如何输出乘以用户创建的类c ++

Question

我正在为一个用户创建的有理数类重写一个赋值重载操作符，但不能在同一行中输出两个“Rational”对象。例如：

std::cout<<5*10<<std::endl;

输出50就好了。但是，当我尝试做的时候，

std::cout<<Rational_1*Rational_2<<std::endl;

我收到一个错误。如果我改为将值赋给第三个理性，例如

Rational_3=Rational_1*Rational_2;
std::cout<<Rational_3<<std::endl;

然后程序输出就好了。我把它带到了我的教授那里，但即使他不知道如何解决它。关于为什么会发生这种情况的任何解释都会有所帮助。我想知道为什么这是一个问题，而不仅仅是简单地获得一段有用的代码。

#include <iostream>
using namespace std;
class Rational{
    public:
        Rational();
        Rational(int whole_number);
        Rational(int numerator_input,int denominator_input);
        friend Rational operator *(Rational rat_1, Rational rat_2);
    `   friend ostream& operator <<(ostream& out,Rational& output);
        void simplify();
    private:
        int numerator,denominator;
};

int main(){
    Rational r1(2,3),r2(3,4),r3;
    r3=r1*r2;
    cout<<r3<<endl;
    //cout<<r1*r2<<endl;

return 0;
}

Rational::Rational(){
    numerator=0;
    denominator=1;
}

Rational::Rational(int whole_number){
    numerator=whole_number;
    denominator=1;
}

Rational::Rational(int numerator_input,int denominator_input){
    numerator=numerator_input;
    if(denominator_input==0){
        cout<<"A rational number can not have a 0 in the denominator\n";
        exit (5);
    }
    denominator=denominator_input;
    simplify();
}

ostream& operator <<(ostream& out,Rational& output){
    out<<output.numerator<<"/"<<output.denominator; 
    return out;
}

Rational operator *(Rational rat_1, Rational rat_2){
    Rational rat_3;
    rat_1.simplify();
    rat_2.simplify();
    rat_3.numerator=rat_1.numerator*rat_2.numerator;
    rat_3.denominator=rat_1.denominator*rat_2.denominator;
    rat_3.simplify();
    return rat_3;
}

void Rational::simplify(){
    //Flip negative sign to numerator
    for(int counter=1000000;counter>0;counter--){
        if((numerator%counter==0)&&(denominator%counter==0))
        {
            numerator=numerator/counter;
            denominator=denominator/counter;    
        }   
    }   
}

Answer 1

采用operator<<的int与你的第一个片段之间的区别在于它是按值进行的。

您的重载operator<<通过引用获取Rational对象，Rational_1*Rational_2返回临时对象，不允许临时对象绑定到非const引用。

要么通过值或const&来解决这个问题：

friend ostream& operator <<(ostream& out, const Rational& output);

要么

friend ostream& operator <<(ostream& out, Rational output);

Answer 2

声明运算符就像

friend ostream& operator <<(ostream& out, const Rational& output);
                                          ^^^^^

您不能将临时对象绑定到非常量左值引用。