Instantiate different specialization with the same parameters

Question

Consider the class below. I would like to use sometimes the first specialization and sometimes the second, but with the same template arguments.

#include <vector>

using namespace std;

// 1
template < typename U, typename V >
class C
{
public: C() {}
};

// 2
template < typename U, typename VT, typename VA >
class C< U, vector<VT, VA> >
{
public: C() {}
};

// 3
template < typename UT, typename UA, typename VT, typename VA >
class C< vector<UT, UA>, vector<VT, VA> >
{
public: C() {}
};

int main() 
{
  C<int, int> a; // 1st
  C<int, vector<int>> b; // 2nd
  C<vector<int>, vector<int>> c; // 3rd
  C<vector<int>, vector<int>> d; // how do I force the instantiation of 2nd?
}

Answer 1

You could define a holder type that is convertible to the held instance

template <typename T>
struct whole
{
    operator T() const { return t; }
    T t;
};

and then change your last line to

 C<whole<vector<int>>, vector<int>> d;

https://godbolt.org/z/479x9qo6v

Answer 2

As already discussed in the comments, if you wish to create different types there must be something different in your definition. It can be either the template arguments, or in a way you can achieve that with different constructor arguments, relying on deduction guides to lead to a specific type (but then without providing the template arguments).

A possible way to achieve what you aim for might be with a factory method that gets your arguments and returns an object of the desired type.