/*
* Copyright (c) 2014, 烟台大学计算机学院
* All rights reserved.
* 文件名称:test.cpp
* 作 者:刘畅
* 完成日期:2015 年 5 月 6 日
* 版 本 号:v1.0
*
* 问题描述:在第八周项目3上
(1)定义分数的一目运算+和-,分别代表分数取正和求反,将“按位取反运算符”~重载为分数的求倒数运算。
(2)定义分数类中<<和>>运算符重载,实现分数的输入输出,改造原程序中对运算结果显示方式,使程序读起来更自然。
* 输入描述: 输入c2;
* 程序输出: 按要求输出。
代码如下:
#include <iostream>
using namespace std;
int gcd(int x, int y);
class CFraction
{
private:
int nume;
int deno;
public:
CFraction(int n=0,int d=1)
{
nume=n;
deno=d;
}
void simplify();
CFraction operator+(const CFraction &c);
CFraction operator-(const CFraction &c);
CFraction operator*(const CFraction &c);
CFraction operator/(const CFraction &c);
bool operator>(const CFraction &c);
bool operator<(const CFraction &c);
bool operator==(const CFraction &c);
bool operator!=(const CFraction &c);
bool operator>=(const CFraction &c);
bool operator<=(const CFraction &c);
friend istream &operator>>(istream &input,CFraction &c);
friend ostream &operator<<(ostream &output,CFraction &c);
CFraction operator+();
CFraction operator-();
CFraction operator~();
};
void CFraction::simplify()
{
int n=gcd(nume,deno);
nume/=n;
deno/=n;
}
int gcd(int x, int y)
{
int z;
while (y!=0)
{
z=x;
x=y;
y=z%y;
}
return x;
}
CFraction CFraction::operator+(const CFraction &c)
{
CFraction c1;
c1.nume=nume*c.deno+c.nume*deno;
c1.deno=deno*c.deno;
c1.simplify();
return c1;
}
CFraction CFraction::operator-(const CFraction &c)
{
CFraction c1;
c1.nume=nume*c.deno-c.nume*deno;
c1.deno=deno*c.deno;
c1.simplify();
return c1;
}
CFraction CFraction::operator*(const CFraction &c)
{
CFraction c1;
c1.nume=nume*c.nume;
c1.deno=deno*c.deno;
c1.simplify();
return c1;
}
CFraction CFraction:: operator/(const CFraction &c)
{
CFraction c1;
if (c.nume == 0)
return *this;
c1.nume=nume*c.deno;
c1.deno=deno*c.nume;
c1.simplify();
return c1;
}
bool CFraction::operator>(const CFraction &c)
{
int t_nume,c_nume,common_deno;
t_nume=nume*c.deno;
c_nume=c.nume*deno;
common_deno=deno*c.deno;
if ((t_nume-c_nume)*common_deno>0)
return true;
return false;
}
bool CFraction::operator<(const CFraction &c)
{
int t_nume,c_nume,common_deno;
t_nume=nume*c.deno;
c_nume=c.nume*deno;
common_deno=deno*c.deno;
if ((t_nume-c_nume)*common_deno<0)
return true;
return false;
}
bool CFraction::operator>=(const CFraction &c)
{
if (*this<c)
return false;
return true;
}
bool CFraction::operator<=(const CFraction &c)
{
if (*this>c)
return false;
return true;
}
bool CFraction::operator!=(const CFraction &c)
{
if (*this<c||*this>c)
return true;
return false;
}
bool CFraction::operator==(const CFraction &c)
{
if (*this!=c)
return false;
return true;
}
istream &operator>>(istream &input,CFraction &c)
{
char ch;
while(1)
{
cin>>c.nume>>ch>>c.deno;
if (c.deno==0)
cerr<<"分母为0, 请重新输入\n";
else if(ch!='/')
cerr<<"格式错误(形如m/n)! 请重新输入\n";
else
break;
}
return input;
}
ostream &operator<<(ostream &output,CFraction &c)
{
if(c.deno<0)
{
c.deno=-c.deno;
c.nume=-c.nume;
}
cout<<c.nume<<'/'<<c.deno<<endl;
return cout;
}
CFraction CFraction::operator+()
{
return *this;
}
CFraction CFraction::operator-()
{
CFraction c;
c.nume=-nume;
c.deno=deno;
return c;
}
CFraction CFraction::operator~()
{
CFraction c;
c.nume=deno;
c.deno=nume;
if(c.deno<0)
{
c.deno=-c.deno;
c.nume=-c.nume;
}
return c;
}
int main()
{
CFraction c1(10,5),c2,c3;
cout<<"c1:"<<c1;
c3=+c1;
cout<<"+c1:";
cout<<c3;
c3=-c1;
cout<<"-c1:";
cout<<c3;
c3=~c1;
cout<<"~c1:";
cout<<c3;
cout<<"c2:";
cin>>c2;
cout<<c2;
cout<<endl;
c3=c1+c2;
cout<<"c1+c2=";
cout<<c3;
c3=+c1+c2;
cout<<"+c1+c2=";
cout<<c3;
c3=c1-c2;
cout<<"c1-c2=";
cout<<c3;
c3=-c1-c2;
cout<<"-c1-c2=";
cout<<c3;
c3=c1*c2;
cout<<"c1*c2=";
cout<<c3;
c3=~c1*c2;
cout<<"~c1*c2=";
cout<<c3;
c3=c1/c2;
cout<<"c1/c2=";
cout<<c3;
cout<<endl;
if (c1>c2)
cout<<"c1>c2"<<endl;
if (c1<c2)
cout<<"c1<c2"<<endl;
if (c1>=c2)
cout<<"c1>=c2"<<endl;
if (c1<=c2)
cout<<"c1<=c2"<<endl;
if (c1==c2)
cout<<"c1==c2"<<endl;
if (c1!=c2)
cout<<"c1!=c2"<<endl;
return 0;
}
运行结果:
学习心得:
好不容易理顺了之前写的代码,然后把要添加的内容写了进去,在定义求反,和求分数的导数时一定要重新定义一个对象,我之前用的原对象修改,然后return *this,对后面的其他计算造成了很大的影响,完全变了。