c++ - 无法解释不明确的模板特化

Question

给定

template <typename...> struct Pack;
using T1 = std::tuple<int, char, double>;
using T2 = std::tuple<bool, double, int, char>;

TupleTree<Pack, T1, T2>应该是

Pack<
    Pack<int, bool>, Pack<int, double>, Pack<int, int>, Pack<int, char>,
    Pack<char, bool>, Pack<char, double>, Pack<char, int>, Pack<char, char>,
    Pack<double, bool>, Pack<double, double>, Pack<double, int>, Pack<double, char>
>

这扩展到任意数量的元组。定义足够容易理解吗？好的，我的程序运行正常:

http://ideone.com/OumeZK

但是现在我想把定义扩展到

`TupleTreeWithRepeats<P, std::index_sequence<Is...>, Tuples...>`

哪里Is...将指示每个元组被重复使用的次数，直到移动到下一个元组。注意 <Is...> = <1,1,...,1>将减少到与 TupleTree<P, Tuples...> 相同.我遇到的歧义在于这两个专业:

TupleTreeWithRepeatsHelper<P, std::index_sequence<Is...>, LoopNumber, P<Ts...>, First, Rest...>
TupleTreeWithRepeatsHelper<P, std::index_sequence<I, Is...>, I, P<Ts...>, First, Rest...>

由于某种原因，P<Ts...> 的存在导致歧义，因为当我用单命名类型替换它时，歧义就被消除了。即使我更换 std::index_sequence<Is...>与 std::index_sequence<I, Is...> ，歧义仍然存在。这是怎么回事？以及如何解决这个问题？这是代码，与 TupleTree 几乎相同:

#include <iostream>
#include <tuple>
#include <type_traits>

template <typename T> struct Identity { using type = T; };

// Merging packs of types.
template <typename...> struct MergePacks;

template <typename Pack>
struct MergePacks<Pack> : Identity<Pack> {};

template <template <typename...> class P, typename... Types1, typename... Types2, typename... Packs>
struct MergePacks<P<Types1...>, P<Types2...>, Packs...> : MergePacks<P<Types1..., Types2...>, Packs...> {};

// Appending a type to a pack.
template <typename Pack, typename T> struct AppendType;

template <template <typename...> class P, typename... Ts, typename T>
struct AppendType <P<Ts...>, T> {
    using type = P<Ts..., T>;
};

// ExpandPackWithTuple takes a pack, and creates N packs that each end with the tuple's elements, N is the size of the tuple.
template <template <typename...> class P, typename Pack, typename Tuple, typename Indices> struct ExpandPackWithTupleHelper;

template <template <typename...> class P, typename Pack, typename Tuple, std::size_t... Is>
struct ExpandPackWithTupleHelper<P, Pack, Tuple, std::index_sequence<Is...>> {
    using type = P<typename AppendType<Pack, typename std::tuple_element<Is, Tuple>::type>::type...>;
};

template <template <typename...> class P, typename Pack, typename Tuple>
using ExpandPackWithTuple = typename ExpandPackWithTupleHelper<P, Pack, Tuple, std::make_index_sequence<std::tuple_size<Tuple>::value>>::type;

// TupleTreeWithRepeats.
template <template <typename...> class P, typename NumRepeats, std::size_t LoopNumber, typename OutputPack, typename... Tuples> struct TupleTreeWithRepeatsHelper;

template <template <typename...> class P, std::size_t I, std::size_t... Is, std::size_t LoopNumber, typename... Ts, typename First, typename... Rest>
struct TupleTreeWithRepeatsHelper<P, std::index_sequence<I, Is...>, LoopNumber, P<Ts...>, First, Rest...> :
    TupleTreeWithRepeatsHelper<P, std::index_sequence<I, Is...>, LoopNumber + 1, typename MergePacks<ExpandPackWithTuple<P, Ts, First>...>::type, First, Rest...> {};

template <template <typename...> class P, std::size_t I, std::size_t... Is, typename... Ts, typename First, typename... Rest>
struct TupleTreeWithRepeatsHelper<P, std::index_sequence<I, Is...>, I, P<Ts...>, First, Rest...> :
    TupleTreeWithRepeatsHelper<P, std::index_sequence<Is...>, 0, typename MergePacks<ExpandPackWithTuple<P, Ts, First>...>::type, Rest...> {};

template <template <typename...> class P, std::size_t... Is, std::size_t LoopNumber, typename OutputPack>
struct TupleTreeWithRepeatsHelper<P, std::index_sequence<Is...>, LoopNumber, OutputPack> {
    using type = OutputPack;
};

template <template <typename...> class P, typename NumRepeats, typename... Tuples> struct TupleTreeWithRepeats;

template <template <typename...> class P, std::size_t I, std::size_t... Is, typename... Tuples>
struct TupleTreeWithRepeats<P, std::index_sequence<I, Is...>, Tuples...> : TupleTreeWithRepeatsHelper<P, std::index_sequence<Is...>, 0, P<P<>>, Tuples...> {};

// Testing
template <typename...> struct Pack;
using T1 = std::tuple<int, char, double>;
using T2 = std::tuple<bool, double, int, char>;
using T3 = std::tuple<double, int>;

int main() {
    std::cout << std::is_same<
        TupleTreeWithRepeats<Pack, std::index_sequence<1,1,1>, T1, T2, T3>::type,
        Pack<
            Pack<int, bool, double>, Pack<int, bool, int>, Pack<int, double, double>, Pack<int, double, int>, Pack<int, int, double>, Pack<int, int, int>, Pack<int, char, double>, Pack<int, char, int>,
            Pack<char, bool, double>, Pack<char, bool, int>, Pack<char, double, double>, Pack<char, double, int>, Pack<char, int, double>, Pack<char, int, int>, Pack<char, char, double>, Pack<char, char, int>,
            Pack<double, bool, double>, Pack<double, bool, int>, Pack<double, double, double>, Pack<double, double, int>, Pack<double, int, double>, Pack<double, int, int>, Pack<double, char, double>, Pack<double, char, int>
        >
    >::value << '\n';  // ambiguous
}

Answer 1

据我所知，这两个应该没有歧义。

无论如何，一个简单的解决方法是根据“无重复”来实现“有重复”:

// TupleTreeWithRepeats.
template <template <typename...> class P, typename Tuples>
struct TupleTreeWithRepeatsHelper;

template <template <typename...> class P, typename... Tuples>
struct TupleTreeWithRepeatsHelper<P, Pack<Tuples...>> :
    Identity<TupleTree<P, Tuples...>> {};

template <template <typename...> class P, typename NumRepeats, typename... Tuples> 
struct TupleTreeWithRepeats;

template <template <typename...> class P, std::size_t... Is, typename... Tuples>
struct TupleTreeWithRepeats<P, std::index_sequence<Is...>, Tuples...> :
    TupleTreeWithRepeatsHelper<P, typename MergePacks<repeat<Tuples, Is>...>::type> {};

repeat 在哪里

template<class T, std::size_t> using id = T;

template<class T, std::size_t... Is>
Pack<id<T, Is>...> do_repeat(std::index_sequence<Is...>);

template<class T, std::size_t I>
using repeat = decltype(do_repeat<T>(std::make_index_sequence<I>()));

Demo .有人可能会争辩说这也是一种更好的设计。

---

最小化为:

template<class> class Purr { };

template <template <class> class, class>
struct Meow;

template <template <class> class P>
struct Meow<P, P<int>> { };

template <template <class> class P, class T>
struct Meow<P, P<T>> { };

Meow<Purr, Purr<int>> c;

GCC 报告了一个歧义，这对我来说绝对像是一个错误。 Clang 正确地处理了这个问题。