如何避免“模板参数在部分专业化中无法推导”

Posted 2021-04-30

我想使用模板访问嵌套类，但无法弄清楚如何操作：示例代码：

template <typename T> class mything {
  typedef unsigned key;
  T data;
public:
  template <typename subT> class mysubthing { typedef subT value_type; };
  using subthing = mysubthing<T>;
  using subthing_ro = mysubthing<const T>;
};

template<typename T> struct X; // a container for all value_types

#ifdef MAKE_IT_FAIL
// this should automatically set the X<mything<T>::mysubthing<subT>
template<typename T,typename subT> struct X<typename mything<T>::template mysubthing<subT>> {
  using value_type = subT;
};
#endif

typedef mything<int> intthing;

#ifndef MAKE_IT_FAIL
template<> struct X<mything<int>::subthing> { using value_type = int; };
template<> struct X<mything<int>::subthing_ro> { using value_type = const int; };
#endif

int main(void) {
  intthing t;
  X<intthing::subthing>::value_type data = 1; // a data object
  X<intthing::subthing_ro>::value_type data_ro = 1; // read-only data object

  return 0;
}

这没有-DMAKE_IT_FAIL编译，但当然它完全忽略了模板的要点，因为我想要的是手动输入。如何使用-DMAKE_IT_FAIL使其工作？

答案

你不能像那样专攻：

template<typename T,typename subT> 
struct X<typename mything<T>::template mysubthing<subT>> {

因为在C ++中不可能（不支持）从outer<T>::anything_after这样的类型中推导出T.

在这个一般情况下，你根本不需要专业化。只需定义默认的X，然后只专门化其他情况：

template <typename T> class mything {
  typedef unsigned key;
  T data;
public:
  template <typename subT> struct mysubthing 
  { 
      typedef subT value_type; 

  };
  using subthing = mysubthing<T>;
  using subthing_ro = mysubthing<const T>;
};

template<typename T> struct X
{
   using value_type = typename T::value_type;
};
// this should automatically set the X<mything<T>::mysubthing<subT>

typedef mything<int> intthing;

template<> struct X<mything<int>::subthing> { using value_type = int; };
template<> struct X<mything<int>::subthing_ro> { using value_type = const int; };

int main(void) {
  intthing t;
  X<intthing::subthing>::value_type data = 1; // a data object
  X<intthing::subthing_ro>::value_type data_ro = 1; // read-only data object

  return 0;
}

Addendum

根据其中一条评论，X实际上是std::iterator_traits，已经定义了。在这种情况下，唯一的方法是在mything类之外定义迭代器类：

template <typename T, typename subT>
class mything_iterator {
   typedef subT value_type; 
};
template <typename T> class mything {
      typedef unsigned key;
      T data;
    public:
      using iterator = mything_iterator<T, T>;
      using const_iterator = mything_iterator<T, const T>;
};
namespace std {
   template<typename T, class subT>
   class iterator_traits<mything_iterator<T, subT>>{
       using value_type =typename  mything_iterator<T, subT>::value_type;
       // etc...
   };
   template<> struct iterator_traits<mything<int>::iterator>
   { using value_type = int; };
   template<> struct iterator_traits<mything<int>::const_iterator>
   { using value_type = int; };
}