2つの可変個引数テンプレートがあるとします。typename... T, typename... U、どうやって彼らを見つけようか。
- 連結
- 最大の共通サブシーケンス
- 最大の共通サブシーケンスの逆
したがって、私が理解していることから、連結は単純です。(t..., u...)、しかし、2つの最大の共通サブシーケンスを見つけるのはどうですか？-それも可能ですか？
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12
これは、タプルタイプのペアに対する集合演算を計算するソリューションです。タプルを使用して可変引数パックを保持できると思います。そのため、とを取得したらTs...、Us...次のようにします。
typename tuple_intersect<std::tuple<Ts...>, std::tuple<Us...>>::type
そして、これによりタプルが得られます。ここで、はとのVs...交点です。引数パックとして再度抽出する必要がある場合は、 ：を受け入れる関数の入力としてタプルを指定するだけです。Ts...Us...Vs...tuple<Ts...>
template<typename... Vs>
void func(std::tuple<Vs...>)
{
// Here, you have Vs... (= Us... & Ts...) as an argument pack
}
フレームワーク：
以下に、以下のすべての主要なメタ関数に共通するいくつかの単純なメタ関数を示します。
template<typename T, typename... Ts>
struct is_member_of_type_seq { static const bool value = false; };
template<typename T, typename U, typename... Ts>
struct is_member_of_type_seq<T, U, Ts...>
{
static const bool value = std::conditional<
std::is_same<T, U>::value,
std::true_type,
is_member_of_type_seq<T, Ts...>
>::type::value;
};
template<typename, typename>
struct append_to_type_seq { };
template<typename T, typename... Ts>
struct append_to_type_seq<T, std::tuple<Ts...>>
{
using type = std::tuple<Ts..., T>;
};
template<typename, typename>
struct prepend_to_type_seq { };
template<typename T, typename... Ts>
struct prepend_to_type_seq<T, std::tuple<Ts...>>
{
using type = std::tuple<T, Ts...>;
};
1- 連結
これは非常に簡単です。
template<typename, typename>
struct concat_type_seq { };
template<typename... Ts, typename... Us>
struct concat_type_seq<std::tuple<Ts...>, std::tuple<Us...>>
{
using type = std::tuple<Ts..., Us...>;
};
そしていくつかのテスト：
static_assert(
std::is_same<
concat_type_seq<
std::tuple<char, int, bool>,
std::tuple<double, double, int>
>::type,
std::tuple<char, int, bool, double, double, int>
>::value,
"Error"
);
2-最長共通部分列
これは少し複雑です：
namespace detail
{
// Meta-function that returns, given two sequences S1 and S2, the longest
// subsequence of S1 in S2 that starts with the first element of S1 and
// begins at the first element of S2 (in other words, it returns the
// subsequence S2[0]..S2[N] such that S1[i] = S2[i] for each 0 <= i <= N.
template<typename, typename>
struct match_seq_in_seq_from_start
{
using type = std::tuple<>;
};
template<typename T, typename U, typename... Ts, typename... Us>
struct match_seq_in_seq_from_start<std::tuple<T, Ts...>, std::tuple<U, Us...>>
{
using type = typename std::conditional<
std::is_same<T, U>::value,
typename prepend_to_type_seq<
T,
typename match_seq_in_seq_from_start<
std::tuple<Ts...>,
std::tuple<Us...>
>::type
>::type,
std::tuple<>
>::type;
};
// Some testing...
static_assert(
std::is_same<
match_seq_in_seq_from_start<
std::tuple<int, double, char>,
std::tuple<int, double, long>
// ^^^^^^^^^^^
>::type,
std::tuple<int, double>
>::value,
"Error!"
);
// Meta-function that returns the same as the meta-function above,
// but starting from the first element of S2 which is identical to
// the first element of S1.
template<typename, typename>
struct match_first_seq_in_seq
{
using type = std::tuple<>;
};
template<typename T, typename U, typename... Ts, typename... Us>
struct match_first_seq_in_seq<std::tuple<T, Ts...>, std::tuple<U, Us...>>
{
using type = typename std::conditional<
std::is_same<T, U>::value,
typename match_seq_in_seq_from_start<
std::tuple<T, Ts...>,
std::tuple<U, Us...>
>::type,
typename match_first_seq_in_seq<
std::tuple<T, Ts...>,
std::tuple<Us...>
>::type
>::type;
};
// Some testing...
static_assert(
std::is_same<
match_first_seq_in_seq<
std::tuple<int, double, char>,
std::tuple<bool, char, int, double, long, int, double, char>
// ^^^^^^^^^^^
>::type,
std::tuple<int, double>
>::value,
"Error!"
);
// Meta-function that returns, given two sequences S1 and S2, the longest
// subsequence of S1 in S2 that starts with the first element of S1.
template<typename T, typename U>
struct match_seq_in_seq
{
using type = std::tuple<>;
};
template<typename U, typename... Ts, typename... Us>
struct match_seq_in_seq<std::tuple<Ts...>, std::tuple<U, Us...>>
{
using type1 = typename match_first_seq_in_seq<
std::tuple<Ts...>,
std::tuple<U, Us...>
>::type;
using type2 = typename match_seq_in_seq<
std::tuple<Ts...>,
std::tuple<Us...>
>::type;
using type = typename std::conditional<
(std::tuple_size<type1>::value > std::tuple_size<type2>::value),
type1,
type2
>::type;
};
// Some testing...
static_assert(
std::is_same<
match_seq_in_seq<
std::tuple<int, double, char>,
std::tuple<char, int, double, long, int, double, char>
// ^^^^^^^^^^^^^^^^^
>::type,
std::tuple<int, double, char>
>::value,
"Error!"
);
}
// Meta-function that returns, given two sequences S1 and S2, the longest
// subsequence of S1 in S2 (longest common subsequence).
template<typename T, typename U>
struct max_common_subseq
{
using type = std::tuple<>;
};
template<typename T, typename... Ts, typename... Us>
struct max_common_subseq<std::tuple<T, Ts...>, std::tuple<Us...>>
{
using type1 = typename detail::match_seq_in_seq<
std::tuple<T, Ts...>,
std::tuple<Us...>
>::type;
using type2 = typename max_common_subseq<
std::tuple<Ts...>,
std::tuple<Us...>
>::type;
using type = typename std::conditional<
(std::tuple_size<type1>::value > std::tuple_size<type2>::value),
type1,
type2
>::type;
};
そしていくつかのテスト：
// Some testing...
static_assert(
std::is_same<
max_common_subseq<
std::tuple<int, double, char>,
std::tuple<char, int, char, double, char, long, int, bool, double>
>::type,
std::tuple<double, char>
>::value,
"Error!"
);
// Some more testing...
static_assert(
std::is_same<
max_common_subseq<
std::tuple<int, double, char, long, long, bool>,
// ^^^^^^^^^^^^^^^^
std::tuple<char, long, long, double, double, char>
// ^^^^^^^^^^^^^^^^
>::type,
std::tuple<char, long, long>
>::value,
"Error!"
);
3-反転
タイプシーケンスを反転するための特性は次のとおりです（反転されたタイプリストを持つタプルを返します）。
template<typename... Ts>
struct revert_type_seq
{
using type = std::tuple<>;
};
template<typename T, typename... Ts>
struct revert_type_seq<T, Ts...>
{
using type = typename append_to_type_seq<
T,
typename revert_type_seq<Ts...>::type
>::type;
};
そしていくつかのテスト：
// Some testing...
static_assert(
std::is_same<
revert_type_seq<char, int, bool>::type,
std::tuple<bool, int, char>
>::value,
"Error"
);
4-交差点
これはリクエストされていませんが、ボーナスとして提供されています。
template<typename, typename>
struct intersect_type_seq
{
using type = std::tuple<>;
};
template<typename T, typename... Ts, typename... Us>
struct intersect_type_seq<std::tuple<T, Ts...>, std::tuple<Us...>>
{
using type = typename std::conditional<
!is_member_of_type_seq<T, Us...>::value,
typename intersect_type_seq<
std::tuple<Ts...>,
std::tuple<Us...>>
::type,
typename prepend_to_type_seq<
T,
typename intersect_type_seq<
std::tuple<Ts...>,
std::tuple<Us...>
>::type
>::type
>::type;
};
そしていくつかのテスト：
// Some testing...
static_assert(
std::is_same<
intersect_type_seq<
std::tuple<char, int, bool, double>,
std::tuple<bool, long, double, float>
>::type,
std::tuple<bool, double>
>::value,
"Error!"
);
于 2013-03-07T15:47:07.613 に答える