高分求助,<com技术内幕>里的例子,运行就是和预期不一致

Question

#include <iostream.h>
#include <objbase.h>void trace(const char *pMsg)
{
cout<<pMsg<<endl;
}interface IX : IUnknown
{
virtual void _stdcall fx()=0;
};interface IY : IUnknown
{
virtual void _stdcall fy()=0;
};interface IZ : IUnknown
{
virtual void _stdcall fz()=0;
};extern const IID IID_IX;
extern const IID IID_IY;
extern const IID IID_IZ; class CA : public IX, public IY
{
public:
//IUnknown implementation
virtual HRESULT _stdcall QueryInterface(const IID& iid,void** ppv);
virtual ULONG _stdcall AddRef(){return 0;}
virtual ULONG _stdcall Release(){return 0;} virtual void _stdcall fx() {cout<<"CA::fx"<<endl;} virtual void _stdcall fy() {cout<<"CA::fy"<<endl;}
};HRESULT _stdcall CA::QueryInterface (const IID &iid,void** ppv)
{
if(iid == IID_IUnknown)
{
trace("QueryInterface:Return pointer to IUnknown.");
*ppv = static_cast<IX*>(this);
}
else if(iid==IID_IX)
{
trace("QueryInterface:Return pointer to IX.");
*ppv = static_cast<IY*>(this);
}
else if(iid == IID_IY)
{
trace("QueryInterface:Return pointer to IY.");
*ppv = static_cast<IY*>(this);
}
else
{
trace("QueryInterface:Interface not supported.");
*ppv = NULL;
return E_NOINTERFACE;
}
reinterpret_cast<IUnknown*>(*ppv)->AddRef ();
return S_OK;
}IUnknown * CreateInstance()
{
IUnknown *pI = static_cast<IX*>(new CA);
pI->AddRef ();
return pI;
}static const IID IID_IX =
{0x32bb8320,0xb41b,0x11cf,
{0xa6,0xbb,0x0,0x80,0xc7,0xb2,0xd6,0x82}};static const IID IID_IY =
{0x32bb8321,0xb41b,0x11cf,
{0xa6,0xbb,0x0,0x80,0xc7,0xb2,0xd6,0x82}};static const IID IID_IZ =
{0x32bb8322,0xb41b,0x11cf,
{0xa6,0xbb,0x0,0x80,0xc7,0xb2,0xd6,0x82}};
int main()
{
HRESULT hr; trace("Client:Get an IUnknown pointer.");
IUnknown *pIUnknown = CreateInstance(); trace("Client:Get interface IX."); IX *pIX = NULL;
hr = pIUnknown->QueryInterface (IID_IX,(void**)&pIX);
if(SUCCEEDED(hr))
{
trace("Client:Succeeded getting IX.");
//为什么这里最终调用的是fy(),正确的应该是fx(),不知道哪里错了
pIX->fx ();
} trace("Client:Get interface IY."); IY * pIY = NULL;
hr = pIUnknown->QueryInterface (IID_IY,(void**)&pIY);
if(SUCCEEDED(hr))
{
trace("Client:Succeeded getting IY.");
pIY->fy ();
} trace("Client:Ask for an unsupported interface."); IZ *pIZ = NULL;
hr = pIUnknown->QueryInterface (IID_IZ,(void**)&pIZ);
if(SUCCEEDED(hr))
{
trace("Client:Succeeded getting IZ.");
pIZ->fz ();
}
else
{
trace("Client:Could not get interface Iz.");
} trace("Client:Get interface IY from interface IX.");
IY *pIYfromIX = NULL;
hr = pIX->QueryInterface (IID_IY,(void**)&pIYfromIX);
if(SUCCEEDED(hr))
{
trace("Client:Succeeded getting IY.");
pIYfromIX->fy ();
} trace("Client:Get interface IUnknown from IY."); IUnknown *pIUnknownFromIY = NULL;
hr = pIY->QueryInterface (IID_IUnknown,(void**)&pIUnknownFromIY);
if(SUCCEEDED(hr))
{
cout<<"Are the IUnknown pointers equal?";
if(pIUnknownFromIY == pIUnknown)
{
cout<<"Yes,pIUnknownFromIY == pIUnknown."<<endl;
}
else
{
cout<<"No,pIUnknownFromIY != pIUnknown."<<endl;
}
} delete pIUnknown; return 0;
}这是全部的代码,在对IX->fx()调用的时候发现每次都是调用的IY->fy(),
实在是搞不明白了,高手们指点一下阿
先谢谢了

Answer 1

HRESULT _stdcall CA::QueryInterface (const IID &iid,void** ppv)
{
...
else if(iid==IID_IX)
{
trace("QueryInterface:Return pointer to IX.");
*ppv = static_cast<IY*>(this);
}
...
}
 hr = pIUnknown->QueryInterface (IID_IX,(void**)&pIX)你这里pIX得到的是IY的接口, pIX->fx()实际上是访问的是IY的fy

Answer 2

仔细看了半天，才发现你这儿错了。
HRESULT _stdcall CA::QueryInterface (const IID &iid,void** ppv)
{
............
else if(iid==IID_IX)
{
trace("QueryInterface:Return pointer to IX.");
*ppv = static_cast<IY*>(this);  //这儿类型转成IY了，你取的是IX的指针，所以应该为*ppv = static_cast<IX*>(this); 
}