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

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 
#include 
#include 
#include 

template 
  struct is_iterator {
  static char test(...);

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

  constexpr static bool value = std::is_same())),long>::value;
};

struct Foo {};

//Returns true
bool f() { return is_iterator::iterator>::value; }
//Returns true    
bool fc() { return is_iterator::const_iterator>::value; }
//Returns true
bool fr() { return is_iterator::reverse_iterator>::value; }
//Returns true
bool fcr() { return is_iterator::const_reverse_iterator>::value; }
//Returns true
bool g() { return is_iterator::value; }
//Returns true
bool gc() { return is_iterator::value; }
//Returns false
bool h() { return is_iterator::value; }
//Returns false
bool i() { return is_iterator::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 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 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 for some type T and does not provide all of the required typedefs.