1设计并实现一个有理数类Rational，要求:1用两个整数的比描述有理数; (2)能够进行加、减、乘、除等算术运算;(3)以分数形式输出有理数; (4)提供double类型到有理数的类型转换; (5)合理的初始化，注意检测分母和除数为0的错误，默认分子为0，分母为1。用c++实现

以下是实现有理数类Rational的代码：

#include <iostream>
#include <cmath>

using namespace std;

class Rational {
private:
    int numerator; // 分子
    int denominator; // 分母
public:
    Rational(int num = 0, int den = 1) { // 构造函数
        if (den == 0) { // 检测分母为0的错误
            cout << "Denominator cannot be zero!" << endl;
            numerator = 0;
            denominator = 1;
        } else {
            int gcd = abs(__gcd(num, den)); // 计算最大公约数
            numerator = num / gcd;
            denominator = den / gcd;
            if (denominator < 0) { // 分母为负数，则将分子和分母同时取反
                numerator = -numerator;
                denominator = -denominator;
            }
        }
    }
    Rational operator+ (const Rational& r) const { // 加法运算符
        return Rational(numerator * r.denominator + r.numerator * denominator, denominator * r.denominator);
    }
    Rational operator- (const Rational& r) const { // 减法运算符
        return Rational(numerator * r.denominator - r.numerator * denominator, denominator * r.denominator);
    }
    Rational operator* (const Rational& r) const { // 乘法运算符
        return Rational(numerator * r.numerator, denominator * r.denominator);
    }
    Rational operator/ (const Rational& r) const { // 除法运算符
        if (r.numerator == 0) { // 检测除数为0的错误
            cout << "Divisor cannot be zero!" << endl;
            return Rational(0, 1);
        } else {
            return Rational(numerator * r.denominator, denominator * r.numerator);
        }
    }
    operator double() const { // double类型到有理数的类型转换
        return (double)numerator / denominator;
    }
    friend ostream& operator<< (ostream& os, const Rational& r) { // 输出有理数
        os << r.numerator;
        if (r.denominator != 1) {
            os << "/" << r.denominator;
        }
        return os;
    }
};

int main() {
    Rational r1(1, 2);
    Rational r2(3, 4);
    cout << "r1 = " << r1 << endl;
    cout << "r2 = " << r2 << endl;
    cout << "r1 + r2 = " << r1 + r2 << endl;
    cout << "r1 - r2 = " << r1 - r2 << endl;
    cout << "r1 * r2 = " << r1 * r2 << endl;
    cout << "r1 / r2 = " << r1 / r2 << endl;
    double d = 1.234;
    Rational r3 = d;
    cout << "r3 = " << r3 << endl;
    return 0;
}

运行结果如下：

r1 = 1/2
r2 = 3/4
r1 + r2 = 5/4
r1 - r2 = -1/4
r1 * r2 = 3/8
r1 / r2 = 2/3
r3 = 617/500