c++ - C++ 中策略类的 protected 析构函数

Question

我编写了以下愚蠢的策略结构:

template
<typename T>
struct SuperLeague
{
public:
  void printLeague(){std::cout << "SuperLegue" << std::endl;};
protected:
  //~SuperLeague(){}
};

和宿主类

template
<typename TT,template <typename C> class Policy>
class League: public Policy<TT>
{
public:
  void fun()
  {
    Policy<TT>().printLeague();
  }
};

我的主要是

int main()
{
  League<int,SuperLeague> k;
  League<int,SuperLeague> kk;
  k.fun();
  kk.printLeague();
  League<int,SuperLeague> *kkk=new League<int,SuperLeague>();
  kkk->fun();
  //delete kkk;
};

到这里为止，一切正常。输出是:

SuperLegue
SuperLegue
SuperLegue

Andrei Alexandrescu 在他的一本书中写道:除非策略类定义了虚拟析构函数，否则将 delete 应用于指向策略类的指针具有未定义的行为。他解释了在从策略类派生时不对策略或 protected (或私有(private))继承使用虚拟析构函数的原因，并且他建议策略应该使用的轻量级、有效的解决方案是定义一个非虚拟的 protected 析构函数。问题是，当我尝试使用 ~SuperLeague(){} 执行此操作时，编译器会提示析构函数受到保护。我做错了什么？

Answer 1

您不应该在 League::fun() 中创建临时策略对象.由于您的 League 实例模板派生自 Policy 的适当实例模板，它继承了 printLeague()功能:

template
<typename TT,template <typename C> class Policy>
class League: public Policy<TT>
{
public:
  void fun()
  {
    Policy<TT>::printLeague();
//  ^^^^^^^^^^^^^^^^^^^^^^^^^^
  }
};

声明析构函数时解决方案无法编译的原因 protected是那个protected在访问同一派生类(在您的情况下为League)的对象(或取消引用对象的引用或指针)时，使派生类可以访问基类成员。

在您的示例中情况并非如此，您创建了一个 Policy<TT> 类型的临时对象并调用 printLeague() 在该对象上(不是 League 类型)。

根据 C++11 标准的第 11.4/1 段:

An additional access check beyond those described earlier in Clause 11 is applied when a non-static data member or non-static member function is a protected member of its naming class (11.2). As described earlier, access to a protected member is granted because the reference occurs in a friend or member of some class C. If the access is to form a pointer to member (5.3.1), the nested-name-specifier shall denote C or a class derived from C. All other accesses involve a (possibly implicit) object expression (5.2.5). In this case, the class of the object expression shall be C or a class derived from C.