Passing std::array as arguments of template variadic function

Question

I am trying to learn about variadic templates in C++11. I have a class which is basically a wrapper around a std::array. I want to be able to pass function obje

Answer 1

You can use this utility template to create a sequence of indices at compile time:

template< std::size_t... Ns >
struct indices {
    typedef indices< Ns..., sizeof...( Ns ) > next;
};

template< std::size_t N >
struct make_indices {
    typedef typename make_indices< N - 1 >::type::next type;
};

template<>
struct make_indices< 0 > {
    typedef indices<> type;
};

Then make a caller function that takes indices as a parameter so you got a way of deducing the indices sequence:

template<typename... Ts, size_t...Is>
void call(std::function<void(Ts...)>&& func, indices<Is...>)
{
    func( data[Is]... );
}

And then you can call it like this:

template< typename... Ts>
void doOperation( std::function<void(Ts...)>&& func )
{
    static_assert(sizeof...(Ts)==N,"Size of variadic template args does not match array length");
    call( std::forward<std::function<void(Ts...)>>(func), typename  make_indices<N>::type() );
}

Answer 2

Given the well-known indices infrastructure:

namespace detail
{
    template<int... Is>
    struct seq { };

    template<int N, int... Is>
    struct gen_seq : gen_seq<N - 1, N - 1, Is...> { };

    template<int... Is>
    struct gen_seq<0, Is...> : seq<Is...> { };
}

You could redefine your class template this way:

template<int N, typename T>
struct Container {
    template<typename... Ts>
    Container(Ts&&... vs) : data{{std::forward<Ts>(vs)...}} {
        static_assert(sizeof...(Ts)==N,"Not enough args supplied!");
    }

    template<typename F>
    void doOperation(F&& func)
    {
        doOperation(std::forward<F>(func), detail::gen_seq<N>());
    }

    template<typename F, int... Is>
    void doOperation(F&& func, detail::seq<Is...>)
    {
        (std::forward<F>(func))(data[Is]...);
    }

    std::array<T,N> data;
};

Here is a live example.

Notice, that you do not need to construct an std::function object in main(): the std::function can be implicitly constructed from the lambda. However, you do not even need to use std::function at all here, possibly incurring in an unnecessary run-time overhead.

In the solution above, I just let the type of the callable object to be a template parameter that can be deduced by the compiler.