c++ - 谁放狗出来的？ - 当 "Is-A"在类层次结构中遇到 "Has-A"时对象销毁

Question

考虑一下:

class FluffyThing
{
public:
  FluffyThing()
  {
    m_pMyFur = new Fur;
  }
  virtual ~FluffyThing();

protected:
  Fur * m_pMyFur;
};

class ClawedFluffyThing : public FluffyThing
{
public:
  ClawedFluffyThing()
    : FluffyThing()
  {
    m_pMyClaws = new Claws;
  }
  virtual ~ClawedFluffyThing();

protected:
  Claws * m_pMyClaws;
};

class ScaryFluffyThing : public ClawedFluffyThing
{
public:
  ScaryFluffyThing()
    : ClawedFluffyThing()
  {
    m_pMyTeeth = new Teeth;
    m_pMyCollar = new SpikedCollar;
  }
  virtual ~ScaryFluffyThing();

protected:
  Teeth * m_pMyTeeth;
  SpikedCollar * m_pMyCollar;
};

希望其中没有太多错误 - 我想你明白了。关键是有 3 个类，它们之间存在 IS-A 关系，每个类也有一个或两个属性，在销毁时间时需要进行一些清理。如果我没有声明虚拟析构函数，编译器会自动为我生成以下内容吗？声明了析构函数并因此被迫实现它们(假设使用了类)是进行销毁的正确的长期方法吗？

FluffyThing::~FluffyThing()
{
  delete m_pMyFur;
}

ClawedFluffyThing::~ClawedFluffyThing()
{
  delete m_pMyClaws;
  FluffyThing::~FluffyThing();
}

ScaryFluffyThing::~ScaryFluffyThing()
{
   delete m_pMyTeeth;
   delete m_pMyCollar;
   ClawedFluffyThing::~ClawedFluffyThing();
}

肯定已经有一个明确的答案了……但我无法根据自己的喜好快速掌握它。

Answer 1

没有。您不需要手动调用基类的析构函数，它们会按照与继承相反的顺序自动调用。除此之外，没关系。

FluffyThing::~FluffyThing()
{
  delete m_pMyFur;
}

ClawedFluffyThing::~ClawedFluffyThing()
{
  delete m_pMyClaws;
} //will call ~FluffyThing

ScaryFluffyThing::~ScaryFluffyThing()
{
   delete m_pMyTeeth;
   delete m_pMyCollar;
}  //will call ~ClawedFluffyThing

当然，如果您使用 RAII(智能指针而不是原始指针)，您甚至不需要 delete。

If I had not declared the virtual destructors, would the compiler have generated the following for me automatically?

没有。 :) virtual 析构函数用于在您通过基类指针删除派生类实例的情况下实现正确的行为。以下内容:

FluffyThing* p = new ScaryFluffyThing;
delete p;

仅当 FluffyThing 的析构函数是虚拟的时才合法。否则，它是未定义的行为。