c++ - 从模板类多重继承

Question

我在从同一模板类的不同实例进行多重继承时遇到问题。具体来说，我正在尝试这样做:

template <class T>
class Base
{

public:

    Base() : obj(NULL)
    {
    }

    virtual ~Base()
    {
        if( obj != NULL ) delete obj;
    }

    template <class T>
    T* createBase()
    {
        obj = new T();

        return obj;
    }

protected:

    T* obj;

};

class Something
{
    // ...
};

class SomethingElse
{
    // ...
};

class Derived : public Base<Something>, public Base<SomethingElse>
{

};

int main()
{
    Derived* d = new Derived();
    Something* smth1 = d->createBase<Something>();
    SomethingElse* smth2 = d->createBase<SomethingElse>();

    delete d;

    return 0;
}

当我尝试编译上面的代码时，出现以下错误:

1>[...](41) : error C2440: '=' : cannot convert from 'SomethingElse *' to 'Something *'
1>        Types pointed to are unrelated; conversion requires reinterpret_cast, C-style cast or function-style cast
1>        [...](71) : see reference to function template instantiation 'T *Base<Something>::createBase<SomethingElse>(void)' being compiled
1>        with
1>        [
1>            T=SomethingElse
1>        ]
1>[...](43) : error C2440: 'return' : cannot convert from 'Something *' to 'SomethingElse *'
1>        Types pointed to are unrelated; conversion requires reinterpret_cast, C-style cast or function-style cast

由于成员 obj 是从 Base< Something > 和 Base< SomethingElse > 继承的，所以这个问题似乎是不明确的，我可以通过消除对 createBase 的调用来解决这个问题:

Something* smth1 = d->Base<Something>::createBase<Something>();
SomethingElse* smth2 = d->Base<SomethingElse>::createBase<SomethingElse>();

但是，从句法上讲，这个解决方案非常不切实际，我更喜欢更优雅的方法。此外，我对第一条错误消息感到困惑。这似乎意味着在 Base 中有一个实例化 createBase，但这怎么可能呢？非常感谢有关此问题的任何信息或建议。

Answer 1

It seems to imply that there is an instantiation createBase< SomethingElse > in Base< Something >, but how is that even possible?

肯定有，因为你的 createBase<T>()是成员模板函数(该函数中的T与周围类中的T无关)。

我会做类似的事情:

// in Derived, or you could make some class (eg. MultiBase) for it

template <class T>
T* createBase()
{
  return Base<T>::createBase();
}