C++ 类型转换与隐式构造函数

Question

我正在实现一个表示分数的 C++ 类。这是我的代码。

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;

};

我现在有两个小问题。现在正在尝试调用

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

只会导致这个错误:

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>

我真正想要的是:

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

但显然，这是行不通的。如果我使用强制转换符号，我希望这两个强制转换运算符被称为。例如 Fraction f(1, 2);双 d = (双)(f);

Answer 1

根据定义，转换运算符是隐式。你不能让他们明确。

正常的解决方案是命名成员函数来进行转换。我认为您还可以创建一个看起来像 static_cast 的专用模板方法，然后调用显式类方法。