c++类型转化

More Effective C++的笔记

4.1 Item M5：谨慎定义类型转换函数 

#include <iostream>


class Rational {
public:
    explicit Rational(int numerator = 0, int denominator = 1);
    //拷贝构造函数不需要explicit修饰，否则需要再定义一个
    //非const构造参数的构造函数，此处允许的非const转const是无害转换
    Rational(const Rational &);
    ~Rational();
public:
    //Rational a, b, c;
    //a = b = c; 这个表达式依赖于这个operator = 重载函数返回一个自身的引用
    Rational& operator = (const Rational &);

    //使用asDouble函数替代double()可以防止编译器的隐式的类型转换
    //Rational r; std::cout << r;
    //如果没有定义operator <<, 将会打印一个double值而不是一个期待的分数值
    //因为r被编译器隐式的转换成了double值
    operator double () const;
    double asDouble() const { return static_cast<double>(m_numerator) / m_denominator; }
    int numerator(void) const { return m_numerator; }
    int denominator(void) const { return m_denominator; }
private:
    int m_numerator;
    int m_denominator;
};

Rational operator *(const Rational &lhs, const Rational &rhs)
{
    int numerator = lhs.numerator() * rhs.numerator();
    int denominator = lhs.denominator() * rhs.denominator();

    Rational result(numerator, denominator);

    return result;
}

Rational::Rational(int numerator, int denominator)
    :m_numerator(numerator), m_denominator(denominator)
{

}

Rational::Rational(const Rational &rhs)
    :m_numerator(rhs.numerator()), m_denominator(rhs.denominator())
{

}

Rational::~Rational()
{

}

Rational::operator double() const 
{
    return static_cast<double>(m_numerator) / m_denominator; 
}

Rational& Rational::operator = (const Rational &rhs)
{
    m_numerator = rhs.numerator();
    m_denominator = rhs.denominator();

    return *this;
}

class Base{
public:
    Base() { }
    virtual ~Base() { }
private:
    int GetNum(void) const { return m_num; }
    
    int m_num;
};

class Derived : public Base {
public:
    Derived() { }
    virtual ~Derived() { }

    //GetNum函数是Base类的私有成员，无法被Derived继承
//    void printNum(void) { std::cout << GetNum() << std::endl; }
};

int main(void)
{
    Rational a;
    double d = 0.5 * a;
    int i = 2 * a;

    Rational b, c;
    c = b * a;

    return 0;
}