How to avoid "template parameters not deducible in partial specialization"

Question

I would like to have access to a nested class using templates, and can't figure out how to do it: A sample code:

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;
}

This compiles without -DMAKE_IT_FAIL, but of course it completely misses the point about templates, since what I wanted was entered manually. How can I make it work with -DMAKE_IT_FAIL?

Answer 1

You can't specialize like that:

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

since deduction of T from types like outer<T>::anything_after is impossible (not supported) in C++.

You don't really need specialization in this general case at all. Just define the default X, and then only specialize the other cases:

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

According to one of the comments, X is actually std::iterator_traits , which is already defined. In that case, the only way around it is to define the iterator class outside the mything class:

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; };
}