C ++运算符重载无法按预期工作

Question

Here is the whole class (copy/paste should work):

#include <cstdio>
#include <iostream>
using namespace std;

class Rational
{
    int _n = 0;
    int _d = 1;
public:
    Rational (int numerator = 0, int denominator = 1) : _n(numerator), _d(denominator) {};
    Rational (const Rational & rhs) : _n(rhs._n), _d(rhs._d) {};
    ~Rational();
    int numerator() const {
        return _n;
    };
    int denominator() const {
        return _d;
    };
    Rational & operator = (const Rational &);

};

Rational operator + (const Rational & lhs, const Rational & rhs)
{
    return ((lhs.numerator() * rhs.denominator()) + (lhs.denominator() * rhs.numerator()), lhs.denominator() * rhs.denominator());
}

Rational::~Rational()
{
    _n = 0;
    _d = 1;
}

std::ostream & operator << (std::ostream & o, const Rational & r)
{
    if (r.denominator() == 1) {
        return o << r.numerator();
    } else {
        return o << r.numerator() << '/' << r.denominator();
    }
}


int main()
{
    Rational a = 7;
    cout << "a is: " << a << endl;

    Rational b(5,3);
    cout << "b is: " << b << endl;

    cout << a << " + " << b << " = " << a + b << endl;
    cout << 14 << " + " << b << " = " << 14 + b << endl << endl;

    return 0;
}

I want to use non-member operator overload. Code compiles but it displays wrong results. The last two lines should display:

7 + 5/3 = 26/3

14 + 5/3 = 47/3

but instead displays:

7 + 5/3 = 3

14 + 5/3 = 3

The problem is most likely this:

Rational operator + (const Rational & lhs, const Rational & rhs)
{
    return ((lhs.numerator() * rhs.denominator()) + (lhs.denominator() * rhs.numerator()), lhs.denominator() * rhs.denominator());
}

because it works properly if I change it to create object first and then return:

Rational operator + (const Rational & lhs, const Rational & rhs)
{
    Rational r((lhs.numerator() * rhs.denominator()) + (lhs.denominator() * rhs.numerator()), lhs.denominator() * rhs.denominator());
    return r;
}

I would expect implicit conversion and both solutions give the same results. Could someone explain to me what is the difference here?

Answer 1

return statement contains comma operator invocation discarding result of first expression (that was supposed to be numerator) and invoking constructor with just one paramter. You should use proper initialization syntax:

return Rational
(
    (lhs.numerator() * rhs.denominator()) + (lhs.denominator() * rhs.numerator())
,   lhs.denominator() * rhs.denominator()
);

or

return
{
    (lhs.numerator() * rhs.denominator()) + (lhs.denominator() * rhs.numerator())
,   lhs.denominator() * rhs.denominator()
};

or (better)

return Rational
{
    (lhs.numerator() * rhs.denominator()) + (lhs.denominator() * rhs.numerator())
,   lhs.denominator() * rhs.denominator()
};

Answer 2

This line causes your bug:

return ((lhs.numerator() * rhs.denominator()) + (lhs.denominator() * rhs.numerator()), lhs.denominator() * rhs.denominator());

The problem is that it gets evaluated as a whole, which gives you an integer. This integer is then fed to the implicitly invoked constructor of your Rational class. Note that this line uses the comma-operator. For code like a, b it first evaluates a , discards the result and then evaluates b and keeps that result as result of the overall expression. That's not what you intended.