程序设计操作符的重载实验报告

1、实验报告二 类和对象一、 实验目的 本实验的目的是熟练学习和掌握类的定义及操作符的重载以及datatime结构。二、 实验内容及结果（一）、第1个题目的要求：定义完整的complexnumber类，在其中重载加减乘除等基本操作符、相等和不等操作符、大小比较操作符以及自增自减操作符 1、程序编程思路：首先定义一个complexnumber类，然后重载各种操作2、程序源码： static void main(string args) console.writeline("请输入第一个复数c1："); complexnumber c1 = new complexnumber(do

2、uble.parse(console.readline(), double.parse(console.readline(); console.writeline("请输入第二个复数c2："); complexnumber c2 = new complexnumber(double.parse(console.readline(), double.parse(console.readline(); complexnumber c3 = c1 + c2; console.writeline("这两个复数的和为 0", c3); complexnumber

3、c4 = c1 - c2; console.writeline("这两个复数的差为 0", c4); complexnumber c5 = c1 * c2; console.writeline("这两个复数的乘积为 0", c5); complexnumber c6 = c1 / c2; console.writeline("这两个复数相除为 0", c6); if (c1 = c2) console.writeline("这两个复数相等"); else if (c1 != c2) console.writelin

4、e("这两个复数不相等"); if (c1 > c2) console.writeline("c1>c2"); else console.writeline("c1<c2"); console.writeline("c1自增结果为 0", c1+); console.writeline("c1自减结果为 0", c1-); console.readline(); public class complexnumber private double x, y; public dou

5、ble x get return x; set x = value; public double y get return y; set y = value; public complexnumber(double x, double y) this.x = x; this.y = y; public static complexnumber operator +(complexnumber c1, complexnumber c2) return new complexnumber(c1.x + c2.x, c1.y + c2.y); public static complexnumber

6、operator -(complexnumber c1, complexnumber c2) return new complexnumber(c1.x - c2.x, c1.y - c2.y); public static complexnumber operator *(complexnumber c1, complexnumber c2) return new complexnumber(c1.x * c2.x - c1.y * c2.y, c1.y * c2.x + c1.x * c2.y); public static complexnumber operator /(complex

7、number c1, complexnumber c2) return new complexnumber(c1.x * c2.x + c1.y * c2.y) / (c2.x * c2.x + c2.y * c2.y), (c1.y * c2.x - c1.x * c2.y) / (c2.x * c2.x + c2.y * c2.y); public static bool operator =(complexnumber c1, complexnumber c2) return (c1.x = c2.x) && (c1.y = c2.y); public static bo

8、ol operator !=(complexnumber c1, complexnumber c2) return (c1.x != c2.x) | (c1.y != c2.y); public static bool operator >(complexnumber c1, complexnumber c2) double a = c1.x * c1.x + c1.y * c1.y; double b = c2.x * c2.x + c2.y * c2.y; return a > b; public static bool operator <(complexnumber

9、c1, complexnumber c2) double a = c1.x * c1.x + c1.y * c1.y; double b = c2.x * c2.x + c2.y * c2.y; return a < b; public static complexnumber operator +(complexnumber c1) return new complexnumber(c1.x+, c1.y+); public static complexnumber operator -(complexnumber c1) return new complexnumber(c1.x-,

10、 c1.y-); public override string tostring() return string.format("0+1i",this.x,this.y); 3、实验结果： （二）、第2个题目的要求：知道一个人的生日，如何计算他的年龄（严格按照周岁计算） 1、程序编程思路： 首先考虑的是怎么来计算年龄，想到用当前日期减去生日日期，用年减去年，月减去月，日减去日，然后思考，各种情况下的年龄计算方法，画出程序的流程图，根据流程图写出源程序；接着就想到如何让操作者方便的一次性将生日的年月日输入，就考虑用字符数组，然后用string类的split方法将输入的字符串劈

11、开。2、程序源码：static void main(string args) int a, b, c, d, e, f; datetime s = datetime.today; d = s.year; e = s.month; f = s.day; console.writeline("请输入您的生日（格式为0000-00-00）"); string g = console.readline(); char sep1 = '-' string h= g.split(sep1); int x = new inth.length; for (int i = 0

12、; i < h.length; i+) xi = int.parse(hi); a = d - x0; b = e - x1; c = f - x2; if (a < 0) | (a = 0 && b < 0) | (a = 0 && b = 0 && c < 0) console.writeline("您输入的生日有误！"); else if (a = 0) console.writeline("您不满一岁！"); else if (a > 0) if (b > 0) console.writeline("您今年0岁！",a); else if (b < 0) console.writeline("您今年0岁！", a - 1); else if (b = 0) if (c<0) console.writeline("您今年0岁！", a-1); else console.writeline("您今年0岁！", a); console.readline(); 3、实验结果： 三、 实验效果对第一个题目,要熟记操作符重载的方法与格式