c-运算符重载中的类数据封装(私有数据)

下面是代码

编码：

#include <iostream>
using namespace std;

class Rational {
  int num;  // numerator
  int den;  // denominator
  friend istream& operator>> (istream & , Rational&);
  friend ostream& operator<< (ostream &  , const Rational&);
 public:
  Rational (int num = 0, int den = 1)
    :     num(num), den(den) {}
  void getUserInput() {
    cout << "num = ";
    cin >> num;
    cout << "den = ";
    cin >> den;
  }
  Rational operator+(const Rational &);
};

Rational Rational::operator+ (const Rational& r) { //HERE 
  int n = num * r.den + den * r.num;
  int d = den * r.den;
  return Rational (n, d);
}

istream& operator>> (istream & is , Rational& r)
{
    is >> r.num >> r.den;
}

ostream& operator<< (ostream & os , const Rational& r)
{
    os << r.num << " / " <<  r.den << endl;;
}
int main() {
  Rational r1, r2, r3;
  cout << "Input r1:\n";
  cin >> r1;
  cout << "Input r2:\n";
  cin >> r2;
  r3 = r1 + r2;
  cout << "r1 = " << r1;
  cout << "r2 = " << r2;
  cout << "r1 + r2 = " << r3;
  return 0;
}

问题

上面的代码有一个运算符重载,在运算符定义中我们可以看到参数r访问私有数据(r.num和r.den).为什么C允许参数访问类外部的私有数据？这是一种特殊情况吗？

谢谢.

解决方法:

访问说明符适用于类级别,而不是实例级别,因此Rational类可以查看任何其他Rational实例的私有数据成员.由于您的Rational运算符是成员函数,因此可以访问其Rational参数的私有数据.

注意：规范方法是定义一个成员运算符=,然后使用它来实现一个非成员运算符

struct Foo
{
  int i;

  Foo& operator+=(const Foo& rhs) 
  { 
    i += rhs.i;
    return *this;
  }

};

Foo operator+(Foo lhs, const Foo& rhs)
{
  return lhs += rhs;
}

标签：c,operator-overloading,encapsulation
来源： https://codeday.me/bug/20191013/1906528.html