在std :: vector <std :: tuple <Ts ... >>中查看所有T的视图

Question

Imagine I have a container which is similar to std::vector<std::tuple<Ts...>> .

From that, I would like to get a " view " (non-copying) to all T 's such that I can operate on that view as if it would be a standard-like container.

So what I would like to have:

using tuple_vector = std::vector<std::tuple<int,float,double>>;
tuple_vector tuple_vec = {{1, 4.f, 8.},
                          {2, 5.f, 9.},
                          {3, 6.f, 10.},
                          {4, 7.f, 11.}}

auto int_view = view<int>(tuple_vec);
^^^
type should be some kind of non-owning reference 

// what I would like to do
int_view[0] = 10; // modify
assert(int_view[0] == std::get<int>(tuple_vec[0])); // modification should modify tuple_vec as well

I have tried std::transform , but then I am getting a owning copy of all int 's.

std::vector<int> int_vec(tuple_vec.size());
std::transform(tuple_vec.begin(), tuple_vec.end(), int_vec.begin(), [&](const auto& elem) {
    return std::get<int>(elem);
});

I am not sure if this is possible at all, but if so, I would appreciate any hints or directions.

Answer 1

Well, if you using Eric Niebler's ranges-v3 library (which is in the process of getting adopted into the standard - there's now a Ranges TS ), you can apply your intuition to use something like std::transform , but with a view:

#include <range/v3/view/transform.hpp>

// ... etc. ...

auto int_view = tuple_vector | ranges::view::transform(
    [](auto& t)->auto&{ return std::get<int>(t);} );

See this in action (Coliru), with modification of one of the elements even.

Note: This view becomes un-modifiable if we remove the -> decltype(auto) ; the change is due to @deduplicator's answer to this question .

Answer 2

You could create a vector of std::reference_wrapper :

template <typename T, typename ContainerOfTuples>
auto make_refs_to(ContainerOfTuples& tuples) {
    using RefType = std::reference_wrapper<T>;

    std::vector<RefType> refs;
    refs.reserve(std::size(tuples));
    std::transform(std::begin(tuples), std::end(tuples), std::back_inserter(refs),
                   [](auto& tup) -> RefType { return {std::get<T>(tup)}; });
    return refs;
}

auto double_view = make_refs_to<double>(tuple_vec);
double_view[1].get() += 3.14;  // Caveat: must access through .get().

Live Example

---

Going further... here's how we can get references to many types:

namespace detail {

// When many types are asked for, return a tuple of references.
template <typename... T> struct RefTypeImpl {
    using type = std::tuple<std::reference_wrapper<T>...>;
};

// When a single type is asked for, return a single reference.
template <typename T> struct RefTypeImpl<T> {
    using type = std::reference_wrapper<T>;
};

// When two types are asked for, return a pair for more convenient access.
template <typename T, typename U> struct RefTypeImpl<T, U> {
    using type = std::pair<std::reference_wrapper<T>, std::reference_wrapper<U>>;
};

}  // namespace detail

template <typename... Ts, typename ContainerOfTuples>
auto make_refs_to(ContainerOfTuples& tuples) {
    using RefType = typename detail::RefTypeImpl<Ts...>::type;

    std::vector<RefType> refs;
    refs.reserve(std::size(tuples));
    std::transform(std::begin(tuples), std::end(tuples), std::back_inserter(refs),
                   [](auto& tup) -> RefType { return {std::get<Ts>(tup)...}; });
    return refs;
}

auto int_float_view = make_refs_to<int, float>(tuple_vec);
std::cout << (int_float_view[2].first.get() == 3) << '\n';

Live Example