如何为 class 的每个接口单独定义一些函数?

How can I define some functions individually for each interface of my class?

我正在学习COM,也是C++的初学者。我定义了三个接口:

#include <objbase.h>

// pure abstract base class
interface IX : IUnknown{  
  virtual void __stdcall Fx() = 0; // pure virtual function
};

// pure abstract base class
interface IY : IUnknown{
  virtual void __stdcall Fy() = 0; // pure virtual function
};

// pure abstract base class
interface IZ : IUnknown{
  virtual void __stdcall Fz() = 0; // pure virtual function
};

extern "C"{
  extern const IID IID_IX;
  extern const IID IID_IY;
  extern const IID IID_IZ;
}

然后我创建一个 class 实现这些接口:

#include <iostream>
#include "IFace.h"

// COM component
class CA :
  public IX,
  public IY,
  public IZ
{
public:
  CA();
  ~CA();
  virtual void __stdcall Fx(); // IX
  virtual void __stdcall Fy(); // IY
  virtual void __stdcall Fz(); // IZ
  // IUnknown (this interface implemented by IX, IY and IZ)
  virtual HRESULT __stdcall QueryInterface(const IID& iid, void** ppv);
  virtual ULONG __stdcall AddRef();
  virtual ULONG __stdcall Release();
private:
  // Each interface has own counter (for convenience of debugging)
  ULONG ix_counter;
  ULONG iy_counter;
  ULONG iz_counter;
};
...

我想为每个接口使用单独的实例计数器(为了调试方便)。在这种情况下我必须如何定义 AddRef()Release() 函数?

我可以这样做:

ULONG __stdcall CA::AddRef(){
  // increase interface counter here
}

但是对于我的 CA class 的 IXIYIZ 实例,此代码是 common ].如果实例是 IX 那么我只需要增加 ix_counter 计数器。如果实例是 IY 那么我只需要增加 iy_counter 计数器,等等

我试过这样的方法:

virtual HRESULT __stdcall QueryInterface(const IID& iid, void** ppv); // common implementation
virtual ULONG __stdcall IX::AddRef(); // for IX instances
virtual ULONG __stdcall IX::Release(); // for IX instances

virtual ULONG __stdcall IY::AddRef(); // for IY instances
virtual ULONG __stdcall IY::Release(); // for IY instances

virtual ULONG __stdcall IZ::AddRef(); // for IZ instances
virtual ULONG __stdcall IZ::Release(); // for IZ instances

而不是

virtual ULONG __stdcall AddRef();
virtual ULONG __stdcall Release();

我也试过:

// This is wrong definitions:
ULONG __stdcall CA::IX::AddRef(){
  return InterlockedIncrement(&ix_counter);
}

ULONG __stdcall CA::IX::Release(){
  InterlockedDecrement(&ix_counter);
  ULONG count = ix_counter + iy_counter + iz_counter;
  if (0 == count) delete this;
  return count;
}

而不是

ULONG __stdcall CA::AddRef(){
      // increase interface counter here
    }

但是我最后一次替换错了。

如何为每个接口单独定义 AddRef()Release() 函数?

我认为您应该从 class 实现 IUnknown 继承通用实现。这将与 COM 定义的内容平行。

I want to use the separate counter of instances for each interface (for convenience of debugging). How I must define AddRef() and Release() function at this case?

这完全违反了 COM 规则。其中一项要求是,每当您通过指向特定对象的任何类型的任何指针检索 IUnknown* 时,您都会获得相同的指针值。这就是为什么您根本无法以任何有意义的方式拥有单独的计数器。

您可以有一个特定于接口的计数器。您的代码可能如下所示:

// (interfaces defined)

template <typename T, typename Base>
class CUnknownWithCounterT :
    public Base
{
public:
    ULONG m_nCounter;

public:
    CUnknownWithCounterT() :
        m_nCounter(0)
    {
    }
    ULONG __stdcall AddRef()
    {
        m_nCounter++;
        return static_cast<T*>(this)->InternalAddRef();
    }
    ULONG __stdcall Release()
    {
        m_nCounter--;
        return static_cast<T*>(this)->InternalRelease();
    }
};

class CA :
    public CUnknownWithCounterT<CA, IX>,
    public CUnknownWithCounterT<CA, IY>,
    public CUnknownWithCounterT<CA, IZ>
{
public:
    ULONG m_nMainCounter;

public:
    // QueryInterface as stuff...
    ULONG __stdcall InternalAddRef()
    {
        return ++m_nMainCounter;
    }
    ULONG __stdcall InternalRelease()
    {
        return --m_nMainCounter;
    }
};

准备出发:

CA A;
IX* pX = &A;
IY* pY = &A;
IZ* pZ = &A;
pY->AddRef();
pZ->AddRef();
pZ->AddRef();
_tprintf(_T("%d %d %d\n"), 
    A.CUnknownWithCounterT<CA, IX>::m_nCounter, 
    A.CUnknownWithCounterT<CA, IY>::m_nCounter, 
    A.CUnknownWithCounterT<CA, IZ>::m_nCounter, 
    0);
// 0 1 2

但是,您还需要分别覆盖 QueryInterface,因为它也是 returns 相应计数器需要递增的指针。

实施将与 COM 兼容(m_nMainCounter 有 COM 的计数器),问题是各个计数器在帮助您查找引用泄漏方面是否准确。如果它们在大多数情况下都有用,那么您当然可以将它们用于故障排除。