get part of std::tuple

Question

I have a tuple of unknown size (it's template parametr of method)

Is it way to get part of it (I need throw away first element of it)

For example, I have tuple<int,int,int>(7,12,42). I want tuple<int,int>(12,42) here

Answer 1

I made some modifications to Adam's code that would strip off the first N arguments of the tuple, as well as create a new tuple with only the last N types ... Here is the complete code (note: if anyone decides to +1 my answer, also please +1 Adam's answer since that is what this code is based on, and I don't wish to take any credit away from his contribution):

//create a struct that allows us to create a new tupe-type with the first
//N types truncated from the front

template<size_t N, typename Tuple_Type>
struct tuple_trunc {};

template<size_t N, typename Head, typename... Tail>
struct tuple_trunc<N, std::tuple<Head, Tail...>>
{
    typedef typename tuple_trunc<N-1, std::tuple<Tail...>>::type type;
};

template<typename Head, typename... Tail>
struct tuple_trunc<0, std::tuple<Head, Tail...>>
{
    typedef std::tuple<Head, Tail...> type;
};

/*-------Begin Adam's Code-----------

Note the code has been slightly modified ... I didn't see the need for the extra
variadic templates in the "assign" structure.  Hopefully this doesn't break something
I didn't forsee

*/

template<size_t N, size_t I>
struct assign 
{
    template<class ResultTuple, class SrcTuple>
    static void x(ResultTuple& t, const SrcTuple& tup) 
    {
        std::get<I - N>(t) = std::get<I>(tup);  
        assign<N, I - 1>::x(t, tup);  //this offsets the assignment index by N
    }
};

template<size_t N>
struct assign<N, 1> 
{
    template<class ResultTuple, class SrcTuple>
    static void x(ResultTuple& t, const SrcTuple& tup) 
    {
        std::get<0>(t) = std::get<1>(tup);
    }
};


template<size_t TruncSize, class Tup> struct th2;

//modifications to this class change "type" to the new truncated tuple type
//as well as modifying the template arguments to assign

template<size_t TruncSize, class Head, class... Tail>
struct th2<TruncSize, std::tuple<Head, Tail...>> 
{
    typedef typename tuple_trunc<TruncSize, std::tuple<Head, Tail...>>::type type;

    static type tail(const std::tuple<Head, Tail...>& tup) 
    {
        type t;
        assign<TruncSize, std::tuple_size<type>::value>::x(t, tup);
        return t;
    }
};

template<size_t TruncSize, class Tup>
typename th2<TruncSize, Tup>::type tail(const Tup& tup) 
{
    return th2<TruncSize, Tup>::tail(tup);
}

//a small example
int main()
{
    std::tuple<double, double, int, double> test(1, 2, 3, 4);
    tuple_trunc<2, std::tuple<double, double, int, double>>::type c = tail<2>(test);
    return 0;
}

Answer 2

With C++17, you can use std::apply:

template <typename Head, typename... Tail>
std::tuple<Tail...> tuple_tail(const std::tuple<Head, Tail...>& t)
{
    return apply([](auto head, auto... tail) {
        return std::make_tuple(tail...)};
    }, t);
}

Answer 3

Please don't use!

• This is [most probably] unspecified behaviour. It could stop working at any time.
• Furthermore, there is a possibility of padding issues (i.e. it may work for int but may fail for your type!).

See comments for discussion. I'm leaving this answer just for reference.

---

Even simpler:

tuple<int,int,int> origin{7,12,42};
tuple<int, int> &tail1 = (tuple<int, int>&)origin;
tuple<int> &tail2 = (tuple<int>&)origin;
cout << "tail1: {" << get<0>(tail1) << ", " << get<1>(tail1) << "}" << endl;
cout << "tail2: {" << get<0>(tail2) << "}" << endl;

I got:

tail1: {12, 42}
tail2: {42}

I'm not certain that this is not an unspecified behaviour. Works for me: Fedora 20 and

❯ clang --version
clang version 3.3 (tags/RELEASE_33/final)
Target: x86_64-redhat-linux-gnu
Thread model: posix
❯ gcc --version
gcc (GCC) 4.8.2 20131212 (Red Hat 4.8.2-7)
Copyright (C) 2013 Free Software Foundation, Inc.
This is free software; see the source for copying conditions.  There is NO
warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.

References: article on voidnish.wordpress.com/.

Answer 4

There may be an easier way, but this is a start. The "tail" function template returns a copied tuple with all values of the original except the first. This compiles with GCC 4.6.2 in C++0x-mode.

template<size_t I>
struct assign {
  template<class ResultTuple, class SrcTuple>
  static void x(ResultTuple& t, const SrcTuple& tup) {
    std::get<I - 1>(t) = std::get<I>(tup);
    assign<I - 1>::x(t, tup);
  }
};

template<>
struct assign<1> {
  template<class ResultTuple, class SrcTuple>
  static void x(ResultTuple& t, const SrcTuple& tup) {
    std::get<0>(t) = std::get<1>(tup);
  }
};


template<class Tup> struct tail_helper;

template<class Head, class... Tail>
struct tail_helper<std::tuple<Head, Tail...>> {
  typedef typename std::tuple<Tail...> type;
  static type tail(const std::tuple<Head, Tail...>& tup) {
    type t;
    assign<std::tuple_size<type>::value>::x(t, tup);
    return t;
  }
};

template<class Tup>
typename tail_helper<Tup>::type tail(const Tup& tup) {
  return tail_helper<Tup>::tail(tup);
}

Answer 5

With help of a compile-time integer list:

#include <cstdlib>

template <size_t... n>
struct ct_integers_list {
    template <size_t m>
    struct push_back
    {
        typedef ct_integers_list<n..., m> type;
    };
};

template <size_t max>
struct ct_iota_1
{
    typedef typename ct_iota_1<max-1>::type::template push_back<max>::type type;
};

template <>
struct ct_iota_1<0>
{
    typedef ct_integers_list<> type;
};

We could construct the tail simply by parameter-pack expansion:

#include <tuple>

template <size_t... indices, typename Tuple>
auto tuple_subset(const Tuple& tpl, ct_integers_list<indices...>)
    -> decltype(std::make_tuple(std::get<indices>(tpl)...))
{
    return std::make_tuple(std::get<indices>(tpl)...);
    // this means:
    //   make_tuple(get<indices[0]>(tpl), get<indices[1]>(tpl), ...)
}

template <typename Head, typename... Tail>
std::tuple<Tail...> tuple_tail(const std::tuple<Head, Tail...>& tpl)
{
    return tuple_subset(tpl, typename ct_iota_1<sizeof...(Tail)>::type());
    // this means:
    //   tuple_subset<1, 2, 3, ..., sizeof...(Tail)-1>(tpl, ..)
}

Usage:

#include <cstdio>

int main()
{
    auto a = std::make_tuple(1, "hello", 7.9);
    auto b = tuple_tail(a);

    const char* s = nullptr;
    double d = 0.0;
    std::tie(s, d) = b;
    printf("%s %g\n", s, d);
    // prints:   hello 7.9

    return 0;
}

(On ideone: http://ideone.com/Tzv7v; the code works in g++ 4.5 to 4.7 and clang++ 3.0)

Answer 6

A tuple slice operation (that also works for std::array and std::pair) can be defined like this (C++14 required):

namespace detail
{
    template <std::size_t Ofst, class Tuple, std::size_t... I>
    constexpr auto slice_impl(Tuple&& t, std::index_sequence<I...>)
    {
        return std::forward_as_tuple(
            std::get<I + Ofst>(std::forward<Tuple>(t))...);
    }
}

template <std::size_t I1, std::size_t I2, class Cont>
constexpr auto tuple_slice(Cont&& t)
{
    static_assert(I2 >= I1, "invalid slice");
    static_assert(std::tuple_size<std::decay_t<Cont>>::value >= I2, 
        "slice index out of bounds");

    return detail::slice_impl<I1>(std::forward<Cont>(t),
        std::make_index_sequence<I2 - I1>{});
}

And an arbitrary subset of a tuple t can be obtained like so :

tuple_slice<I1, I2>(t); 

Where [I1, I2) is the exclusive range of the subset and the return value is a tuple of either references or values depending on whether the input tuple is an lvalue or an rvalue respectively (a thorough elaboration can be found in my blog).