Question

c++分数类

定义分数类:
//fraction.h
#include <iostream.h>
class fraction{
private:
int above;
int below;
public:
fraction(int a=0,int b=1){above=a;below=b;}
friend ostream& operator<<(ostream& os, fraction& com);
friend istream& operator>>(istream& is, fraction& com);
friend fraction operator+(fraction& x, fraction& y);
};

ostream& operator<<(ostream& os,fraction&com)
{
os<<com.above<<"/"<<com.below<<endl;
return os;
}

istream& operator>>(istream& is, fraction&com)
{
is >> com.above >>com.below;
return is;
}
fraction operator+(fraction& x, fraction& y)
{
fraction temp(x.above*y.below+y.above*x.below,y.below*x.below);
return temp;
}//使用以下代码编程:
#include "fraction.h"
#include <iostream.h>
int main()
{
fraction f1(-2,-3),f2,f3,f4,f5(6);
cin>>f2; //input: -2 3
cout << "f1=" << f1 << endl;
cout << "f2=" << f2 << endl;
f3=f1+f2;
f4=f1+1;
f5=3+f2;
cout << "f3=" << f3 << endl;
cout << "f4=" << f4 << endl;
cout << "f5=" << f5 << endl;
return 0;
}
//运行程序，输入: -2 3 将得到以下结果：:
f1=2/3
f2=-2/3
f3=0
f4=5/3
f5=7/3
看我哪里出错了

Answer 1

#include <iostream.h>
class fraction{
private:
int above;
int below;
public:
fraction(int a=0,int b=1){above=a;below=b;}
friend ostream& operator<<(ostream& os, fraction& com);
friend istream& operator>>(istream& is, fraction& com);
friend fraction operator+(fraction& x, fraction& y);
friend fraction operator+(fraction& x, int y);
friend fraction operator+( int y,fraction& x);
};

ostream& operator<<(ostream& os,fraction&com)
{
os<<com.above<<"/"<<com.below<<endl;
return os;
}

istream& operator>>(istream& is, fraction&com)
{
is >> com.above >>com.below;
return is;
}
fraction operator+(fraction& x, fraction& y)
{
fraction temp(x.above*y.below+y.above*x.below,y.below*x.below);
return temp;
}
fraction operator+(fraction& x, int y)
{
return fraction(x.above+y*x.below,x.below);
}
fraction operator+( int y,fraction& x)
{
return fraction(x.above+y*x.below,x.below);
}
int main()
{
fraction f1(-2,-3),f2,f3,f4,f5(6);
cin>>f2; //input: -2 3
cout << "f1=" << f1 << endl;
cout << "f2=" << f2 << endl;
f3=f1+f2;
f4=f1+1;
f5=3+f2;
cout << "f3=" << f3 << endl;
cout << "f4=" << f4 << endl;
cout << "f5=" << f5 << endl;
return 0;
}
帮你加了两个函数，楼主在做3+f2这种运算的时候没有对应的重载函数，似乎不能能编译通过，所以加了两个整形与分数的运算，加两个是为了满足加法的交换律，至于最后的结果，都是没有化简的，楼主可以自己修改一下！让他更完美