如何在两个不同对象的基类中使用模板类
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21-12-2019 - |
题
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,例如:
class MediaSample
{
public:
MediaSample(ObjectManager<?>* manager){}
private:
ObjectManager<?> mMediaSampleManager;
}
现在我从 MediaSample 声明另外两个驱动类,例如:
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, ,输入参数 MediaSample
班级。
必须放置哪种类型而不是问号字符?(音频、视频还是什么?)
有关更多澄清,我发布了ObjectManager的完整来源,请注意,对象管理器是媒体样本对象的民意调查,媒体样本可以是音频或视频
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;
};
音频样本类:
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;
};
解决方案
混合编译时(模板)和运行时多态性(继承)可能会变得相当困难。你可以做什么:
a) 将基类作为模板
template <class T>
class MediaSample {
ObjectManager<T> mMediaSampleManager;
};
class AudioSample : public MediaSample<AudioSample>
在这种情况下 AudioSample
和 VideoSample
不会有一个共同的基类,这可能不是你想要的。或者,您可以
b) 给ObjectManager一个非模板基类(一个接口)
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>()){}
另请注意,如果您的 MediaSample
构造函数接受一个指向新构造的指针 ObjectManager
并且您可能将其复制到成员对象,您必须记住删除指针,否则会泄漏。您可能应该改为按值传递。
其他提示
如何将 MediaSample 设为模板:
template <typename T>
class MediaSample
{
public:
MediaSample(ObjectManager<T>* manager){}
private:
ObjectManager<T> mMediaSampleManager;
};
这样您就可以拥有不同类型的 MediaSample,而无需派生类 AudioSample 和 VideoSample。
在这种情况下 MediaSample
需要是一个模板
template<typename T>
class MediaSample
{
public:
MediaSample(ObjectManager<T>* manager): mMediaSampleManager(manager) {}
private:
ObjectManager<T>* mMediaSampleManager;
}
和 XxxSample
必须继承专业 MediaSample
class AudioSample : public MediaSample<AudioSample>
{
public:
AudioSample() : MediaSample(new ObjectManager<AudioSample>()){}
...
}
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