Element-wise tuple addition

Question

I have some values held in a tuple, and I am looking to add another tuple to it element-wise. So I would like functionality like this:

std::tuple         


        

Answer 1

The solution by @lubgr is satisfactory for your specific use-case. I offer you a generic solution which will work for tuples of different types, as well tuples of different (equal) sizes.

#include <tuple>
#include <utility>
#include <iostream>

template<typename... T1, typename... T2, std::size_t... I>
constexpr auto add(const std::tuple<T1...>& t1, const std::tuple<T2...>& t2, 
                   std::index_sequence<I...>)
{
    return std::tuple{ std::get<I>(t1) + std::get<I>(t2)... };
}

template<typename... T1, typename... T2>
constexpr auto operator+(const std::tuple<T1...>& t1, const std::tuple<T2...>& t2)
{
    // make sure both tuples have the same size
    static_assert(sizeof...(T1) == sizeof...(T2));

    return add(t1, t2, std::make_index_sequence<sizeof...(T1)>{});
}

I used a few C++17 features (mainly template-related) without which the code would become a bit more complicated. A possible improvement would be to make use of move-semantics.

Obviously, this applies for the first "possible syntax" you provided.

Answer 2

You could also consider using std::valarray since it allows exactly the things that you seem to want.

#include <valarray>

int main()
{
    std::valarray<int> a{ 1, 2 }, b{ 2, 4 }, c;
    c = a - b; // c is {-1,-2}
    a += b; // a is {3,6}
    a -= b; // a is {1,2} again
    a += {2, 4}; // a is {3,6} again
    return 0;
}

Answer 3

Here is an operator definition for syntax #1:

template <class S, class T> std::tuple<S, T> operator + (const std::tuple<S, T>& lhs, const std::tuple<S, T>& rhs)
{
   return std::make_tuple(std::get<0>(lhs) + std::get<0>(rhs), std::get<1>(lhs) + std::get<1>(rhs));
}

Syntax #2 and #3 are not possible without creating a custom struct, because they can only be defined as members of the classes they operate on (and you can't touch existing classes in namespace std).

Answer 4

You could use something like this, which supports all three of your syntax proposals:

#include <tuple>
#include <utility>

namespace internal
{
    //see: https://stackoverflow.com/a/16387374/4181011
    template<typename T, size_t... Is>
    void add_rhs_to_lhs(T& t1, const T& t2, std::integer_sequence<size_t, Is...>)
    {
        auto l = { (std::get<Is>(t1) += std::get<Is>(t2), 0)... };
        (void)l; // prevent unused warning
    }
}

template <typename...T>
std::tuple<T...>& operator += (std::tuple<T...>& lhs, const std::tuple<T...>& rhs)
{
    internal::add_rhs_to_lhs(lhs, rhs, std::index_sequence_for<T...>{});
    return lhs;
}

template <typename...T>
std::tuple<T...> operator + (std::tuple<T...> lhs, const std::tuple<T...>& rhs)
{
   return lhs += rhs;
}

Working example:

http://coliru.stacked-crooked.com/a/27b8cf370d44d3d5

http://coliru.stacked-crooked.com/a/ff24dae1c336b937

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I would still go with named structs in most cases. Tuples are seldom the correct choice.