条件类型 - 特征通用参考的问题

时间:2015-02-24 23:39:25

标签: c++ templates c++11 universal-reference forwarding-reference

所以我有一个函数用于检查值并在值无效时抛出异常,否则在收到值时将值传回。我试图用普遍的崇敬和类型特征来概括这个例程。我觉得我很接近,因为我的例子在某些情况下起作用,但不是全部。它似乎只适用于右值。

#include <iostream>
#include <utility>
#include <type_traits>
#include <string>
#include <vector>

using namespace std;

struct error_type{};

template <class T>
class has_empty
{
   template<class U, class = typename std::enable_if<std::is_member_pointer<decltype(&U::empty)>::value>::type>
      static std::true_type check(int);
   template <class>
      static std::false_type check(...);
public:
   static constexpr bool value = decltype(check<T>(0))::value;
};

template <bool invalid>
struct valid {};


template <>
struct valid<false>
{
   template<typename U, typename E>
   static inline U&& check(U&& toCheck, const E& toThrow)
   {
      if (!toCheck)
        std::cout << "NoEmpty Throw" << '\n';
      else
        std::cout << "NoEmpty valid" << '\n';
      return std::forward<U>(toCheck);
   }
};

template <>
struct valid<true>
{
   template<typename U, typename E>
   static inline U&& check(U&& toCheck, const E& toThrow)
   {
      if (toCheck.empty())
        std::cout << "HasEmpty Throw" << '\n';
      else
        std::cout << "HasEmpty valid" << '\n';
      return std::forward<U>(toCheck);
   }
};

template<typename T
   , typename E
   , typename  = typename std::enable_if<std::is_base_of<error_type, E>::value>::type>
   inline T&& do_check(T&& toCheck, const E& toThrow)
{
   return valid<has_empty<T>::value>::check(std::forward<T>(toCheck), toThrow);
}

struct HasEmpty
{
    bool empty() {return false;}
};

struct NoEmpty
{
};


int main()
{
    error_type e;

    cout << has_empty<std::wstring>::value << '\n';
    cout << has_empty<std::vector<std::wstring>>::value << '\n';
    cout << has_empty<int>::value << '\n';
    cout << has_empty<HasEmpty>::value << '\n';
    cout << has_empty<NoEmpty>::value << '\n';

    do_check(true, e);
    do_check(false, e);

    do_check(std::string("45"), e);
    do_check(HasEmpty(), e);
    do_check(std::vector<bool>(), e);
    HasEmpty he;
    do_check(std::move(he), e);
    //do_check(he, e); // does not work, has_empty<T>::value is 0
}

生成输出

1
1
0
1
0
NoEmpty valid
NoEmpty Throw
HasEmpty valid
HasEmpty valid
HasEmpty Throw
HasEmpty valid

如果我取消注释最后一行,我会收到以下错误:

prog.cpp: In instantiation of 'static T&& valid<false, T, E>::check(T&&, const E&) [with T = HasEmpty&; E = error_type]':
prog.cpp:56:84:   required from 'T&& do_check(T&&, const E&) [with T = HasEmpty&; E = error_type; <template-parameter-1-3> = void]'
prog.cpp:87:19:   required from here
prog.cpp:30:11: error: no match for 'operator!' (operand type is 'HasEmpty')
       if (!toCheck)
           ^
prog.cpp:30:11: note: candidate is:
prog.cpp:30:11: note: operator!(bool) <built-in>
prog.cpp:30:11: note:   no known conversion for argument 1 from 'HasEmpty' to 'bool'

has_empty<T>::value似乎正在评估false。我相信我可以做一个不同的工作来使这个工作,所以在这一点上它是一种学术。不过,任何帮助都会受到赞赏。

1 个答案:

答案 0 :(得分:3)

在您的示例中将左值传递给do_check deduces T as HasEmpty&时。当然,引用类型没有名为empty的成员函数,表达式decltype(&U::empty)格式错误,导致static std::true_type check(int)重载导致SFINAE出现。

如果您更换

static constexpr bool value = decltype(check<T>(0))::value;

static constexpr bool value = decltype(check<typename std::remove_reference<T>::type>(0))::value;

因此U永远不是引用类型,您的代码按预期工作。

Live demo