如何正确传播通用引用的类型？

Question

我的代码库中有一个情况，我必须实现 std::get() 的通用形式，它适用于任何类型的类似元组的类型。该函数接受对 Tuple 的通用引用和 return 对 Tuple 的第 I 个元素的引用。我不知道如何命名引用的类型。不幸的是，我无法使用 auto return 类型，只能让编译器自行解决。

这是我的第一次尝试：

#include <type_traits>
#include <tuple>

template<class T, class U>
struct propagate_reference
{
  using type = U;
};

template<class T, class U>
struct propagate_reference<T&,U>
{
  using type = typename std::add_lvalue_reference<U>::type;
};

template<class T, class U>
struct propagate_reference<T&&,U>
{
  using type = typename std::add_rvalue_reference<U>::type;
};

template<size_t I, class TupleReference>
struct get_result
{
  using tuple_type = typename std::decay<TupleReference>::type;

  using type = typename propagate_reference<
    TupleReference,
    typename std::tuple_element<I,tuple_type>::type
  >::type;
};

template<size_t I, class Tuple>
typename get_result<I,Tuple&&>::type my_get(Tuple&& t)
{
  return std::get<I>(std::forward<Tuple>(t));
}

int foo(const std::tuple<int>& t)
{
  return my_get<0>(t);
}

int main()
{
  return 0;
}

Clang 拒绝此程序：

$ clang -std=c++11 test_get.cpp 
test_get.cpp:36:10: error: binding of reference to type 'int' to a value of type 'const __tuple_element_t<0UL, tuple<int> >' (aka 'const int') drops qualifiers
  return std::get<I>(std::forward<Tuple>(t));
         ^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
test_get.cpp:41:10: note: in instantiation of function template specialization 'my_get<0, const std::tuple<int> &>' requested here
  return my_get<0>(t);
         ^
1 error generated.

我怀疑问题出在我实例化 get_result 的方式上。我做错了什么？

Answer 1

问题是 std::decay removes cv-qualifiers, and defines the resulting type as the member typedef type. As already TC mentioned in the comments what you need here is std::remove_reference:

template<size_t I, class TupleReference>
struct get_result
{
  using tuple_type = typename std::remove_reference<TupleReference>::type;

  using type = typename propagate_reference<
    TupleReference,
    typename std::tuple_element<I,tuple_type>::type
  >::type;
};

Live Demo

如何正确传播通用引用的类型？

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