运算符重载关键字operator

operator关键字用来重载内置运算符，使用方法如下：

public class OperatorController : Controller
{
        //
        // GET: /Operator/
        public int num, den;
        public OperatorController(int num, int den)
        {
            this.num = num;
            this.den = den;
        }

        // overload operator +
        public static OperatorController operator +(OperatorController a, OperatorController b)
        {
            return new OperatorController(a.num * b.den + b.num * a.den, a.den * b.den);
        }

        // overload operator *
        public static OperatorController operator *(OperatorController a, OperatorController b)
        {
            return new OperatorController(a.num * b.num, a.den * b.den);
        }
}

上面的OperatorController类提供了一个构造函数和"+"，"*"的重载方法。

下面是一组执行代码：

var a = new OperatorController(, );
var b = new OperatorController(, );
var c = new OperatorController(, );
var d = a + b;
var e = a * b + c;

为了便于理解，我将执行过程中的变量一一输出：

a

Response.Write("a:" + a.num + "--" + a.den + "<br/>");

输出：a:1--2

b

Response.Write("b:" + b.num + "--" + b.den + "<br/>");

输出：b:3--7

c

Response.Write("c:" + c.num + "--" + c.den + "<br/>");

输出：b:2--3

d

Response.Write("d:" + d.num + "--" + d.den + "<br/>");

输出：d:13--14

e

Response.Write("e:" + e.num + "--" + e.den + "<br/>");

输出：e:37--42

接下来，重点说一下重载"+"是如何实现的：

以var d = a + b;为例， 分解一下执行过程，

step1. 执行"+"重载方法

 public static OperatorController operator +(OperatorController a, OperatorController b)
 {
      return new OperatorController(a.num * b.den + b.num * a.den, a.den * b.den);
 }

当a和b做"+"运算时，会把a和b作为参数传入到这个重载方法（重载运算符只支持两个参数），通过运算可以演变成

public static OperatorController operator +(OperatorController a, OperatorController b)
{
      return new OperatorController(13, 14);
}

step2. 执行构造函数

public OperatorController(int num, int den)
{
     this.num = num;
     this.den = den;
}

所以执行Response.Write("d:" + d.num + "--" + d.den + "<br/>");时，会输出d:13--14

在下面重载"=="的方法里，没有自己的业务逻辑，跟Equals(x,y)效果一样

public static bool operator ==(BaseEntity x, BaseEntity y)
{
     return Equals(x, y);
}