Using a template structure as a namespace

Question

I just finished an iteration on a library that I am writing. In order to isolate certain types, I originally put them into a namespace. Then, I realised that this namespace should actually be templated, so I turned it into a struct, ie:

namespace A
{
    typedef _foo foo;
}

became

template <class T>
struct A
{
    typedef _foo<T> foo;
};

This was more convenient to use in say, this situation:

template <class T>
class MyClass
{
public:
    typedef A<T>               nsA;
    typedef typename nsA::foo  foo_type;
};

Now, I'm using that library in another library, and there, I know that the template type T will never change. So I would like to do something like:

namespace B
{
    using namespace A<double>;
}

but obviously, this doesn't work. Of course, I could just stop being lazy, and typedef manually each definition in A<double>, but I like being lazy when I can. Any thoughts on how to do something equivalent to this last step?

Answer 1

Actually, I just found the answer. Sorry, if you think this question was useless, I'll delete it, just let me know in comment.

Since I used a struct as a namespace in the first place, the solution is to keep using a struct as a namespace (!):

struct B
    : public A<double>
{};

Here is a demo.