条款16:在opeartor=中为所有的data members赋值内容

1,先看一个错误的情况:
class Base
{
public:
  Base(int initialValue = 0): x(initialValue) {}
private:
  int x;
};
class Derived: public Base
{
public:
  Derived(int initialValue)
  : Base(initialValue), y(initialValue) {}
  Derived& operator=(const Derived& rhs);
private:
  int y;
};


Derived& Derived::operator=(const Derived& rhs)
{
  if (this == &rhs) return *this;   
  y = rhs.y;                       
  return *this;                   
}

那么执行下面的代码:
void assignmentTester()
{
  Derived d1(0);                      // d1.x = 0, d1.y = 0
  Derived d2(1);                      // d2.x = 1, d2.y = 1
  d1 = d2;       // d1.x = 0, d1.y = 1!
}
注意:d1的base成分并未因assignment动作而改变内容.

正确的做法:
Derived& Derived::operator=(const Derived& rhs)
{
  if (this == &rhs) return *this;
  Base::operator=(rhs);    // call this->Base::operator=
  y = rhs.y;
  return *this;
}

3,复制构造函数存在一样的问题:

class Base
{
public:
  Base(int initialValue = 0): x(initialValue) {}
  Base(const Base& rhs): x(rhs.x) {}
private:
  int x;
};
class Derived: public Base
{
public:
  Derived(int initialValue)
  : Base(initialValue), y(initialValue) {}
  Derived(const Derived& rhs)     
  : y(rhs.y) {}             // 这里基类成分,采用默认构造函数
private:
  int y;
};

为了避免这一问题,派生类的copy constructor必须确保调用基类的copy constructor
class Derived: public Base
{
public:
  Derived(const Derived& rhs): Base(rhs), y(rhs.y) {}
  ...
};