c++ - 如何覆盖父类的静态变量

Question

如何覆盖父类的静态变量。

所以我有父类

class DatabaseItem
{
   static int instanceCount;

   DatabaseItem()
   {
     instanceCount++;
   }
};

如果我有 2 个继承自 DatabaseItem 的类，我希望每个类都记录它们的类只存在多少个实例。我该怎么做？

所以:

class Person : public DatabaseItem
{
  // how do I make sure when I make the call  int numOfpeople = Person::instanceCount;
  // that I only get the number of people objects that exist & not all the DatabaseItem
  // objects that exist?
};

class FoodItem : public DatabaseItem
{
  // how do I make sure when I make the call  int numOffoodItems = FoodItem::instanceCount;
  // that I only get the number of FoodItem objects that exist & not all the DatabaseItem
  // objects that exist?
};

---

编辑回应评论

是的，但是，上面只是一个例子，如果我这样做，那么我有很多重复的代码......

所以:

    class DatabaseItem
{
    public:
        static unsigned int instanceCount;
        static Vector <unsigned int> usedIDs;

        unsigned int ID;

        DatabaseItem()
        {
            ID = nextAvailableID();
            usedIDs.add( ID );
            DatabaseItem::instanceCount++;
        }

        DatabaseItem( unsigned int nID )
        {
            if ( isIDFree( nID ) )
            {
                ID = nID;
            }
            else ID = nextAvailableID();

            usedIDs.add( ID );
            DatabaseItem::instanceCount++;
        }

        bool isIDFree( unsigned int nID )
        {
            // This is pretty slow to check EVERY element

            for (int i=0; i<usedIDs.size(); i++)
            {
                if (usedIDs[i] == nID)
                {
                    return false;
                }
            }

            return true;
        }

        unsigned int nextAvailableID()
        {
            unsigned int nID = 0;

            while ( true )
            {
                if ( isIDFree( ID ) )
                {
                    return nID;
                }
                else nID++;
            }
        }
};


class Person    {
    public:
        static unsigned int instanceCount;
        static Vector <unsigned int> usedIDs;

        unsigned int ID;

        Person()
        {
            ID = nextAvailableID();
            usedIDs.add( ID );
            Person::instanceCount++;
        }

        Person( unsigned int nID )
        {
            if ( isIDFree( nID ) )
            {
                ID = nID;
            }
            else ID = nextAvailableID();

            usedIDs.add( ID );
            Person::instanceCount++;
        }

        bool isIDFree( unsigned int nID )
        {
            // This is pretty slow to check EVERY element

            for (int i=0; i<usedIDs.size(); i++)
            {
                if (usedIDs[i] == nID)
                {
                    return false;
                }
            }

            return true;
        }

        unsigned int nextAvailableID()
        {
            unsigned int nID = 0;

            while ( true )
            {
                if ( isIDFree( ID ) )
                {
                    return nID;
                }
                else nID++;
            }
        }
};

.. 然后我必须为 FoodItem、coffeeRun 重写相同的代码....

Answer 1

使用模板来避免在每个构造函数中更新实例计数等所有此类问题。

template<class T>
struct Instance
{
  static unsigned int count;
  Instance () { count ++; }
  Instance (const Instance& o) { count ++; }
};
template<class T>
unsigned int Instance<T>::count = 0;

现在这个模板可以被任何类继承，你需要计算它们的实例:

class DatabaseItem : public Instance<DatabaseItem> {};
class Person : public DatabaseItem, Instance<Person> {};
class FoodItem : public DatabaseItem, Instance<FoodItem> {};

就是这样!

只要声明了一个类对象，它就会完成它的工作。它可以按如下方式使用:

DatabaseItem *pD = new DatabaseItem[5];
Person obj[10];
cout<<Instance<DatabaseItem>::count<<endl;  // 5 + 10 = 15
cout<<Instance<Person>::count<<endl;        // 10

您不必在任何地方更新计数。