#include <iostream>
#include <memory>

class Parent {
 public:
  virtual void Print() { std::cout << "Parent" << std::endl; }
};

class Child : public Parent {
 public:
  void Print() override { std::cout << "Child" << std::endl; }
};

template <typename ty>
class A {
 public:
  A(ty t) : t_(t) {}
  void Print() { t_.Print(); }

 private:
  ty t_;
};

int main() {
  std::shared_ptr<A<Child>> a = std::make_shared<A<Child>>(Child());
  std::shared_ptr<A<Parent>> b = std::reinterpret_pointer_cast<A<Parent>>(a);

  std::cout << "a.use_count: " << a.use_count()
            << ", b.use_count: " << b.use_count() << std::endl;

  std::cout << "a address: " << a.get() << ", b address: " << b.get()
            << std::endl;

  b->Print();
  getchar();
  return 0;
}

上图这种情况，常规的转换不能保证“转换以后，引用计数是正确的”。

用reinterpret_pointer_cast转换以后，可以保证这一点。