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Home/ Questions/Q 5976923
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Editorial Team
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Editorial Team
Asked: May 22, 20262026-05-22T21:17:49+00:00 2026-05-22T21:17:49+00:00

I’m implementing a c++-class representing a fraction. Here goes my code. class Fraction {

  • 0

I’m implementing a c++-class representing a fraction. Here goes my code.

class Fraction
{
    public:
        Fraction(char i);
        Fraction(int i);
        Fraction(short i);
        Fraction(long int l);
#ifdef __LP64__
        Fraction(long long l);
#endif
        Fraction(float f);
        Fraction(double d);
        Fraction(double x, double y);

        Fraction operator +() const;
        Fraction operator -() const;

        Fraction& operator +=(const Fraction& other);
        Fraction& operator -=(const Fraction& other);
        Fraction& operator *=(const Fraction& other);
        Fraction& operator /=(const Fraction& other);

        bool operator ==(const Fraction& other);
        bool operator !=(const Fraction& other);
        bool operator  >(const Fraction& other);
        bool operator  <(const Fraction& other);
        bool operator >=(const Fraction& other);
        bool operator <=(const Fraction& other);

        operator double();
        operator float();

        static void commonize(Fraction& a, Fraction& b);
        void shorten();

        double getNumerator();
        double getDenominator();

        friend Fraction operator +(Fraction const& a, Fraction const& b);
        friend Fraction operator -(Fraction const& a, Fraction const& b);
        friend Fraction operator *(Fraction const& a, Fraction const& b);
        friend Fraction operator /(Fraction const& a, Fraction const& b);
        friend ostream& operator <<( ostream& o, const Fraction f);

    protected:
        double numerator, denominator;

};

I now have two little problems.
Now trying to call

Fraction a(1, 2);
cout << (3 + a) << endl;

simply results in this error:

fractiontest.cpp:26: error: ambiguous overload for ‘operator+’ in ‘3 + a’
fractiontest.cpp:26: note: candidates are: operator+(int, double) <built-in>
fractiontest.cpp:26: note:                 operator+(int, float) <built-in>

All I’d really want is this:

explicit operator double();
explicit operator float();

But apparently, this doesn’t work. I’d like these two cast-operators to be called iff I use the cast notation. For example Fraction f(1, 2); double d = (double)(f);

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  1. Editorial Team
    Editorial Team
    2026-05-22T21:17:50+00:00Added an answer on May 22, 2026 at 9:17 pm

    By definition the conversion operators are implicit. You can’t make them explicit.

    The normal solution is named member functions to do the conversion. I think you could also create a specialized template method that would look just like a static_cast and call through to the explicit class method.

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