Question

我在两个对象的基础上存储模板类时遇到问题。

假设我有一个名为ObjectManager的通用类定义为：

template<typename T>
class ObjectManager
{}

我有一个基类：

class MediaSample
{
public:
    MediaSample(ObjectManager<?>* manager){}
private:
    ObjectManager<?> mMediaSampleManager;
}

现在我宣布另外两个来自MediaSample的课程，说：

class AudioSample : public MediaSample
{
public:
    AudioSample() : MediaSample(new ObjectManager<AudioSample>()){}
    //...
}
public class VideoSample : public MediaSample
{
public:
    VideoSample() : MediaSample(new ObjectManager<VideoSample>())
    //...
}

我的问题是经理的类型，MediaSample类的输入参数。

必须使用哪种类型而不是问号字符？（音频，视频或什么？）

为了更多说明，我发布了完整的ObjectManager源代码请注意，对象管理器是对媒体示例对象的轮询，而媒体示例可以是音频或视频

class BaseMediaSampleManager
{
public:
    ~BaseMediaSampleManager(){}

    virtual INT initialize(INT objectCount) = 0;

protected:
private:
};

template<typename T>
class MediaSamppleManager : public BaseMediaSampleMnager //(based on user2079303 advise)
{
protected:
    MediaSamppleManager(void);

    MediaSamppleManager(const MediaSamppleManager& manager);

    MediaSamppleManager& operator = (const MediaSamppleManager& rhs);

public: 

    virtual ~MediaSamppleManager(void);

    static MediaSamppleManager<T>* instance();

    INT initialize(INT objectCount);

    // take existing object from object pool and return that object
    INT aquireObject(T** object);

    // take back in use object to object pool
    INT release(T* object);

private:
    std::list<T*> mPooledSamples;
    INT32 mPooledObjectCount;
    INT32 mInUseObjects;
    Mutex mPooledSamplesMutex;
    static Mutex mSampleManagerObjectMutex;
    static MediaSamppleManager<T>* mSampleManagerObject;
};

template<typename T>
Mutex MediaSamppleManager<T>::mSampleManagerObjectMutex;

template<typename T>
MediaSamppleManager<T>* MediaSamppleManager<T>::mSampleManagerObject = NULL;

template<typename T>
MediaSamppleManager<T>* MediaSamppleManager<T>::instance()
{
    if (mSampleManagerObject == NULL)
    {
        ScopedLock<Mutex> guard(mSampleManagerObjectMutex);
        if (mSampleManagerObject == NULL)
        {
            MediaSamppleManager<T>* temp = new MediaSamppleManager<T>();
            mSampleManagerObject = temp;
        }
    }
    return mSampleManagerObject;
}

template<typename T>
MediaSamppleManager<T>& MediaSamppleManager<T>::operator=(const MediaSamppleManager<T>& rhs)
{

}

template<typename T>
MediaSamppleManager<T>::MediaSamppleManager(const MediaSamppleManager<T>& manager)
{

}

template<typename T>
INT MediaSamppleManager<T>::release(T* object)
{
    ScopedLock<Mutex> guard(mPooledSamplesMutex);
    mPooledSamples.push_back(object);
    mInUseObjects--;
}

template<typename T>
INT MediaSamppleManager<T>::aquireObject(T** object)
{
    ScopedLock<Mutex> guard(mPooledSamplesMutex);
    if (mInUseObjects == mPooledObjectCount)
    {
        // do we need waiting until new sample become available? or 
        // is it required to create new sample when no free sample exist in pool? 
        return 2;
    } 
    else
    {
        T* temp = 0;
        temp = mPooledSamples.front();
        *object = temp;
        mPooledSamples.pop_front();
        mInUseObjects++;
    }
    return 1;
}

template<typename T>
INT MediaSamppleManager<T>::initialize(INT objectCount)
{
    if (objectCount<=0)
    {
        return -1;
    }
    mPooledObjectCount = objectCount;
    mPooledSamples.resize(objectCount);
    for (int i = 0 ; i< objectCount ; i++)
    {
        T* temp = new T(this);
        mPooledSamples.push_back(temp);
    }
    return 1;
}

template<typename T>
MediaSamppleManager<T>::~MediaSamppleManager(void)
{
    std::cout << "MediaSampleManager Destroyed \n";
}

template<typename T>
MediaSamppleManager<T>::MediaSamppleManager(void)
    :mPooledObjectCount(0)
    ,mInUseObjects(0)
{

}

类MediaSample：

class MediaSample
{
public:
    // create empty media sample
    MediaSample(BaseMediaSampleManager* manager);

    virtual ~MediaSample(void);

    virtual INT32 rate() = 0;

    virtual double frameSize() = 0;

    // data size of media sample
    virtual INT size();

    UINT* data();

    ULONG samplingTime();

    INT32 addRef();

    INT32 release();    

private:
    UINT8* mMediaSampleBuffer;
    // The time sample captured
    ULONG mTimestamp;
    // length of data
    INT mDataLength;

    INT mRefCount;

    BaseMediaSampleManager* mMediaSampleManager;
};

类AudioSample：

class PMCAPI AudioSample 
    :public MediaSample/*<AudioSample>*/
{
public:
    AudioSample(MediaSamppleManager<AudioSample>* manager);

    ~AudioSample();

    virtual INT32 rate();

    virtual double frameSize();

    virtual INT size();

protected:
private:
    // in 8,16 KHZ
    INT32 mSampleRate;
    // Mono or Stereo
    INT mChannels;
    // Used format ex: PCM,G729,G723,G711A,G711U
    INT mCodec;
    // ex : 8 bit , 16 bit
    INT mBitsPerSample;
};

Answer 1

混合编译时间（模板）和运行时多态（继承）会很困难。你能做什么：

a）使基类成为模板

template <class T>
class MediaSample {
    ObjectManager<T> mMediaSampleManager;
};

class AudioSample : public MediaSample<AudioSample>

在这种情况下，AudioSample和VideoSample将没有共同的基类，这可能不是您想要的。或者，你可以

b）为ObjectManager提供非模板基类（接口）

class BaseManager{
    virtual ~BaseManager(){}
    // pure virtual functions that ObjectManager implements
};

template<typename T>
class ObjectManager: public BaseManager{
    // implementation
};

class MediaSample {
    BaseManager mMediaSampleManager;
    // ...
};

class AudioSample: public MediaSample {
public:
    AudioSample() : MediaSample(new ObjectManager<AudioSample>()){}

另外，请注意，如果您的MediaSample构造函数接受指向新构造的ObjectManager的指针，并且您可能将其复制到成员对象，则必须记住删除指针或者它会泄漏。你可能应该通过值传递。

Answer 2

如何将MediaSample设为模板：

template <typename T>
class MediaSample
{
public:
    MediaSample(ObjectManager<T>* manager){}
private:
    ObjectManager<T> mMediaSampleManager;
};

通过这种方式，您可以使用不同类型的MediaSample，而无需使用派生类AudioSample和VideoSample。

Answer 3

在这种情况下，MediaSample需要是模板

template<typename T>
class MediaSample
{
public:
    MediaSample(ObjectManager<T>* manager): mMediaSampleManager(manager) {}
private:
    ObjectManager<T>* mMediaSampleManager;
}

并且XxxSample必须继承专门的MediaSample

class AudioSample : public MediaSample<AudioSample>
{
public:
    AudioSample() : MediaSample(new ObjectManager<AudioSample>()){}
    ...
}

如何在两个不同对象的基类中使用模板类

3 个答案: