C++有理数运算符重载问题
#include <iostream>#include <string>
#include <stdlib.h>
class Rational
{
public:
Rational(int num, int denom);// num = 分子, denom = 分母
Rational operator+(Rational rhs); // rhs == right hand side
Rational operator-(Rational rhs);
Rational operator*(Rational rhs);
Rational operator/(Rational rhs);
void print();
private:
void normalize(); // 负责对分数的简化处理
int numerator; // 分子
int denominator;// 分母
};
Rational::Rational(int num, int denom)
{
numerator = num;
denominator = denom;
normalize();
}
// normalize() 对分数进行简化操作包括：
// 1. 只允许分子为负数，如果分母为负数则把负数挪到分子部分，如 1/-2 == -1/2
// 2. 利用欧几里德算法（辗转求余原理）将分数进行简化：2/10 => 1/5
void Rational::normalize()
{
// 确保分母为正
if( denominator < 0 )
{
numerator = -numerator;
denominator = -denominator;
}
// 欧几里德算法
int a = abs(numerator);
int b = abs(denominator);
// 求出最大公约数
while( b > 0 )
{
int t = a % b;
a = b;
b = t;
}
// 分子、分母分别除以最大公约数得到最简化分数
numerator /= a;
denominator /= a;
}
// a c a*d c*b a*d + c*b
// - + - = --- + --- = ---------
// b d b*d b*d = b*d
Rational Rational::operator+(Rational rhs)
{
int a = numerator;
int b = denominator;
int c = rhs.numerator;
int d = rhs.denominator;
int e = a*b + c*d;
int f = b*d;
return Rational(e, f);
}
// a c a -c
// - - - = - + --
// b d b d
Rational Rational::operator-(Rational rhs)
{
rhs.numerator = -rhs.numerator;
return operator+(rhs);
}
// a c a*c
// - * - = ---
// b d b*d
Rational Rational::operator*(Rational rhs)
{
int a = numerator;
int b = denominator;
int c = rhs.numerator;
int d = rhs.denominator;
int e = a*c;
int f = b*d;
return Rational(e, f);
}
// a c a d
// - / - = - * -
// b d b c
Rational Rational::operator/(Rational rhs)
{
int t = rhs.numerator;
rhs.numerator = rhs.denominator;
rhs.denominator = t;
return operator*(rhs);
}
void Rational::print()// 1/8
{
if( numerator % denominator == 0 )
std::cout << numerator / denominator;
else
std::cout << numerator << "/" << denominator;
}
int main()
{
Rational f1(2, 16);
Rational f2(7, 8);
// 测试有理数加法运算
Rational res = f1 + f2;
f1.print();
std::cout << " + ";
f2.print();
std::cout << " = ";
res.print();
std::cout << "\n";
// 测试有理数减法运算
res = f1 - f2;
f1.print();
std::cout << " - ";
f2.print();
std::cout << " = ";
res.print();
std::cout << "\n";
// 测试有理数乘法运算
res = f1 * f2;
f1.print();
std::cout << " * ";
f2.print();
std::cout << " = ";
res.print();
std::cout << "\n";
// 测试有理数除法运算
res = f1 / f2;
f1.print();
std::cout << " / ";
f2.print();
std::cout << " = ";
res.print();
std::cout << "\n";
return 0;
}
return operator*(rhs);
return operator+(rhs); 返回值res= f1 / f2 , res= f1 - f2, 打印怎么出来的？
厉害的
页:
[1]