("Invoking delegate c:");
        c("C");
        System.Console.WriteLine("Invoking delegate d:");
        d("D");
    }
}
/* Output:
Invoking delegate a:
  Hello, A!
Invoking delegate b:
  Goodbye, B!
Invoking delegate c:
  Hello, C!
  Goodbye, C!
Invoking delegate d:
  Goodbye, D!
*/
于 2010-02-03T13:40:38.977 回答
55

C# 规范规定所有委托类型都必须可转换为System.Delegate. 事实上,实现的方式是所有委托类型都派生自System.MulticastDelegate,而委托类型又派生自System.Delegate

明白了吗?我不确定这是否回答了你的问题。

于 2010-02-03T15:57:25.600 回答
15

“所有委托实例都具有多播功能。” - http://msdn.microsoft.com/en-us/library/orm-9780596527570-03-04.aspx

“在 C# 中,所有委托类型都支持多播” - http://msdn.microsoft.com/en-us/library/orm-9780596516109-03-09.aspx

于 2013-01-10T18:14:37.857 回答
7

澄清一点:所有委托都是类的实例MulticastDelegate,无论它们是否具有一个或多个目标方法。原则上,具有单个目标或多个目标的委托之间没有区别,尽管运行时针对具有单个目标的常见情况进行了一些优化。(但不可能有 0 个目标的委托,它是一个或多个。)

当您实例化像 之类的委托new MyMethodHandler(Method1)时,您会创建一个具有单个目标(Method1方法)的委托。

具有多个目标的委托是通过组合两个现有委托来创建的。生成的委托将具有目标的总和。委托可以与 显式组合Delegate.Combine(),也可以通过+=在现有委托上使用运算符来隐式组合,如您的示例所示。

调用委托依次调用委托中的每个目标。因此,在您的示例handler(someObject);中将调用两个方法(Method1Method2),因为您已经使用这两个目标创建了一个委托。

于 2015-07-03T10:44:36.003 回答
4

很抱歉添加到其他人的答案,但我认为代表是按添加顺序调用的。

“多播代表”部分

http://msdn.microsoft.com/en-us/library/orm-9780596527570-03-04.aspx

于 2013-08-21T08:45:32.033 回答
-2

多播委托是引用多个函数的委托。当您调用多播委托时,将调用委托所指向的所有函数。

类型 1:

0 参数和 void 返回类型委托 -

方法1 -

using System;

delegate void SampleDelegate ();    //A delegate with 0 argument and void     return type is declared

class MainClass
{
    public static void Main ()
    {
        SampleDelegate Del1 = new SampleDelegate (Message1);         //Del1 declared which points to function Message1
        SampleDelegate Del2 = new SampleDelegate (Message2);        //Del2 declared which points to function Message2
        SampleDelegate Del3 = new SampleDelegate (Message3);        //Del3 declared which points to function Message3
        SampleDelegate Del4 = Del1 + Del2 + Del3;                   //Del4 declared which points to function Message4

        //Del4 is then initialized as sum of Del1 + Del2 + Del3

        Del4 ();        //Del4 is invoked;

        //Del4 in turn invokes Del1, Del2 and Del3 in the same order they were initialized to Del4
        //Del1, Del2, Del3 in turn invokes their respective functions to which they point to
        //The three functions Message1, Message2 and Message3 gets executed one after another

    }

        //Output:
        //This is message 1
        //This is message 2
        //This is message 3

        Del4 - Del1;    //Removes Del1 from Del4
        Del4();           

        //New Output:
        //This is message 2
        //This is message 3

        Del4 + Del1;    //Again adds Del1 to Del4
        Del4();

        //New Output:
        //This is message 1
        //This is message 2
        //This is message 3


    public static void Message1 ()      //First sample function matching delegate signature
    {
        Console.WriteLine ("This is message 1");
    }

    public static void Message2 ()      //Second sample function
    {
         Console.WriteLine ("This is message 2");
    }

    public static void Message3 ()      //Third sample function
    {
        Console.WriteLine ("This is message 3");
    }
}

方法2 -

using System;

delegate void SampleDelegate ();

class MainClass
{
    public static void Main ()
    {
        SampleDelegate del = new SampleDelegate (Message1);         //Declares del and initializes it to point to method Message1
        del += Message2;                                        //Now method Message2 also gets added to del. Del is now pointing to two methods, Message1 and Message2. So it is now a MultiCast Delegate
        del += Message3;                                        //Method Message3 now also gets added to del

        del ();                                                 //Del invokes Message1, Message2 and Message3 in the same order as they were added

        /*
        Output:
        This is Message1
        This is Message2
        This is Message3
        */

        del -= Message1;                                        //Method     Message1 is now removed from Del. It no longer points to Message1
                                                                //Del invokes the two remaining Methods Message1 and Message2 in the same order
        del ();
        /*
        New Output:
        This is Message2
        This is Message3
        */

        del += Message4;                                        //Method Message4 gets added to Del. The three Methods that Del oints to are in the order 1 -> Message2, 2 -> Message3, 3 -> Message4
                                                                //Del invokes the three methods in the same order in which they are present.
        del ();
        /*
        New Output:
        This is Message2
        This is Message3
        This is Message4
        */

    }

    public static void Message1 ()
    {
        Console.WriteLine ("This is Message1");
    }

    public static void Message2 ()
    {
        Console.WriteLine ("This is Message2");
    }

    public static void Message3 ()
    {
        Console.WriteLine ("This is Message3");
    }

    public static void Message4 ()
    {
        Console.WriteLine ("This is Message4");
    }
}

类型 2:

0 个参数和 int 返回类型委托

方法1-

using System;

delegate int SampleDelagate ();

class MainClass
{
    public static void Main ()
   {
        SampleDelagate del1 = new SampleDelagate (Method1);
        SampleDelagate del2 = new SampleDelagate (Method2);
        SampleDelagate del3 = new SampleDelagate (Method3);
        SampleDelagate del4 = del1 + del2 + del3;

        int ValueReturned = del4 ();

        //Del4 invokes Del1, Del2, Del3 in the same order. Here the return type is int. So the return of last delegate del3 is returned. Del3 points to Method3. So returned value is 3.

        Console.WriteLine (ValueReturned);

        //Output: 3
    }

    public static int Method1 ()
    {
        return 1;
    }

    public static int Method2 ()
    {
        return 2;
    }

    public static int Method3 ()
    {
        return 3;
    }
}

方法2-

与类型 1 相同的过程

因此,当有 MultiCast Delegate 的返回类型时,返回值是最后一个委托的返回值。

类型 3:

int、int、ref int 参数和 void 返回类型委托 -

using System;

delegate void SampleDelegate (ref int SampleReferenceParameter);

class MainClass
{
    public static void Main ()
    {
        SampleDelegate del1, del2, del3, del4;
        del1 = new SampleDelegate (SampleMethodOne);
        del2 = new SampleDelegate (SampleMethodTwo);
        del3 = new SampleDelegate (SampleMethodTwo);
        del4 = del1 + del2 + del3 - del3;

        int SampleReferenceParameterValue = 0;
        del4 (ref SampleReferenceParameterValue);

        Console.WriteLine (SampleReferenceParameterValue); 
    }

    public static void SampleMethodOne (ref int SampleReferenceParameter)
    {
        SampleReferenceParameter = 1;
    }

    public static void SampleMethodTwo (ref int SampleReferenceParameter)
    {
        SampleReferenceParameter = 2;
    }

    public static void SampleMethodThree (ref int SampleReferenceParameter)
    {
        SampleReferenceParameter = 3;
    }
}

/*
Here del4 is first set as sum of del1, del2 and del3. Then del3 is subtracted from del4. So del4 now has del1, del2.

When del4 is invoked, first del1 and then del2 is invoked.

del1 sets reference parameter to 1. del2 sets reference parameter to 2.

But since del2 is called last final value of reference parameter is 2
*/
于 2017-12-27T13:34:39.610 回答