我有一个像这样的对象缓存类:

#include "boost/thread/mutex.hpp"
#include "boost/unordered_map.hpp"

template <typename type1, typename type2>
class objectCache
{
public:
    objectCache()
    {
        IDCounter = 0;
    }
    ~objectCache()
    {
        for ( it=free_objects.begin() ; it != free_objects.end(); it++ )
            delete (*it).second;
        for ( it=busy_objects.begin() ; it != busy_objects.end(); it++ )
            delete (*it).second;
    }
    type1* objectCache::Get()
    {
        boost::mutex::scoped_lock(io_mutex);
        if(free_objects.size() > 0)
        {
            it = free_objects.begin();
            type1 *temp = (*it).second;
            busy_objects[(*it).first] = temp;
            free_objects.erase(free_objects.begin());
            return temp;
        }
        type1 * temp = new type2;
        ++IDCounter;
        busy_objects[IDCounter] = temp;
        return temp;
    }
    void objectCache::Pushback(type1)
    {
        boost::mutex::scoped_lock(io_mutex);
        free_objects[ID] = socket;
        it = busy_objects.find(ID);
        busy_objects.erase(it);
    }
protected:
private:
    boost::mutex io_mutex;
    long long IDCounter;
    boost::unordered_map<long long, type1*> free_objects;
    boost::unordered_map<long long, type1*> busy_objects;
    typename boost::unordered_map<long long, type1*>::iterator it;
};

class A{
public:
    A(int num){
        number = num;
    }
    int number;
};
int main(int argc, char* argv[])
{
    objectCache<a, a(1)> intcache;
    A* temp = intcache.Get();
    cout <<temp->number <<endl;
    system("pause");
    return 0;
}


我知道“ typename type2”是不必要的,但是我需要一种将具有构造函数(如类A)的类对象传递给模板的方法。
还是他们这样做的另一种方式?请帮忙。

最佳答案

传递一个为您创建实例的对象,而不是传递一个显式值:

template <typename type1>
struct DefaultInstanceCreator {
  type1 * operator ()() const {
    return new type1;
  }
};

template < typename type1
         , typename InstanceCreator = DefaultInstanceCreator<type1> >
class objectCache {
public:
  objectCache (InstanceCreator const & instCreator)
    : instCreator_ (instCreator)  {
  }
  type1* Get() {
    type1 * temp = instCreator_ ();
  }
private:
  InstanceCreator instCreator_;
};


然后,您的对象可以具有自己的特定创建者:

class A {
public:
    A(int num){
        number = num;
    }
    int number;

public:
  struct CreateInstance  {
    CreateInstance (int value)
      : value_ (value) {
    }
    A * operator ()() const {
      return new A(value_);
    }
    int value_;
  };
};

int main(int argc, char* argv[]) {
  objectCache<A, A::CreateInstance> intcache ( A::CreateInstance (1) );
  A* temp = intcache.Get();
  return 0;
}


这种方法的优点是,可以将缓存与具有不同数量构造函数的对象一起使用。

07-24 19:01
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