我如何在不考慮類型順序的情況下比較等效類型的元組？

Question

我正在尋找一種比較兩個元組以查看它們是否包含相同類型的方法。
類型的順序無關緊要。 只要兩個元組的類型之間存在一對一的映射，我就會認為它們是等效的。

這是我設置的一個小測試。
我在實現equivalent_types()時遇到了麻煩：

#include <iostream>
#include <utility>
#include <tuple>
#include <functional>

template <typename T, typename U>
bool equivalent_types(T t, U u){
    return (std::tuple_size<T>::value == std::tuple_size<U>::value);
    //&& same types regardless of order
}


int main() {

    //these tuples have the same size and hold the same types.
    //regardless of the type order, I consider them equivalent.  
    std::tuple<int,float,char,std::string> a;
    std::tuple<std::string,char,int,float> b;

    std::cout << equivalent_types(a,b) << '\n'; //should be true
    std::cout << equivalent_types(b,a) << '\n'; //should be true

    //examples that do not work:  

    //missing a type (not enough types)
    std::tuple<std::string,char,int> c;

    //duplicate type (too many types)
    std::tuple<std::string,char,int,float,float> d;

    //wrong type
    std::tuple<bool,char,int,float> e;

    std::cout << equivalent_types(a,c) << '\n'; //should be false
    std::cout << equivalent_types(a,d) << '\n'; //should be false
    std::cout << equivalent_types(a,e) << '\n'; //should be false
}

Answer 1

通過計算兩個元組的類型，您可以執行以下操作：

template <typename T, typename Tuple>
struct type_counter;

template <typename T, typename ... Ts>
struct type_counter<T, std::tuple<Ts...>> :
    std::integral_constant<std::size_t, (... + std::is_same<T, Ts>::value)> {};

template <typename Tuple1, typename Tuple2, std::size_t... Is>
constexpr bool equivalent_types(const Tuple1&, const Tuple2&, std::index_sequence<Is...>)
{
    return (...
            && (type_counter<std::tuple_element_t<Is, Tuple1>, Tuple1>::value
               == type_counter<std::tuple_element_t<Is, Tuple1>, Tuple2>::value));
}

template <typename Tuple1, typename Tuple2>
constexpr bool equivalent_types(const Tuple1& t1, const Tuple2& t2)
{
    constexpr auto s1 = std::tuple_size<Tuple1>::value;
    constexpr auto s2 = std::tuple_size<Tuple2>::value;

    return s1 == s2
      && equivalent_types(t1, t2, std::make_index_sequence<std::min(s1, s2)>());
}

演示C ++ 17
演示C ++ 14

我使用c ++ 17來折疊表達式，但是可以很容易地將其重寫為constexpr函數。

Answer 2

使用Hana（與最新的Boost版本打包在一起），我們可以將每個元組類型轉換成一個映射，從類型到它們發生的次數，然后比較這些映射是否相等：

template <typename T, typename U>
bool equivalent_types(T t, U u) {
    namespace hana = boost::hana;
    auto f = [](auto m, auto&& e) {
        auto k = hana::decltype_(&e);
        return hana::insert(hana::erase_key(m, k),
            hana::make_pair(k, hana::find(m, k).value_or(0) + 1));
    };
    return hana::fold(t, hana::make_map(), f) == hana::fold(u, hana::make_map(), f);
}

例子

請注意，必須使用&e作為hana::decltype_的參數，以確保將例如int和int&視為不同類型（通過通用引用傳遞e的同上）。

Answer 3

該代碼似乎可以按任何順序與參數一起使用。 false結果是編譯器錯誤。 我對TMP不太滿意，但是它是100％的編譯時。. 我很喜歡關於如何清理它的一些建議。 直播： https ： //godbolt.org/g/3RZaMQ

#include <tuple>
#include <type_traits>
using namespace std;

// This struct removes the first instance of TypeToRemove from the Tuple or 'returns' void if it isn't present
template<class TypeToRemove, class ProcessedTupleParts, class RemainingTuple, class=void>
struct RemoveType;

template<class T, class... ProcessedTupleParts, class TupleHead, class... TupleTail>
struct RemoveType<T, std::tuple<ProcessedTupleParts...>, std::tuple<TupleHead, TupleTail...>, enable_if_t<std::is_same<T, TupleHead>::value>> {
    using RemovedType = std::tuple<ProcessedTupleParts..., TupleTail...>;
};

template<class T, class... ProcessedTupleParts, class TupleHead, class... TupleTail>
struct RemoveType<T, std::tuple<ProcessedTupleParts...>, std::tuple<TupleHead, TupleTail...>, enable_if_t<!std::is_same<T, TupleHead>::value>> {
    using RemovedType = typename RemoveType<T, std::tuple<ProcessedTupleParts..., TupleHead>, std::tuple<TupleTail...>>::RemovedType;
};

template<class T, class... Anything>
struct RemoveType<T, std::tuple<Anything...>, std::tuple<>> {
    using RemovedType = void;
};

template<class T1, class T2>
struct CompareTuples;

template<class T1Head, class... T1Tail, class T2>
struct CompareTuples<std::tuple<T1Head, T1Tail...>, T2> {
    using Result = typename CompareTuples<std::tuple<T1Tail...>, typename RemoveType<T1Head, std::tuple<>, T2>::RemovedType>::Result;
};

template<>
struct CompareTuples<std::tuple<>, std::tuple<>> {
    using Result = std::tuple<>;
};


template<class... T2Body>
struct CompareTuples<std::tuple<>, std::tuple<T2Body...>> {
    using Result = void;
};

template<class T1>
struct CompareTuples<T1, void> {
    using Result = void;
};



int main() {
    RemoveType<int, std::tuple<>,
    RemoveType<char, std::tuple<>, std::tuple<int, char>>::RemovedType>::RemovedType aa;

    CompareTuples<std::tuple<int>, std::tuple<int>>::Result a;
    CompareTuples<std::tuple<char, int>, std::tuple<int, char>>::Result b;
    CompareTuples<std::tuple<char, int>, std::tuple<int, char, double>>::Result e;
    CompareTuples<std::tuple<char, double, int>, std::tuple<int, char, double>>::Result f;
    CompareTuples<std::tuple<char, double, int>, std::tuple<int, char>>::Result g;
    CompareTuples<std::tuple<char>, std::tuple<int>>::Result c;
    CompareTuples<std::tuple<int>, std::tuple<int, char>>::Result d;

}