C++面对对象：实现Fraction类，explicit ctor

 1 //Fraction.h
 2 #pragma once
 3 #include <iostream>
 4 #include <string>
 5 
 6 using namespace std;
 7 
 8 class Fraction;
 9 int __gcd(int, int);
10 
11 class Fraction {
12 public:
13     explicit Fraction(int num, int den = 1): m_numerator(num), m_denominator(den) {
14         int t = __gcd(num, den); 
15         //约分
16         m_numerator /= t;
17         m_denominator /= t;
18     }
19 
20     operator double() const { return (double)m_numerator / m_denominator; } //conversion function 转换函数
21     double operator +(const double& d) const { return (double)m_numerator / m_denominator + d; }
22     Fraction operator +(const Fraction&) const;
23     operator string() const { return to_string(m_numerator) + '/' + to_string(m_denominator); } //转换函数
24 
25 private:
26     int m_numerator;
27     int m_denominator;
28 
29     friend int __gcd(int, int);
30 };
31 
32 Fraction Fraction::operator +(const Fraction& f) const {
33     //a/c + b/d = (ad + bc) / cd
34     int num = m_numerator * f.m_denominator + m_denominator * f.m_numerator;
35     int den = m_denominator * f.m_denominator;
36     return Fraction(num, den);
37 }
38 
39 inline int __gcd(int a, int b) {
40     /*
41     while (a % b != 0) {
42         int t = a % b;
43         a = b;
44         b = t;
45     }
46     return b;
47     */
48     return b == 0 ? a : __gcd(b, a % b);
49 }

 1 //Fraction.cpp
 2 #include "Fraction.h"
 3 
 4 int main() {
 5     //conversion function
 6     Fraction f0(3);
 7     Fraction f1(3, 2);
 8     Fraction f2(9, 6);
 9     cout << f0 << " " << f1 << " " << f2 << " " << string(f1) << " " << string(f2) << endl;
10     cout << f0 + f1 << endl;
11     cout << f1 + 2.7 << endl;
12 
13     Fraction f3(f0 + f2);
14     Fraction f4 = f0 + f2;
15     double f5 = f0 + 2.1; 
16     //Fraction f6 = f0 + 2.1; //non-explicit ctor时, 情况一: 2.1由double转换为Fraction, 情况二: f6由double转换为Fraction, 会引起ambiguous Err
17     //Fraction f7 = f0 + 2.1; //explicit ctor时, double转化为Fraction的隐式转换被禁用, 本行报错
18     cout << f3 << " " << f4 << " " << f5 << endl;
19     //cout << f6 << " " << f7 << endl;
20 
21     return 0;
22 }