通过指向Derived的指针删除，而不是Base [英] delete via a pointer to Derived, not Base

问题描述

我实现了一个基本的智能指针类。它适用于以下类型的代码。
（考虑Base1有一个公共构造函数）

I implemented a basic Smart pointer class. It works for the following type of code. (considering Base1 has a public constructor)

Sptr<Base1> b(new Base1);
b->myFunc();
{
    Sptr<Base1> c = b;
    Sptr<Base1> d(b);
    Sptr<Base1> e;
    e = b;
}

但在测试代码中它有一个受保护的构造函数这条路）。和代码

But in the test code it has a protected constructor(I need it to be this way). and the code

Sptr<Base1> sp(new Derived);

产生以下错误（注意导出）：

Produces the following error (notice the Derived):

Sptr.cpp: In instantiation of ‘my::Sptr<T>::~Sptr() [with T = Base1]’:
Sptr.cpp:254:39:   required from here
Sptr.cpp:205:9: error: ‘Base1::~Base1()’ is protected
Sptr.cpp:97:17: error: within this context

问题是我必须确保你通过指向Derived而不是Base1的指针删除。

The problem is I have to Make sure that you delete via a pointer to Derived, not Base1. How can I do that?

这是类代码（修改为显示构造函数和分词和类成员）

Here is the class code (clipped to show constructor and distructor and class members)

template <class T>
class Sptr {
private:
   T* obj; // The actual object pointed by
       RC* ref;// A reference object to keep track of count
public:
  //declarations


template <typename T>
Sptr<T>::Sptr():obj(NULL),ref(NULL) {
    //do something
    std::cout<<"()\n";
    ref = new RC();
    ref->AddRef();
}

template <typename T>
Sptr<T>::Sptr(const Sptr &a) : obj(a.obj),ref(a.ref) {
    //do something
    std::cout<<"const Sptr\n";
    ref->AddRef();
}

template <typename T>
Sptr<T>::~Sptr() {
    //do something
    if(ref->Release() == 0) {
        if(obj)
            delete obj;

        delete ref;
    }
}

template <typename T>
template <typename U>
Sptr<T>::Sptr(U* u) : obj(u),ref(NULL) {
    //do something
    ref = new RC();
    ref->AddRef();
}

template <typename T>
template <typename U> 
Sptr<T>::Sptr(const Sptr<U> &u) : obj(u.obj),ref(u.ref) {
    std::cout<<"const Sptr<U>\n";
    ref->AddRef();
}

EDIT

析构函数不是虚拟的。这是我必须解决的情况。以下是 Base1 和派生类

The destructor is not virtual. That is the case I have to solve. Below is the Base1 and Derived classes

class Base1 {
    protected:
        Base1() : derived_destructor_called(false) {
            printf("Base1::Base1()\n");
        }
    private:
        Base1(const Base1 &); // Disallow.
        Base1 &operator=(const Base1 &); // Disallow.
    protected:
        ~Base1() {
            printf("Base1::~Base1()\n");
            assert(derived_destructor_called);
        }
    protected:
        bool derived_destructor_called;
};

class Derived : public Base1 {
        friend void basic_tests_1();
    private:
        Derived() {}
        Derived(const Derived &); // Disallow.
        Derived &operator=(const Derived &); // Disallow.
    public:
        ~Derived() {
            printf("Derived::~Derived()\n");
            derived_destructor_called = true;
        }
        int value;
};

推荐答案

如果使构造函数成为模板，在构造时传递的指针类型，并将该信息保存在智能指针中（例如，在多态删除对象中）。这是（我相信） shared_ptr<> 如何。如果传入的指针类型没有可访问的析构函数，您也可以使用SFINAE产生编译器错误。

If you make your constructor a template, you can detect the type of pointer passed in at construction time and save that information in your smart pointer (for example, in a polymorphic deleter object). This is (I believe) how shared_ptr<> does it. You can probably also use SFINAE to produce a compiler error if a pointer type is passed in that doesn't have an accessible destructor.

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