Constructing templated tuple types

Question

I am trying to write a function like this

template<
        bool b, RT = std::conditional_t<b,
               std::tuple<int, int, int, int>,
               std::tuple<int, int, int, int, double, double, double, double>
        >
RT function()
{
    int i1, i2, i3, i4;

    if constexpr(b)
    {
        double i5, i6, i7, i8;
        return { i1, i2, i3, i4, i5, i6, i7, i8 };
    }
    else
    {
        return { i1, i2, i3, i4 };
    }
}

Is there a way to create a templated typedef for the tuple so that I can simplify the above function

template<typename T, int N>
using tuple_t = std::tuple<T, T, ... N1 times>

template<typename T1, int N1, typename T2, int N2>
using tuple_t = std::tuple<T1, T1, ... N1 times, T2, T2, ... N2 times>

Answer 1

You can use return type deduction and replace the aggregate initialization with a call to make_tuple:

template<bool b>
auto function()
{
    int i1, i2, i3, i4;

    if constexpr(b)
    {
        double i5, i6, i7, i8;
        return std::make_tuple(i1, i2, i3, i4, i5, i6, i7, i8);
    }
    else
    {
        return std::make_tuple(i1, i2, i3, i4);
    }
}

If you still need the return type you can simply make a trait:

template <bool b>
using return_t = decltype(function<b>());