0

我用来std::tuple_cat将参数列表的子集选择到元组中，如下所示：

template <class...>
struct odds;

template <class T1>
struct odds<T1>
{
    typedef std::tuple<T1> type;
    static type value(T1&& t1)
    {
        return std::make_tuple(std::forward<T1>(t1));
    }
};

template <class T1, class T2>
struct odds<T1, T2>
{
    typedef std::tuple<T1> type;
    static type value(T1&& t1, T2&&)
    {
        return std::make_tuple(std::forward<T1>(t1));
    }
};

template <class T1, class T2, class... TTail> 
struct odds<T1, T2, TTail...>
{
        typedef decltype(std::tuple_cat(T1(), typename odds<TTail...>::type())) type; // L32
        static type value(T1&& t1, T2&&, TTail&&... rest)
    {
        return std::tuple_cat(std::forward<T1>(t1), odds<TTail...>::value(std::forward<TTail>(rest)...)); // L35
    }
};

，以下作为测试用例：

// assume <tuple>, <utility> are included at top of file
template <class... T>
auto foo(T... x) -> typename odds<T...>::type
{
         return odds<T...>::value(x...);
         //...
 }       
 int main() {
         auto bar = foo(5, true, 6, false); // L46
         auto baz = odds<int, bool, int, bool>::value(5, true, 6, false); // L47
         // bar, baz should be tuple<int,int> with value { 5, 6 }
 }

但是，在 clang-3.1 和 gcc-4.7.2 中都会出现模板推导问题：

叮当输出：

test.cc:32:19: error: no matching function for call to 'tuple_cat'
        typedef decltype(std::tuple_cat(T1(), typename odds<TTail...>::type())) type;
                         ^~~~~~~~~~~~~~
test.cc:40:30: note: in instantiation of template class 'odds<int, bool, int, bool>' requested here
auto foo(T... x) -> typename odds<T...>::type
                             ^
test.cc:40:6: note: while substituting deduced template arguments into function template 'foo' [with T = <int, bool, int, bool>]
auto foo(T... x) -> typename odds<T...>::type
     ^
/usr/include/c++/v1/tuple:1063:1: note: candidate template ignored: substitution failure [with _Tuple0 = int, _Tuples = <std::__1::tuple<int>>]
tuple_cat(_Tuple0&& __t0, _Tuples&&... __tpls)
^
/usr/include/c++/v1/tuple:987:1: note: candidate function not viable: requires 0 arguments, but 2 were provided
tuple_cat()
^
test.cc:46:13: error: no matching function for call to 'foo'
        auto bar = foo(5, true, 6, false);
                   ^~~
test.cc:40:6: note: candidate template ignored: substitution failure [with T = <int, bool, int, bool>]
auto foo(T... x) -> typename odds<T...>::type
     ^
test.cc:35:10: error: no matching function for call to 'tuple_cat'
                return std::tuple_cat(std::forward<T1>(t1), odds<TTail...>::value(std::forward<TTail>(rest)...));
                       ^~~~~~~~~~~~~~
test.cc:47:41: note: in instantiation of member function 'odds<int, bool, int, bool>::value' requested here
        auto baz = odds<int, bool, int, bool>::value(5,true,6,false);
                                               ^
/usr/include/c++/v1/tuple:1063:1: note: candidate template ignored: substitution failure [with _Tuple0 = int, _Tuples = <std::__1::tuple<int>>]
tuple_cat(_Tuple0&& __t0, _Tuples&&... __tpls)
^
/usr/include/c++/v1/tuple:987:1: note: candidate function not viable: requires 0 arguments, but 2 were provided
tuple_cat()
^
3 errors generated.

海合会输出：

test.cc: In instantiation of ‘struct odds<int, bool, int, bool>’:
test.cc:40:6:   required by substitution of ‘template<class ... T> typename odds<T ...>::type foo(T ...) [with T = {int, bool, int, bool}]’
test.cc:46:34:   required from here
test.cc:32:74: error: no matching function for call to ‘tuple_cat(int, odds<int, bool>::type)’
test.cc:32:74: note: candidate is:
In file included from test.cc:1:0:
/usr/lib/gcc/x86_64-pc-linux-gnu/4.7.2/include/g++-v4/tuple:1027:5: note: template<class ... _Tpls, class> constexpr typename std::__tuple_cat_result<_Tpls ...>::__type std::tuple_cat(_Tpls&& ...)
/usr/lib/gcc/x86_64-pc-linux-gnu/4.7.2/include/g++-v4/tuple:1027:5: note:   template argument deduction/substitution failed:
/usr/lib/gcc/x86_64-pc-linux-gnu/4.7.2/include/g++-v4/tuple:1024:31: error: no type named ‘type’ in ‘struct std::enable_if<false, void>’
test.cc: In function ‘int main()’:
test.cc:46:34: error: no matching function for call to ‘foo(int, bool, int, bool)’
test.cc:46:34: note: candidate is:
test.cc:40:6: note: template<class ... T> typename odds<T ...>::type foo(T ...)
test.cc:40:6: note:   substitution of deduced template arguments resulted in errors seen above
test.cc:46:34: error: unable to deduce ‘auto’ from ‘&lt;expression error>’
test.cc:47:13: error: ‘value’ is not a member of ‘odds<int, bool, int, bool>’
test.cc:47:61: error: unable to deduce ‘auto’ from ‘&lt;expression error>’

Gcc 在这里更有帮助，尤其是在出现错误的情况下

    test.cc:32:74: error: no matching function for call to ‘tuple_cat(int, odds<int, bool>::type)’

目标是调用一个函数，该函数递归地解包参数，将选择收集到一个收集元组中，然后返回它。为了以扁平的方式累积它，我使用std::tuple_cat()扁平递归尾元组，添加一个头，然后返回元组。转发用于在递归期间不丢弃引用限定符。

稍后在代码中，结果元组被解包以调用不同的可变参数函数，但这超出了此错误的范围。

显然，我在某处省略了一些微妙但至关重要的细节，但我发现在追查潜在问题方面存在很大困难。

4

2 回答 2

2

to 的参数tuple_cat必须是元组（或支持std::tuple_size和std::tuple_elementAPI 的“类似元组”的东西，例如std::pairor std::array），但从错误消息来看，您似乎使用非元组类型调用它。为此，您可能需要以下内容：

std::tuple_cat(std::make_tuple(a_non_tuple), a_tuple, another_tuple);

这会将第一个参数转换为 a tuple<decltype(a_non_tuple)>，因此它可以与其他元组连接。

连接std::tuple<T1, std::tuple<T2, std::tuple<T3, ...>>>成std::tuple<T1, T2, T3, ...>

这不是“串联”。我认为正确的术语是“扁平化”。

我不确定您在尝试什么，因为您没有提供完整的示例。你打电话foo的论点是这样的std::tuple<T1, std::tuple<T2, std::tuple<T3, ...>>>吗？

那是行不通的，因为foo将推断T为std::tuple<T1, std::tuple<T2, std::tuple<T3, ...>>>并实例化特化odds<T1>，它只是将参数包装到foo另一个元组中！

于 2012-12-15T23:48:45.467 回答
1

我不知道这对您是否感兴趣，因为它是一种替代实现，而不是您问题的答案。我认为使用tuple_cat这样的方法最终会在运行时进行大量移动/复制，尽管编译器可能会为琐碎的类型避免这些。无论如何，这似乎是一个有趣的 TMP 问题。

这就是我想出的（这很可能过于冗长）；它肯定可以使用一些清理（以及我拥有的 clang 版本——几周前的主干——在尝试编译它时断言）。

有趣的递归位是 in Skip，它使用前一个元组作为模式（计数，真的）来构造下一个；最后，您可以看到 Odds 如何使用 SkipHelper 传递两个参数模式来设置初始调用。（那部分肯定需要一些工作，也许使用数字会更容易）：

namespace skip_args {
namespace detail {

// We use Tuple as a kind of generic typelist.
template<typename ...T>
using Tuple = std::tuple<T...>;

// Utility "functions"
// Pusher<T, Tuple<U...> >::type = Tuple<T, U...>
template<typename A, typename B> struct Pusher;
template<typename T, typename ...U> struct Pusher<T, Tuple<U...>> {
  using type = Tuple<T, U...>;
};
template<typename A, typename B>
using push = typename Pusher<A, B>::type;

// Skip is an intermediate used to skip over ignored elements.
// To allow recursion, we declare the general form first.
// All three arguments are Tuples.
template<typename Next, typename Rest, typename This> struct Skip;

// Node is actually used to store some data item. It also inherits from the next
// following Node (if there is one) so that we end up with an inheritance chain.
// (That's the part similar to libstdc++ tuples; it makes the layout the same,
// too, as long as there's no EBO to deal with, because we don't bother here.)
template<typename Rest, typename ...This> struct Node;

template<typename ...R_, typename T, typename ...T_>
struct Node<Tuple<R_...>, T, T_...>
: Skip<Tuple<>, Tuple<R_...>, Tuple<T, T_...>>::type {
  using self = Node<Tuple<R_...>, T, T_...>;
  using next = typename Skip<Tuple<>, Tuple<R_...>, Tuple<T, T_...>>::type;
  // Recursive construction of node types
  using nodes = push<self, typename next::nodes>;
  using value_type = T;
  // The constructor takes all the arguments, uses the first one, skips some,
  // and passes the rest to the next node.
  constexpr Node<Tuple<R_...>, T, T_...>(T&& t, T_&&...t_, R_&&...r_)
      : next(std::forward<R_>(r_)...), value(std::forward<T>(t)) {
  }
  T&& value;
};

// Base:
template<>
struct Node<Tuple<>> {
  using nodes = Tuple<>;
  constexpr Node<Tuple<>>() {}
};

// Skip (N...) (R R...) (T T...) => Skip (N... R) (R...) (T...)
// In other words, it drops elements from the third tuple, and for each one it
// moves an element from the second tuple to the first tuple. If it runs out of
// the third tuple, it "returns" a new Node. If it runs out of the second tuple,
// then we're done, but to satisfy the node requirements, it actually needs to
// declare a constructor (which drops all its arguments)

// General case:
template<typename ...N_, typename R, typename ...R_, typename T, typename ...T_>
struct Skip<Tuple<N_...>, Tuple<R, R_...>, Tuple<T, T_...>>
  : Skip<Tuple<N_..., R>, Tuple<R_...>, Tuple<T_...>> {
};
// Ran out of pattern
template<typename ...N_, typename R_, typename T_>
struct Skip<Tuple<N_...>, R_, T_> {
  using type = Node<R_, N_...>;
  using nodes = typename type::nodes;
};

template<typename T> struct TupleTyper;
template<typename ...T> struct TupleTyper<Tuple<T...>> {
  using type = Tuple<typename T::value_type...>;
};

template<typename A, typename B, typename C> struct TupleMaker;
template<typename Tup, typename H, typename ...T>
struct TupleMaker<Tup, H, Tuple<T...>> {
  Tup operator()(H&& helper) {
    return Tup(static_cast<T&>(helper).value...);
  }
};

template<typename N> struct SkipHelper {
  using tuple_type = typename TupleTyper<typename N::nodes>::type;
  template<typename ...U>
  tuple_type operator()(U&& ...u) {
    return TupleMaker<tuple_type, N, typename N::nodes>()(N(std::forward<U>(u)...));
  }
};

} // namespace detail

template<typename ...T> struct Odds;

template<typename T1, typename T2, typename ...T_>
struct Odds<T1, T2, T_...>
  : detail::SkipHelper<detail::Node<detail::Tuple<T_...>, T1, T2>> {
};
template<typename T1>
struct Odds<T1> : detail::SkipHelper<detail::Node<detail::Tuple<>, T1>> {
};

// tuple_from_odds takes any number of arguments,
// and returns a tuple of the odd numbered ones.
template<typename...T> auto tuple_from_odds(T&&...t)
    -> typename Odds<T...>::tuple_type {
  return Odds<T...>()(std::forward<T>(t)...);
}

} // namespace skip_args
于 2012-12-16T19:11:08.913 回答