Selecting container type at compile time

Question

I want to select one of the standard containers with one template parameter at compile time. Something like

    template<typename T>
    void foo()
    {
        using Container = std::conditional_t< std::is_same_v<T, int>, 
                std::vector, // T is int
                std::set>;    // any other T
        Container<T> bar;
    }

How to do this properly?

Answer 1

The simplest way seems to be

using Container = std::conditional_t< std::is_same_v<T, int>, 
            std::vector<T>, // T is int
            std::set<T>>;
Container bar;

std::conditional_t allows you to select a type. There is no standard conditional template that would allow to select a template. You can write one if you want but this won't be convenient to use.

template <bool t, template <typename...> typename X, template <typename...> typename Y>
struct conditional_template;

template <template <typename...> typename X, template <typename...> typename Y>
struct conditional_template<true, X, Y>
{
    template <typename... ts> using type = X<ts...>;
};

template <template <typename...> typename X, template <typename...> typename Y>
struct conditional_template<false, X, Y>
{
    template <typename... ts> using type = Y<ts...>;
};

template <typename... ts>
using Container = conditional_template<true, std::list, std::vector>::type<ts...>;

Container<int> x;

It doesn't seem possible to define an analogue of std::conditional_t convenience alias.