第三周任务三之分数类运算符的重载

和任务2遇到了相似的问题，看不出来了...

源程序:

/*(文件注释头部开始） *程序的版权和版本声明部分 *Copyright (c) 2011,烟台大学计算机学院学生 *All rights reserved. *文件名称：分数类运算符的重载*作    者：2011级计114-3张宗佳 *完成日期：2011年4月11号 *版本号：vc* 对任务及求解方法的描述部分 * 输入描术：* 问题描述：实现分数类中的运算符重载，在分数类中可以完成分数的加减乘除（运算后再化简）、求反、比较（6种关系）的运算。* 程序输出：* 程序头部的注释结束 */  #include<iostream>using namespace std;int gcd(int a,int b);class CFraction{private:int nume;  // 分子int deno;  // 分母public://构造函数及运算符重载的函数声明CFraction(int nu = 0,int de = 1);  CFraction operator + (CFraction &c1);CFraction operator - (CFraction &c1);CFraction operator * (CFraction &c1);CFraction operator / (CFraction &c1);bool operator > (CFraction &c1);bool operator < (CFraction &c1);bool operator >= (CFraction &c1);bool operator <= (CFraction &c1);bool operator = (CFraction &c1);bool operator != (CFraction &c1);CFraction operator - ();//求反void Simplify();void output();};//重载函数的实现及用于测试的main()函数CFraction::CFraction(int nu,int de) {nume = nu;deno = de;}CFraction CFraction::operator + (CFraction &c1){CFraction c;c.nume = deno * c1.nume + nume * c1.deno;c.deno = deno * c1.deno;c.Simplify();return c;}CFraction CFraction::operator - (CFraction &c1){CFraction c;c.nume = nume * c1.deno - deno * c1.nume;c.deno = deno * c1.deno;c.Simplify();return c;}CFraction CFraction::operator * (CFraction &c1){CFraction c;c.nume = nume * c1.nume;c.deno = deno * c1.deno;c.Simplify();return c;}CFraction CFraction::operator / (CFraction &c1){CFraction c;c.nume = nume * c1.deno;c.deno = deno * c1.nume;    c.Simplify();return c;}bool CFraction::operator > (CFraction &c1){if(nume * c1.deno > deno * c1.nume){return true;}else {return false;}}bool CFraction::operator < (CFraction &c1){if(nume * c1.deno < deno * c1.nume){return true;}else {return false;}}bool CFraction::operator >= (CFraction &c1){if(nume * c1.deno < deno * c1.nume){return false;}else {return true;}}bool CFraction::operator <= (CFraction &c1){if(nume * c1.deno > deno * c1.nume){return false;}else {return true;}}bool CFraction::operator = (CFraction &c1){if(nume * c1.deno == deno * c1.nume){return true;}else {return false;}}bool CFraction::operator != (CFraction &c1){if(nume * c1.deno != deno * c1.nume){return true;}else {return false;}}//求反CFraction CFraction::operator - (){return CFraction(-nume,deno);}void CFraction::Simplify(){int t,n,d;n = nume;d = deno; while(n > 0){t = d % n;d = n;n = t;}nume = nume / n;deno = deno / n;return;}void CFraction::output(){cout << nume << "/" << deno << endl;}void main(){CFraction c1(3,5),c2(4,7),c3;cout << "c1=";c1.output();cout << "c2=";c2.output();c3 = c1 + c2;cout << "c1+c2=";c3.output();c3 = c1 - c2;cout << "c1-c2=";c3.output();c3 = c1 * c2;cout << "c1*c2=";c3.output();c3 = c1 / c2;cout << "c1/c2=";c3.output();cout<<"下面比较两个分数的大小：\n";if (c1 > c2) cout << "t1>t2" << endl;if (c1 < c2) cout << "t1<t2" << endl;if (c1 = c2) cout << "t1＝t2" << endl; if (c1 != c2) cout << "t1≠t2" << endl;if (c1 >= c2) cout << "t1≥t2" << endl;if (c1 <= c2) cout << "t1≤t2" << endl;c3 = - c1;cout << "c1的相反数是：";c3.output();system("pause");}