Nested C++ template parameters for functions

Question

I want to have a templated function in C++, where one template parameter is itself a template of another template parameter. If that doesn\'t make any sense, take the follow

Answer 1

You're better off not doing that at all; consider just templating on the container

template <typename C>
void print_container(const C& container)
{

    for(auto v: container)
        std::cout << v << " ";
    std::cout << std::endl;
}

If you need the stored type in the function, you can use: `typedef typename C::value_type T;

Answer 2

I am not sure that I understood what you want but you can try this:

template <typename V>
void print_vector(V &vec)
{
    for(auto v: vec)
        std::cout << v << " ";
    std::cout << std::endl;
}
...
std::vector<double> vec(5);
...
print_vector(vec);

The point here is that usually you don't need construct like template < template V< typename T> > because whole template template V< typename T> can be generalized to type V.

Answer 3

std::vector has two parameters, type and allocator. Try this

template <typename T, typename Alloc, template <typename, typename> class V>
void print_container(V<T, Alloc> &con)
{
}

print_container(vec);

This will work for vector, list, etc., but will not work with map, set.

However, since you use auto you can use C++11 and then you can to this:

template <typename T, template <typename, typename...> class V, typename... Args>
void print_container(V<T, Args...> &con)

or

template <template <typename, typename...> class V, typename... Args>
void print_container(V<Args...> &con)

and of course most simple way is to do something like

template<typename C>
void print_container(C& con)

probably with some checks for deduce, that C is really container.

template<typename C>
auto print_container(C& con) -> decltype(con.begin(), void())