c++ - 如何从 std::tuple 中删除元素？

Question

给定一个

struct A{}; struct B{}; struct C{};
std::tuple<A,B,C> tabc;

如何删除第二个元素 B从中得到tuple<A,C> ，比如

std::tuple<A,C> tac = drop<B>(tabc);

或相同

std::tuple<A,C> tac = drop<1>(tabc);

我假设这会产生一个带有元素拷贝的新类型。

Answer 1

这相当简单，可以使用元编程技术进行编码:

template<size_t drop, size_t ...ixs>
constexpr auto calc_drop_sequence_dropper(std::index_sequence<ixs...>)
{
    return std::index_sequence<(ixs >= drop ? ixs + 1 : ixs)...>{};
}

//Creates a monotonically increasing sequence on the range [0, `count`), except
//that `drop` will not appear.
template<size_t count, size_t drop>
constexpr auto calc_drop_copy_sequence()
{
    static_assert(count > 0, "You cannot pass an empty sequence.");
    static_assert(drop < count, "The drop index must be less than the count.");
    constexpr auto count_sequence = std::make_index_sequence<count - 1>();
    return calc_drop_sequence_dropper<drop>(count_sequence);
}

template<typename Tuple, size_t ...ixs>
constexpr auto copy_move_tuple_by_sequence(Tuple &&tpl, std::index_sequence<ixs...>)
{
    using TplType = std::remove_reference_t<Tuple>;

    return std::tuple<std::tuple_element_t<ixs, TplType>...>(
        std::get<ixs>(std::forward<Tuple>(tpl))...);
}

template<size_t drop, typename Tuple>
constexpr auto drop_tuple_element(Tuple &&tpl)
{
    using TplType = std::remove_reference_t<Tuple>;

    constexpr size_t tpl_size = std::tuple_size<TplType>::value;

    constexpr auto copy_seq = calc_drop_copy_sequence<tpl_size, drop>();

    return copy_move_tuple_by_sequence(std::forward<Tuple>(tpl), copy_seq);
}

主要函数是 drop_tuple_element，它执行您假设的 drop 函数的操作。当然，如果您要删除多个元素，您希望一次性删除它们，而不是单独删除它们。因此您需要修改代码。