让"cout << obj1 + obj2 << endl;"发挥作用

因此，我得到了类内运算符重载，用于obj1+obj2:

fraction operator+ (fraction op);

类似地，cout<lt；obj的工作原理是重载操作符重载ostream:

ostream& operator<< (ostream& os, fraction& frac);

然而，如果我试图将两者结合起来，一切都会变得一团糟。

fraction.cpp:77:27: error: no match for ‘operator<<’ in 
‘std::operator<< [with _Traits = std::char_traits<char>]((* & std::cout), ((const char*)"Sum: ")) 
<< f1.fraction::operator+(f2)’

这是代码：

#include <iostream>
using namespace std;
class fraction
{
    private:
        int n, d;
    public:
        fraction ()
        {
            this->n = 1;
            this->d = 0;
        }
        fraction (int n, int d)
        {
            this->n = n;
            this->d = d;
        }
        int getNumerator ()
        {
            return n;
        }
        int getDenominator ()
        {
            return d;
        }
        fraction operator+ (fraction op)
        {
            return *(new fraction(this->n*op.d + op.n*this->d, this->d*op.d));
        }
        fraction operator- (fraction op)
        {
            return *(new fraction(this->n*op.d - op.n*this->d, this->d*op.d));
        }
        fraction operator* (fraction op)
        {
            return *(new fraction(this->n*op.n, this->d*op.d));
        }
        fraction operator/ (fraction op)
        {
            return *(new fraction(this->n*op.d, this->d*op.n));
        }
};
ostream& operator<< (ostream& os, fraction& frac)
{
    int n = frac.getNumerator();
    int d = frac.getDenominator();
    if(d == 0 && n == 0)
        os << "NaN";
    else if(d == 0 && n != 0)
        os << "Inf";
    else if(d == 1)
        os << n;
    else
        os << n << "/" << d;
}
int main ()
{
    fraction f1(2, 3);
    fraction f2(1, 3);
    cout << f1 << " " << f2 << endl;
    /*
    cout << "Sum: " << f1+f2 << endl;
    cout << "Difference: " << f1-f2 << endl;
    cout << "Product: " << f1*f2 << endl;
    cout << "Quotient: " << f1/f2 << endl;
    */
    return 0;
}

帮助。D：