通过按位运算确定返回类型(Determining return type via bitwise operation)

 你的析构函数肯定是错的。 类的析构函数不会也不能delete该对象的内存。 
 
 如果没有虚函数，你对公共继承的反对意见是什么？ （至少）有两种方法可以防止某人通过基指针意外删除派生对象。 一个是protected基础析构函数。 
 
 另一种方法是将派生类的动态分配实例直接填充到shared_ptr 。 这甚至可以是shared_ptr<Base> ： 
 
std::shared_ptr<Base> foo(new DerivedA(...));
 
 由于shared_ptr具有捕获其参数类型的模板构造函数，因此Base*指针将在与shared_ptr关联的删除函数中转换为DerivedA* ，因此会被正确删除。 没有人会如此愚蠢地尝试从shared_ptr提取指针并将其删除为Base* 。 
 
 当然，如果你没有虚函数，那么当衍生类之间的唯一区别就是它们在构造函数中设置的时候，这个技巧才真正有用。 否则，您最终需要从shared_ptr向下转换Base*指针，在这种情况下，您应该首先使用shared_ptr<DerivedA> 。 

Your destructor is definitely wrong. The destructor of a class does not and must not delete the memory for the object.
 
What's your objection to public inheritance without virtual functions? There are (at least) a couple of ways to prevent someone accidentally deleting a derived object through a base pointer. One is to make the base destructor protected.
 
Another is to stuff dynamically-allocated instances of the derived class straight into a shared_ptr. This can even be a shared_ptr<Base>:
 
std::shared_ptr<Base> foo(new DerivedA(...));
 
Because shared_ptr has a template constructor that captures the type of its argument, the Base* pointer will be converted to DerivedA* in the deleter function associated with the shared_ptr, and hence deleted correctly. Nobody should ever be so daft as to try to extract a pointer out of a shared_ptr and delete it as Base*.
 
Of course if you have no virtual functions, then the trick is only really useful when the only difference between the derived classes is what they set up in their constructors. Otherwise you'll end up needing to downcast the Base* pointer from the shared_ptr, in which case you should have used a shared_ptr<DerivedA> to begin with.