委托之Lambda表达式的推导

public delegate void ConsoleWriteStr(string name,DateTime now);

public delegate int DelegateAdd(int x, int y);

//第一步使用匿名函数
            ConsoleWriteStr cws1= new ConsoleWriteStr(
                delegate(string name, DateTime now){
                  Console.WriteLine("我叫{0}，我喜欢学习{1}", name, now);
                }
            );

//第二步将delegate换成=> ， =>的左边是参数列表，右边是函数体
ConsoleWriteStr cws2 = (string name, DateTime now) => { Console.WriteLine("我叫{0}，我喜欢学习{1}", name, now); };

//第三步去掉操作类型，原因是委托要求与方法签名一致
ConsoleWriteStr cws3 = ( name, now) => { Console.WriteLine("我叫{0}，我喜欢学习{1}", name, now); };

//第三步去掉方法体的大括号，如果方法体只有一行，可以将大括号去掉
ConsoleWriteStr cws4 = (name, now) => Console.WriteLine("我叫{0}，我喜欢学习{1}", name, now);

DelegateAdd func = (x, y) => { return x + y; };
DelegateAdd func1 = (x, y) => x + y;

            //系统自带的委托Action，该委托不带返回值
            Action ac1 = () => { }; //无参数无返回值
            Action<string> ac2 = x =>{ }; //泛型委托一个参数可以去掉小括号
            Action<string> ac3 = x => { }; //泛型委托
            Action<string,int,DateTime,Action> ac4 = (x, y, z, a) => { }; //泛型委托

            //系统自带的委托Func，该委托带有返回值，最多16个参数，最后一个参数为返回值
            Func<string> fun1 = () => " ";
            Func<DateTime> fun2 = () => DateTime.Now;
            Func<string, double,long, int> fun3 = (s,l, d) => { return 1+l; };

委托之Lambda表达式的推导

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