如何執行模板參數類從超派生使用匿名模板參數

Question

我有幾個模板類的

template < class Cost > 
class Transition { 
    public: 
    virtual Cost getCost() = 0; 
}; 

template < class TransitionCl, class Cost > 
class State { 
    protected: 
     State(){ 
      static_assert(
       std::is_base_of< Transition<Cost>, TransitionCl >::value, 
       "TransitionCl class in State must be derived from Transition<Cost>" 
      ); 
     } 
    public: 
     virtual void apply(const TransitionCl&) = 0; 
}; 

，我寧可不要通過Cost爲State，因爲State是完全獨立的Cost，但我想確保TransitionCl實現接口Transition。

有沒有辦法在第二個模板中使Cost匿名，以便在聲明新的State類時不必傳遞它？

僅供參考，我使用g++ -std=c++14 [source file]

編輯：我張貼的問題的改寫（希望更清晰）版本，並獲得最佳答案here

Answer 1

從我從你的問題的理解，在這裏是一個快速解決方案，我有：

template < class Cost > 
class Transition { 
    public: 
    virtual Cost getCost() = 0; 
}; 

template <typename T> 
class Der : public Transition<T> 
{ 
public: 
    T getCost() override { 
    } 
}; 

template < class Transition > 
class State; 

template <template <typename> class TransitionCl, typename Cost> 
class State <TransitionCl<Cost>> { 
public: 
     State(){ 
      static_assert(
       std::is_base_of< Transition<Cost>, TransitionCl<Cost> >::value, 
       "TransitionCl class in State must be derived from Transition<Cost>" 
      ); 
     } 
}; 

int main() 
{ 
    Der<int> d; 
    State<decltype(d)> s; 

    return 0; 
} 

在上面的示例中，您不必在創建狀態對象時傳遞'成本'類型。

===== UPDATE ======

template <typename Cost> 
class Transition 
{ 
public: 
    virtual Cost getCost() = 0; 
    virtual ~Transition() {} 
}; 

class TD: public Transition<int> 
{ 
public: 
    int getCost() override { 
     std::cout << "getCost override" << std::endl; 
     return 42; 
    } 
}; 

namespace detail { 
    template <typename T> 
    struct is_base_of_cust { 
     // This is a bit hacky as it is based upon the internal functions 
     // (though public) of the Transition class. 
     using CostType = decltype(std::declval<T>().getCost()); 
     static const bool value = std::is_base_of<Transition<CostType>, T>::value; 
    }; 
}; 

template <class TransitionCl> 
class State 
{ 
protected: 
    State() { 
     static_assert(
      detail::is_base_of_cust<TransitionCl>::value, 
      "TransitionCl class in State must be derived from Transition<Cost>" 
     ); 
    } 
public: 
    virtual void apply(const TransitionCl&) = 0; 
    virtual ~State() {} 
}; 


class StateImpl: public State<TD> 
{ 
public: 
    void apply(const TD&) override { 
     std::cout << "StateImpl::apply" << std::endl; 
    } 
}; 


int main() { 
    StateImpl impl; 
    return 0; 
} 

Answer 2

一種方法是使用getCost()返回類型（但它可能給你一個醜陋的錯誤messgae如果TransactionCl()不有這樣的公共成員功能）。

std::is_base_of< Transition< decltype(TransitionCl().getCost()) >, TransitionCl >::value, 

另一種選擇是加入typedef基類：

template < class Cost > 
class Transition { 
    public: 
    typedef Cost Cost_Type; // <------- 
    virtual Cost getCost() = 0; 
}; 

然後你就可以刪除State的typename Cost參數和使用typedef相反在你的靜態斷言......

std::is_base_of< Transition< typename TransitionCl::Cost_Type >, TransitionCl >::value,