如何用几个较小的数组加上非数组元素构造一个std :: array?
即:
std::array<std::string, 2> strings = { "a", "b" };
std::array<std::string, 2> more_strings = { "c", "d" };
std::string another_string = "e";
std::array<std::string, 5> result = {strings, more_strings, another_string};
同样的问题适用于使用较小的std :: initializer_lists初始化std :: initializer_list。
答案 0 :(得分:1)
如果您可以使用,您可能会发现元组更容易(也更有效),直到最终创建数组。
从概念上讲,你可以这样写: using namespace std;
auto strings = make_tuple( "a"s, "b"s );
auto more_strings = make_tuple( "c"s, "d"s);
auto another_string = make_tuple("e"s);
auto result = to_array(tuple_cat(strings, more_strings, another_string));
当然,您需要为to_array实现一个非常简单的实现。完整的代码如下,在这个答案中,Luc Danton可以获得信用: Convert std::tuple to std::array C++11
完整代码供参考:
#include <iostream>
#include <string>
#include <tuple>
#include <array>
template<int... Indices>
struct indices {
using next = indices<Indices..., sizeof...(Indices)>;
};
template<int Size>
struct build_indices {
using type = typename build_indices<Size - 1>::type::next;
};
template<>
struct build_indices<0> {
using type = indices<>;
};
template<typename T>
using Bare = typename std::remove_cv<typename std::remove_reference<T>::type>::type;
template<typename Tuple>
constexpr
typename build_indices<std::tuple_size<Bare<Tuple>>::value>::type
make_indices()
{ return {}; }
template<typename Tuple, int... Indices>
std::array<
typename std::tuple_element<0, Bare<Tuple>>::type,
std::tuple_size<Bare<Tuple>>::value
>
to_array(Tuple&& tuple, indices<Indices...>)
{
using std::get;
return {{ get<Indices>(std::forward<Tuple>(tuple))... }};
}
template<typename Tuple>
auto to_array(Tuple&& tuple)
-> decltype( to_array(std::declval<Tuple>(), make_indices<Tuple>()) )
{
return to_array(std::forward<Tuple>(tuple), make_indices<Tuple>());
}
auto main() -> int
{
using namespace std;
auto strings = make_tuple( "a"s, "b"s );
auto more_strings = make_tuple( "c"s, "d"s);
auto another_string = make_tuple("e"s);
auto result = to_array(tuple_cat(strings, more_strings, another_string));
for (const auto& e : result)
{
cout << e << endl;
}
return 0;
}
答案 1 :(得分:0)
这是一个可以加入std::array
的函数。
template<typename T>
auto wrap_value(T&& value)
{
return std::tuple<T&&>(std::forward<T>(value));
}
template<typename T, std::size_t N>
std::array<T, N>& wrap_value(std::array<T, N>& value)
{
return value;
}
template<typename T, std::size_t N>
std::array<T, N> const& wrap_value(std::array<T, N> const& value)
{
return value;
}
template<typename T, std::size_t N>
std::array<T, N>&& wrap_value(std::array<T, N>&& value)
{
return std::move(value);
}
template<std::size_t... Is, typename... Ts>
std::array<std::common_type_t<Ts...>, sizeof...(Is)>
join_arrays_impl(std::index_sequence<Is...>, std::tuple<Ts...>&& parts)
{
return {std::get<Is>(std::move(parts))...};
}
template<typename... Ts>
auto join_arrays(Ts&&... parts)
{
auto wrapped_parts = std::tuple_cat((wrap_value)(std::forward<Ts>(parts))...);
constexpr auto size = std::tuple_size<decltype(wrapped_parts)>::value;
std::make_index_sequence<size> seq;
return (join_arrays_impl)(seq, std::move(wrapped_parts));
}
我认为你不能为std::initializer_list
实现类似的功能。 See it working here