Hiding C++ template parameter packs in a tuple

Raymond Chen

Raymond

C++11 introduced variadic templates and template parameter packs.

template<typename... Args>
struct S;

Passing template parameter packs around is a bit of a hassle, because the dots “soak up” parameters, which make them hard to pass to other templated types. You also can’t “save” parameter packs in another type:

template<typename... Args>
struct S
{
    // doesn't compile: cannot create a type that
    // is itself a parameter pack
    using TheArgs = Args...;
};

One workaround for this is to capture the types into a std::tuple.

template<typename... Args>
struct S
{
    using Tuple = std::tuple<Args...>;
};

You can then pass the Tuple around, and if you need to extract the types, you can pull them out of the tuple.

My first thought about how to extract the types was to use std::get:

// code in italics is wrong
template<typename... Args>
struct Traits
{
    using Tuple = std::tuple<Args...>;
    static constexpr auto Size = sizeof...(Args);
    template <std::size_t N>
    using Nth = decltype(std::get<N>(std::declval<Tuple>()));
    using First = Nth<0>;
    using Last  = Nth<Size - 1>;
};

This doesn’t work because std::get returns a reference:

void f()
{
 whatis<Traits<int, double>::First>();
}

    call    void whatis<int&&>()

This behavior is presumably so you can modify the tuple:

std::tuple<int> tuple;
std::get<0>(tuple) = 2;

Fortunately, there’s another way to extract the type from a tuple: std::tuple_element.

template<typename... Args>
struct Traits
{
    using Tuple = std::tuple<Args...>;
    static constexpr auto Size = sizeof...(Args);
    template <std::size_t N>
    using Nth = typename std::tuple_element<N, Tuple>::type;
    using First = Nth<0>;
    using Last  = Nth<Size - 1>;
};

This provides a simpler way to extract the last type from a template parameter pack than writing some horrible recursive template metaprogram, which is what some people did instead of hiding the pack inside a tuple.

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