How to check if an arbitrary type is an iterator?

Question

This is an exact duplicate of this question, except that accepted answer is wrong, so I ask it again:

How do you correctly check to see if

Answer 1

How about something like this?

template<typename T, typename = void>
struct is_iterator
{
   static constexpr bool value = false;
};

template<typename T>
struct is_iterator<T, typename std::enable_if<!std::is_same<typename std::iterator_traits<T>::value_type, void>::value>::type>
{
   static constexpr bool value = true;
};

example:

#include <iostream>
#include <type_traits>
#include <vector>

template<typename T, typename = void>
struct is_iterator
{
   static constexpr bool value = false;
};

template<typename T>
struct is_iterator<T, typename std::enable_if<!std::is_same<typename std::iterator_traits<T>::value_type, void>::value>::type>
{
   static constexpr bool value = true;
};

int main()
{
   static_assert(!is_iterator<int>::value, "ass");
   static_assert(is_iterator<int*>::value, "ass");
   static_assert(is_iterator<std::vector<int>::iterator>::value, "ass");
}

http://liveworkspace.org/code/7dcf96c97fd0b7a69f12658fc7b2693e

Answer 2

I implemented this one some time ago:

template <typename T>
struct is_iterator {  
    template <typename U>
    static char test(typename std::iterator_traits<U>::pointer* x);

    template <typename U>
    static long test(U* x);

    static const bool value = sizeof(test<T>(nullptr)) == 1;
};

It compiles fine using your example. I can't test it on VC though.

Demo here.

Answer 3

I believe this should be a complete solution. Try it on http://gcc.godbolt.org and see the resulting assembly for the test functions.

#include <type_traits>
#include <iterator>
#include <vector>
#include <utility>

template <typename T>
  struct is_iterator {
  static char test(...);

  template <typename U,
    typename=typename std::iterator_traits<U>::difference_type,
    typename=typename std::iterator_traits<U>::pointer,
    typename=typename std::iterator_traits<U>::reference,
    typename=typename std::iterator_traits<U>::value_type,
    typename=typename std::iterator_traits<U>::iterator_category
  > static long test(U&&);

  constexpr static bool value = std::is_same<decltype(test(std::declval<T>())),long>::value;
};

struct Foo {};

//Returns true
bool f() { return is_iterator<typename std::vector<int>::iterator>::value; }
//Returns true    
bool fc() { return is_iterator<typename std::vector<int>::const_iterator>::value; }
//Returns true
bool fr() { return is_iterator<typename std::vector<int>::reverse_iterator>::value; }
//Returns true
bool fcr() { return is_iterator<typename std::vector<int>::const_reverse_iterator>::value; }
//Returns true
bool g() { return is_iterator<int*>::value; }
//Returns true
bool gc() { return is_iterator<const int*>::value; }
//Returns false
bool h() { return is_iterator<int>::value; }
//Returns false
bool i() { return is_iterator<Foo>::value; }

This implementation uses SFINAE and overloading precedence. test(U&&) always has higher precedence than test(...) so it will always be chosen if not removed by SFINAE.

For an iterator type T, std::iterator_traits<T> has all of the above mentioned typedefs present so test(U&&) and test(...) are both overload candidates. Since test(U&&) has higher precedence, its always chosen.

For a non-iterator type T, test(U&&) fails SFINAE because std::iterator_traits<T> does not have the nested typedefs. Therefore the only remaining candidate is test(...).

Note that this trait will also fail if someone specializes std::iterator_traits<T> for some type T and does not provide all of the required typedefs.