Singleton模板专业化编译的奇怪错误

Weird error of compilation of template specialization of Singleton

本文关键字：错误 编译 专业化 Singleton      更新时间：2023-10-16

我对使用类和函数作为参数的模板有问题。问题是，当只声明和定义AAInterceptor时，它可以正常工作。当我添加BBInterceptor时，有一个编译错误，看起来很奇怪。

这是代码

template< class T>
class Singleton
{
protected:
    static T* ms_Singleton;
    Singleton()
    {
        if(!ms_Singleton)ms_Singleton = new T;
    }
public:
    static T& getSingleton( void ){assert( 0); return *ms_Singleton; }
};
struct InterceptorData
{
    unsigned int flag;
    InterceptorData():flag(0){}
};
template <class C, void (C::*TMethod)(const InterceptorData*)>
class InterceptorManager : public Singleton< InterceptorManager<C,TMethod> >
{
};
class AClass
{
public:
    virtual void Amethod(const InterceptorData* p_data = 0){};
};
class AAInterceptor : public InterceptorManager<AClass, &AClass::Amethod>
{
public:
    static AAInterceptor& getSingleton(void)
    {
        if (!ms_Singleton) new AAInterceptor();
        assert( ms_Singleton );  
        return ( *(static_cast< AAInterceptor*>(ms_Singleton)) );
    }
};
class BClass
{
public:
    virtual void Bmethod(const InterceptorData* p_data = 0){};
};
class BBInterceptor : public InterceptorManager<BClass, &BClass::Bmethod>
{
public:
    static BBInterceptor& getSingleton(void)
    {
        if (!ms_Singleton) new BBInterceptor();
        assert( ms_Singleton );  
        return ( *(static_cast< BBInterceptor*>(ms_Singleton)) );  //Here is the error of compilation
    }
};
int main(void)
{
    AAInterceptor a;
    BBInterceptor b;
    return 0;
}

它与GCC配合良好(http://codepad.org/Bi6zbsmq)，但MSVC2008没有。

这是Visual Studio中的错误：

error:  " error C2440: 'static_cast' : cannot convert from 'InterceptorManager<C,TMethod> *' to 'BBInterceptor *' " 58
with
1>        [
           1>            C=BClass,
           1>            TMethod=void AClass::`vcall'{0}'(const InterceptorData *)
           1>        ]
1>        Types pointed to are unrelated; conversion requires reinterpret_cast, C-style cast or function-style cast

编译器似乎把不同的模板混合在一起，不是吗？

提前感谢您的帮助

此代码不会导致编译器错误，只需将虚拟调用封装到非虚拟函数中并获取其地址即可。

template<typename T>
class Singleton
{
protected:
  static T* ms_Singleton;
  Singleton()
  {
    if(!ms_Singleton)ms_Singleton = new T;
  }
public:
  static T& getSingleton( void ){assert(0); return *ms_Singleton; }
};
template<typename T>
T* Singleton<typename T>::ms_Singleton = 0;
struct InterceptorData
{
  unsigned int flag;
  InterceptorData():flag(0){}
};
template <typename C, void (C::*TMethod)(const InterceptorData*)>
class InterceptorManager : public Singleton< InterceptorManager<C,TMethod> >
{
};
class AClass
{
public:
  void Amethod(const InterceptorData* p_data = 0){AmethodImpl(p_data);};
  virtual void AmethodImpl(const InterceptorData* p_data = 0){};
};
class AAInterceptor : public InterceptorManager<AClass, &AClass::Amethod>
{
public:
  static AAInterceptor& getSingleton(void)
  {
    if (!ms_Singleton) new AAInterceptor();
    assert( ms_Singleton );  
    return ( *(static_cast< AAInterceptor*>(ms_Singleton)) );
  }
};
class BClass
{
public:
  void Bmethod(const InterceptorData* p_data = 0){BmethodImpl(p_data);};
  virtual void BmethodImpl(const InterceptorData* p_data = 0){};
};
class BBInterceptor : public InterceptorManager<BClass, &BClass::Bmethod>
{
public:
  static BBInterceptor& getSingleton(void)
  {
    if (!ms_Singleton) new BBInterceptor();
    assert( ms_Singleton );  
    return ( *(static_cast< BBInterceptor* >(ms_Singleton)) );  
  }
};
int main(void)
{
  AAInterceptor a;
  BBInterceptor b;
  return 0;
}
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