如何使用模板模板参数专门化模板类的成员

发布于 2024-12-06 11:18:41 字数 598 浏览 0 评论 0原文

我有一个带有 int 和模板模板参数的模板类。现在我想专门化一个成员函数：

template <int I> class Default{};
template <int N = 0, template<int> class T = Default> struct Class
{
    void member();
};

// member definition
template <int N, template<int> class T> inline void Class<N, T>::member() {}

// partial specialisation, yields compiler error
template <template<int> class T> inline void Class<1, T>::member() {}

谁能告诉我这是否可能以及我在最后一行做错了什么？

编辑：我要感谢大家的意见。由于我还需要对某些 T 进行专门化，因此我选择反对 Nawaz 建议的解决方法，并将整个类专门化，因为无论如何它只有一个成员函数和一个数据成员。

原文

I have a template class with an int and a template template parameter.
Now I want to specialize a member function:

template <int I> class Default{};
template <int N = 0, template<int> class T = Default> struct Class
{
    void member();
};

// member definition
template <int N, template<int> class T> inline void Class<N, T>::member() {}

// partial specialisation, yields compiler error
template <template<int> class T> inline void Class<1, T>::member() {}

Can anyone tell me if this is possible and what I am doing wrong on the last line?

EDIT: I'd like to thank everyone for their input. As I also need a specialization for some T, I opted against the workaround suggested by Nawaz and specialized the whole class, as it had only one member function and one data member anyway.

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山川志 2024-12-13 11:18:41

您不能部分特化单个成员函数，您必须为整个类执行此操作。

template <int I> class Default{};
template <int N = 0, template<int> class T = Default> struct Class
{
    void member();
};

// member definition
template <int N, template<int> class T> inline void Class<N, T>::member() {}

// partial specialization
template <template<int> class T> struct Class<1, T>
{
  void member() {}
};

You can't partially specialize a single member function, you'll have to do it for the entire class.

template <int I> class Default{};
template <int N = 0, template<int> class T = Default> struct Class
{
    void member();
};

// member definition
template <int N, template<int> class T> inline void Class<N, T>::member() {}

// partial specialization
template <template<int> class T> struct Class<1, T>
{
  void member() {}
};

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牵你的手，一向走下去 2024-12-13 11:18:41

由于这是不允许的，这里有一种解决方法：

template <int I> class Default{};

template <int N = 0, template<int> class T = Default> 
struct Class
{
    void member()
    {
         worker(int2type<N>()); //forward the call
    }
 private:
     template<int N> struct int2type {};

     template<int M>
     void worker(const int2type<M>&) //function template
     {
         //general for all N, where N != 1
     }
     void worker(const int2type<1>&) //overload 
     {
         //specialization for N == 1
     }
};

想法是，当 N=1 时，函数调用 worker(int2type()) 将解析为第二个函数（专业化），因为我们传递的是 int2type<1> 类型的实例。否则，第一个（一般）功能将得到解决。

As that is not allowed, here is one workaround:

template <int I> class Default{};

template <int N = 0, template<int> class T = Default> 
struct Class
{
    void member()
    {
         worker(int2type<N>()); //forward the call
    }
 private:
     template<int N> struct int2type {};

     template<int M>
     void worker(const int2type<M>&) //function template
     {
         //general for all N, where N != 1
     }
     void worker(const int2type<1>&) //overload 
     {
         //specialization for N == 1
     }
};

The idea is, when N=1, the function call worker(int2type<N>()) will resolve to the second function (the specialization), because we're passing an instance of the type int2type<1>. Otherwise, the first, the general, function will be resolved.

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