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 

 1 #include <iostream>
 2 
 3 
 4 class Rational {
 5 public:
 6     explicit Rational(int numerator = 0, int denominator = 1);
 7     //拷贝构造函数不需要explicit修饰，否则需要再定义一个
 8     //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
 9     Rational(const Rational &);
10     ~Rational();
11 public:
12     //Rational a, b, c;
13     //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
14     Rational& operator = (const Rational &);

15     //使用asDouble函数替代double()可以防止编译器的隐式的类型转换
16     //Rational r; std::cout << r;
17     //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
18     //因为r被编译器隐式的转换成了double值
19     operator double () const;
20     double asDouble() const { return static_cast<double>(m_numerator) / m_denominator; }
21     int numerator(void) const { return m_numerator; }
22     int denominator(void) const { return m_denominator; }
23 private:
24     int m_numerator;
25     int m_denominator;
26 };
27 
28 Rational operator *(const Rational &lhs, const Rational &rhs)
29 {
30     int numerator = lhs.numerator() * rhs.numerator();
31     int denominator = lhs.denominator() * rhs.denominator();
32 
33     Rational result(numerator, denominator);
34 
35     return result;
36 }
37 
38 Rational::Rational(int numerator, int denominator)
39     :m_numerator(numerator), m_denominator(denominator)
40 {
41 
42 }
43 
44 Rational::Rational(const Rational &rhs)
45     :m_numerator(rhs.numerator()), m_denominator(rhs.denominator())
46 {
47 
48 }
49 
50 Rational::~Rational()
51 {
52 
53 }
54 
55 Rational::operator double() const 
56 {
57     return static_cast<double>(m_numerator) / m_denominator; 
58 }
59 
60 Rational& Rational::operator = (const Rational &rhs)
61 {
62     m_numerator = rhs.numerator();
63     m_denominator = rhs.denominator();
64 
65     return *this;
66 }
67 
68 class Base{
69 public:
70     Base() { }
71     virtual ~Base() { }
72 private:
73     int GetNum(void) const { return m_num; }
74     
75     int m_num;
76 };
77 
78 class Derived : public Base {
79 public:
80     Derived() { }
81     virtual ~Derived() { }
82 
83     //GetNum函数是Base类的私有成员，无法被Derived继承
84 //    void printNum(void) { std::cout << GetNum() << std::endl; }
85 };
86 
87 int main(void)
88 {
89     Rational a;
90     double d = 0.5 * a;
91     int i = 2 * a;
92 
93     Rational b, c;
94     c = b * a;
95 
96     return 0;
97 }

 

posted @ 2013-01-05 17:13  Jojodru  阅读(211)  评论(0)  编辑  收藏  举报