Can I get a const dependent name from a const type template argument when instantiating a template?

Question

I have some code snippets like below,

struct stA
{
    struct stB
    {
        void func() const { std::cout << "const" << std::endl; }
        void func()       { std::cout << "non-cont" << std::endl; }
    };
};

template <typename T>
void test()
{
   typename T::stB b;//can I get a const version of b?
   b.func();
   /*...*/
}

In my test, I found I can't get a const version of b even if I instantiated this function template test with T = const stA argument.

So the question is can I get a const dependent name when instantiating a template?

If the answer is NO, I also want to know why qualifier const is discarded when substituting the template argument?

If the answer is YES, I want to HOW?

BTW, I tested above code in VS2017.

Answer 1

 typename T::stB b;//can I get a const version of b? 

Sure. Use a helper class to select the type.

#include <iostream>

struct stA
{
   struct stB
   {
      void func() const
      {
         std::cout << "const" << std::endl;
      }
      void func()
      {
         std::cout << "non-cont" << std::endl;
      }
   };
};

// Helper class to select the type.
// Base version provides a non-const type.
template <typename T> struct type_selector
{
   using type = typename T::stB;
};

// Specialization provides a const type.
template <typename T> struct type_selector<const T>
{
   using type = const typename T::stB;
};


template <typename T>
void test()
{
   typename type_selector<T>::type b;
   b.func();
}

int main()
{
   test<stA>();
   test<const stA>();
}

Output:

non-cont
const

Answer 2

As alternative, using existing traits:

template <typename T>
void test()
{
   typename std::conditional<std::is_const<T>::value,
                             const typename T::stB,
                             typename T::stB>::type b;
   b.func();
   /*...*/
}

Demo