c++ - 简单引用计数 : smart pointers

标签 c++ templates smart-pointers reference-counting

我想使用智能指针实现一个简单的引用计数。变量pointer表示指向存储对象的指针,reference_count表示对象的拷贝总数。

  • 如果我们使用 NULL 初始化对象:reference_count = -1 else reference_count = 1
  • 复制 ctor 和 operator = increment reference_count
  • 析构函数递减 reference_count,如果没有其他指向对象的引用,则执行删除。

这是我的代码:

#ifndef smart_pointer_H
#define smart_pointer_H

template < typename T > class smart_pointer
{
    private:
        T*    pointer;      
        int reference_count;    

    public:

        smart_pointer() : pointer(0), reference_count(-1) {}

        smart_pointer(T* p) : pointer(p)
        {
            if (p != NULL)
            {
                this->reference_count = 1;
            }

            else
            {
                this->reference_count = -1;
            }
        }

        smart_pointer(const smart_pointer <T> & p) : pointer(p.pointer),     reference_count(p.reference_count + 1) {}
        bool operator == (const smart_pointer <T>& p) { return pointer == p.pointer; }
        bool operator != (const smart_pointer <T>& p) { return pointer != p.pointer; }


        ~ smart_pointer()
        {
            if(-- reference_count == 0)
        {
                std::cout << "Destructing: " << '\n';
                delete pointer;
            }
        }

        T& operator *  () { return *pointer; }
        T* operator -> () { return pointer; }

        smart_pointer <T> & operator = (const smart_pointer <T> & p)
        {
                if (this != &p)
                {
                    if( -- reference_count == 0)
                    {
                        delete pointer;
                    }

                        pointer = p.pointer;
                        reference_count = p.reference_count + 1;
                }

        return *this;
        }
};    

这是我的测试代码,类示例存储 2D 点和两个指向任何其他 2D 点的指针。

template < typename T >
class smart_pointer;

class Point
{
private:
    double x, y;
    smart_pointer <Point> p1;
    smart_pointer <Point> p2;

public:
    Point(double xx, double yy): x(xx), y(yy) {this-> p1 = NULL; this->p2 = NULL;}
    Point(double xx, double yy, smart_pointer <Point> p1, smart_pointer <Point> p2): x(xx), y(yy) {this-> p1 = p1, this->p2 = p2; }
    double getX(){ return x;}
    double getY(){ return y;}
    void setX(double xx)  {this->x = xx;}
    void setY(double yy)  {this->y = yy;}
    void setP1(smart_pointer <Point> p1) {this->p1 = p1;}
    void setP2(smart_pointer <Point> p2) {this->p2 = p2;}

    void print()
    {
         std::cout << "x= " << x << " y= " << y << '\n';
         std::cout << "p1" << '\n';
         if (p1 != NULL)
         {
             p1->print();
         }
         std::cout << "p2" << '\n';
         if (p2 != NULL)
         {
            p2->print();
         }
         std::cout << '\n';
    }

};

二维点列表:

#include "Point.h"

class PointsList
{
private:
    std::vector <smart_pointer <Point> > points;

public:
    smart_pointer <Point> & operator [] ( int index ) {return points[index];}

public:
    void push_back(smart_pointer <Point> p) {points.push_back(p);}
    void erase(unsigned int index) {points.erase(points.begin() += index );}
    void printPoints()
    {
        std::cout << "List of points" << '\n';
        for (unsigned int i = 0; i < points.size();  i++)
        {
            points[i]->print();

        }

    }
};

测试代码:

#include "Point.h"
#include "PointsList.h"

int main()
{
    smart_pointer <Point> pb = NULL;
    pb = (new Point(0,0));
    smart_pointer <Point> p0(new Point(0,0));
    p0->print();
    smart_pointer <Point> p1(new Point(10,10));
    p1->print();
    smart_pointer <Point> p2(new Point(20,20));
    p2->print();
    smart_pointer <Point> p3(new Point(30,30));
    p3->print();

    smart_pointer <Point> pa(p3);
    p0->setP1(p2);
    p0->setP2(p3);
    p0->print();    
    p0 = p1;
    p0->print();
    p0->print();

    PointsList pl1;
    pl1.push_back(p0);
    pl1.push_back(p1);

    PointsList pl2;
    pl2.push_back(p2);
    pl2.push_back(p3);
    pl1.erase(0);
    pl1.printPoints();
    pl2.printPoints();
    return 0;
}

这种解决方案的优点或缺点在哪里?海量数据的运行速度、转换、继承可能出现的问题等。感谢您的帮助。

我对这个例子还有一个问题:哪种类型的智能指针(共享的,作用域的)最适合这样的数据结构:

//Class with cross-references to points p1, p2
class PointTopo
{
private:
    double x, y;
    PointTopo * p1;
    Point * p2;

public:
    PointTopo(double xx, double yy): x(xx), y(yy) {this-> p1 = NULL; this->p2 = NULL;}
    ...

};

//Class  with cross references: topological model for Delaunay triangulation
class Edge
{
   private:
      Point2D * start;
      Edge *next;
      Edge *previous;
      Edge *twin;
...
};

感谢您的帮助...

最佳答案

您的引用计数不起作用。

如果您将两个智能指针一起复制或赋值,它们需要使用相同的位置来执行计数。

目前每个对象都有自己的计数,因此它们可能会不同步。

smart_pointer<int>  x(new x);      // x.pointer: <good> x.reference_count: 1
{
    smart_pointer<int>  y;         // y.pointer: NULL   y.reference_count: -1

    y = x;  // x.pointer: <good> x.reference_count: 1
            // y.pointer: <good> y.reference_count: 2

    smart_pointer<int>  z;
    x = z;  // x.pointer: NULL                        x.reference_count:  0 (BAD)
            // z.pointer: NULL                        z.reference_count: -1
            // y.pointer: <bad> (it was deleted by x) y.reference_count:  2
}

编辑:

按照评论中的要求说明问题。

此时。我们刚刚创建 z 的地方。但还没有完成 x = z;

x { pointer: 0xabcd1234  reference_count: 1  }
y { pointer: 0xabcd1234  reference_count: 2  }
z { pointer: NULL        reference_count: -1 }


    // So here is your assignment operator.
    // Doing the `x = z` we will walk through the following code.
    //
    smart_pointer <T> & operator = (const smart_pointer <T> & p)
    {
            if (this != &p)
            {
                // We get here.
                // Left hand side is 'x' so this condition will be true.
                if( -- reference_count == 0)
                {
                    // Now we are deleting a pointer.
                    // That is held by 'x'
                    delete pointer;

                    // But 'y' is holding a pointer with the same value.
                    // Now y is holding a pointer to a deleted variable.
                }

                // Here we copy 'z' into 'x'
                // Copy the pointer. That happens to be NULL.
                pointer = p.pointer;

                // Now we copy and increment the reference count.
                // So 'x' has a value of 0 while 'z' has a value of -1.
                // This also breaks the invariant on 'x' that NULL values should
                // have a reference count of -1 (as X is NULL and ref-count is 0)
                reference_count = p.reference_count + 1;
            }

    return *this;
    }

如果有人现在尝试使用“y”,我们就会有未定义的行为,因为它包含指向已取消分配的内存的指针。

编辑经典(但过于简单的智能指针:

#include <vector>

template<typename T>
class SP
{
    T*        object;
    size_t*   count;

    public:
        SP(T* data)
        try
        // Use weird try around initializer list to catch new throwing.
        // If it does we delete data to stop it from leaking.
           :object(data)
           ,count(data ? new int(1) : NULL)
        { /* This is the constructor */}
        catch(...)
        {delete data;}

        SP():                 object(NULL),       count(NULL)       {}
      //SP(T* data):          object(data),       count(new int(1)) {}  // Lined up here so it look neat but implemented above to use weird try catch
        SP(SP<T> const& rhs): object(rhs.object), count(rhs.count)  {if (count) {++(*count);}}
        SP<T>& operator=(SP<T> rhs)  // Note implicit copy construction in rhs
        {
            // Using copy swap idiom for assignment.
            // The copy is hidden because the parameter is pass by value.
            this->swap(rhs);
            return *this;
        }
        void swap(SP<T>& rhs) throw()
        {
            std::swap(object, rhs.object);
            std::swap(count,  rhs.count);
        }
        ~SP()
        {
            if ((count) && (--(*count) == 0))
            {
                 delete count;
                 delete object;
            }
        }
};

关于c++ - 简单引用计数 : smart pointers,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/3613299/

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