How to declare/define a class with template template parameters without using an extra template parameter

Question

Consider the following use of template template parameters...

#include <iostream>

template <typename X>
class A
{
    X _t;
public:
    A(X t)
        :_t(t)
    {
    }
    X GetValue()
    {
        return _t;
    }
};

template <typename T, template <typename T> class C >
class B
{
    C<T> _c;
public:
    B(T t)
        :_c(t)
    {
    }
    T GetValue()
    {
        return _c.GetValue();
    }
};

using namespace std;

int main()
{
    B<int, A> b(10);
    cout<<b.GetValue();
    return 0;
}

Is there a way by which the template parameter T can be removed? For example is there a way to make the following work?

//Does not compile
template <template <typename T> class C >
class B
{
    C _c;
public:
    B(T t)
        :_c(t)
    {
    }
    T GetValue()
    {
        return _c.GetValue();
    }
};

int main()
{
    B< A<int> > b(10);
    cout<<b.GetValue();
    return 0;
}

Answer 1

What is wrong with:

template <typename C >
struct B
{
    C c;
};

int main()
{
    B< A<int> > b;
    return 0;
}

Answer 2

This is not possible. Note that this is a common misunderstanding: A<int> is not a class template anymore! So it would not fit to a template-template parameter, but would have to be accepted using a type-parameter:

template<typename C>
struct B {
    C c;
};

B< A<int> > b;

Your way of using a separate parameter is alright.

If you want to accept A<int> but want to re-bind it to another parameter, you can use this pattern, also used by standard-allocators:

template<typename T>
struct A {
    template<typename U>
    struct rebind {
        typedef A<U> type;
    };
};

template<typename C>
struct B {
    typename C::template rebind<float>::type c;
};

B< A<int> > b;

Now, B< A<int> >::c is of type A<float>. The typename before C:: tells the compiler the ::type at the end is a type and not a static non-type member. The template after C:: tells the compiler the rebind<float> is a template instantiation, and not a comparision.

Answer 3

I assume you're after X, as well as A, in your code.

The usual pattern is to have

template<typename C>
struct B
{
   C c;
};

and then, inside classes eligible for substitution:

template<typename X>
class A
{
   typedef X type_name;
   X t;
};

Then you can access the template parameter using C::type_name.

Answer 4

You can nest parameters. That is, the value of a parameter can itself be parameterized.

template <typename X>
struct A
{
  X t;
};

template <typename  C>
struct B
{
  C c;
};

int main()
{
  B< A<int> > b;
  return 0;
}

In this example, the declaration of b in main() creates a specialization of A using int as the parameter, then it creates a specialization of B using A<int> as the parameter. Thus, in the specialization of B, C is A<int>.