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多态和shared_ptr通过引用传递

[英]Polymorphism and shared_ptr passed by reference

I try to send to function a shared_ptr with polymorphic class. 我尝试发送具有多态类的shared_ptr函数。 My objective is to find a best way to send my shared_ptr without increase ref_count. 我的目标是找到一种无需增加ref_count即可发送我的shared_ptr的最佳方法。

EDIT: I don't search solution where my shared_ptr is replaced because I want to call shared_ptr.reset() for example. 编辑:我不搜索替换我的shared_ptr的解决方案,因为例如我想调用shared_ptr.reset()。

Currently, void doGenericTemplate(std::shared_ptr<CLASS>& ptr) is what I want in result BUT I prefer a single function in program. 目前, void doGenericTemplate(std::shared_ptr<CLASS>& ptr)是我想要的结果,但是我更喜欢程序中的单个函数。

  1. Do you have another solution ? 您还有其他解决方案吗?

Moreover, I don't understand why the function void doGeneric(std::shared_ptr<Base>& ptr) doesn't compile (equivalent without shared_ptr work fine: please check doClassic in complete code). 而且,我不明白为什么void doGeneric(std::shared_ptr<Base>& ptr)函数无法编译(等效于不使用shared_ptr的情况也很好:请检查完整代码中的doClassic )。

  1. Do you have an explain ? 你有解释吗?

Thanks you ! 谢谢 !

Partial code 部分代码

#include <iostream>
#include <memory>

class Base
{
    public:
        Base() = default;
        virtual ~Base() = default;
        virtual void run() = 0;
};

class Derived1: public Base
{
    public:
        Derived1() = default;
        virtual ~Derived1() = default;
        void run()
        {
            std::cout << "  Derived1";
        }
};

class Derived2: public Base
{
    public:
        Derived2() = default;
        virtual ~Derived2() = default;
        void run()
        {
            std::cout << "  Derived2";
        }
};

// This function works but increase count
void doGenericCopy(std::shared_ptr<Base> ptr)
{
    ptr->run();
    std::cout << "  Ref count: " << ptr.use_count() << std::endl;
}

// This function works without increase count = OK !
void doSpecificD1(std::shared_ptr<Derived1>& ptr)
{
    ptr->run();
    std::cout << "  Ref count: " << ptr.use_count() << std::endl;
}

// Compilation error = FAILED !
void doGeneric(std::shared_ptr<Base>& ptr)
{
    ptr->run();
    std::cout << "  Ref count: " << ptr.use_count() << std::endl;
}

// Working fine for all Derivate = OK !
template<typename CLASS>
void doGenericTemplate(std::shared_ptr<CLASS>& ptr)
{
    ptr->run();
    std::cout << "  Ref count: " << ptr.use_count() << std::endl;
}

int main()
{
    auto d1 = std::make_shared<Derived1>();
    auto d2 = std::make_shared<Derived2>();

    std::cout << "With copy: " << std::endl;
    doGenericCopy(d1);
    doGenericCopy(d2);

    std::cout << "Specific: " << std::endl;
    doSpecificD1(d1);

    std::cout << "Template: " << std::endl;
    doGenericTemplate(d1);
    doGenericTemplate(d2);

    // Compilation issue
    //doGeneric(d1);
}

Complete code 完整的代码

https://ideone.com/ZL0v7z https://ideone.com/ZL0v7z

Conclusion 结论

Currently in c++, shared_ptr has not in language a specific tools to use polymorphism of class inside template. 当前在c ++中,shared_ptr尚无特定语言的工具来使用模板内类的多态性。

The best way is to refactor my code and avoids to manage shared_ptr (ref_count, reset). 最好的方法是重构我的代码,并避免管理shared_ptr(ref_count,重置)。

Thanks guys ! 多谢你们 !

  1. Do you have another solution ? 您还有其他解决方案吗?

Pass object by reference or const reference instead of shared_ptr . 通过引用或const引用而不是shared_ptr传递对象。

void doGeneric(Base& r)
{
    r.run();
}

Firstly - this shows explicitly that you do not store or cache pointer somwhere. 首先-这明确表明您没有在任何地方存储或缓存指针。 Secondly - you avoid ambiguities like the one you presented here. 其次,您要避免像此处介绍的那样含糊不清。

  1. Do you have an explain ? 你有解释吗?

Passing shared_ptr<Derived> to function causes implicit cast to shared_ptr<Base> . shared_ptr<Derived>传递给函数会导致隐式强制转换为shared_ptr<Base> This new shared_ptr<Base> is temporary, so it can not be cast to shared_ptr<Base> & . 此新的shared_ptr<Base>是临时的,因此无法将其shared_ptr<Base> &shared_ptr<Base> & This implicit cast would increase ref count even if you could pass it. 即使您可以通过此隐式转换,也会增加引用计数。

A shared_ptr<Base> and shared_ptr<Derived> are unrelated types, except you can implicitly create a shared_ptr<Base> from a shared_ptr<Derived> . shared_ptr<Base>shared_ptr<Derived>是不相关的类型,除了可以从shared_ptr<Derived>隐式创建shared_ptr<Base> shared_ptr<Derived>

This creation adds a reference count. 此创建将添加参考计数。

If you really, really want to avoid that reference count... 如果您真的要避免引用计数...

template<class T>
struct shared_ptr_view {
  template<class D>
  shared_ptr_view( std::shared_ptr<D>& sptr ):
    vtable( get_vtable<D>() ),
    ptr( std::addressof(sptr) )
  {}
  shared_ptr_view( shared_ptr_view const& ) = default;
  shared_ptr_view() = default;
  shared_ptr_view& operator=( shared_ptr_view const& ) = delete;
  T* get() const { if(vtable) return vtable->get(ptr); return nullptr; }
  void clear() const { if(vtable) vtable->clear(ptr); }
  std::shared_ptr<T> copy() const { if(vtable) return vtable->copy(ptr); return {} }
  operator std::shared_ptr<T>() const { return copy(); }
  T* operator->() const { return get(); }
  T& operator*() const { return *get(); }
  explicit operator bool() const { return get(); }
  std::size_t use_count() const { if (vtable) return vtable->use_count(ptr); return 0; }
private:
  struct vtable_t {
    T*(*get)(void*) = 0;
    std::shared_ptr<T>(*copy)(void*) = 0;
    void(*clear)(void*) = 0;
    std::size_t(*use_count)(void*) = 0;
  };
  vtable_t const* vtable = 0;
  void* ptr = 0;
  template<class D>
  static vtable_t create_vtable() {
    return {
      [](void* ptr)->T*{ return static_cast<std::shared_ptr<D>*>(ptr)->get(); },
      [](void* ptr)->std::shared_ptr<T>{ return *static_cast<std::shared_ptr<D>*>(ptr); },
      [](void* ptr){ static_cast<std::shared_ptr<D>*>(ptr)->reset(); },
      [](void* ptr){ return static_cast<std::shared_ptr<D>*>(ptr)->use_count(); }
    };
  }
  template<class D>
  static vtable_t const* get_vtable() {
    static const auto vtable = create_vtable<D>();
    return &vtable;
  }
};

then 然后

void doGeneric( shared_ptr_view<Base> ptr ) {
  ptr->run();
  std::cout << "  Ref count: " << ptr.use_count() << std::endl;
}

does not increase the reference count. 不会增加参考计数。 I think it is raw insanity. 我认为这是原始的疯狂。

shared_ptr_view.clear() works, but shared_ptr_view.reset(T*) cannot: a shared_ptr_view<Derived> cannot be reset to point to a Base* . shared_ptr_view.clear()可以工作,但是shared_ptr_view.reset(T*)不能: shared_ptr_view<Derived>不能重置为指向Base*

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