How to define is_iterator type trait?

问题:

• I'm trying to code a is_iterator<T> type trait. Where when T is an iterator type is_iterator<T>::value == true otherwise is is_iterator<T>::value == false.

• What I tried so far:

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template <class, class Enable = void> 
struct is_iterator : std::false_type {};

template <typename T> 
struct is_iterator<T, typename std::enable_if<std::is_pointer<typename
     std::iterator_traits<T>::pointer>::value>::type> : std::true_type {};

LIVE DEMO

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Q: Is there a more proper way to define a is_iterator type trait than the one displayed above?

回答1:

As I said in comments, the solutions presented here rely on non-portable properties of iterators_traits in some implementations. According to the C++03 and C++11 standards, iterator_traits is only defined for iterators (and the special case of pointers) so any other use is undefined behaviour. Specifically, using iterator_traits<T>::pointer in a SFINAE context isn't guaranteed to work, because instantiating iterator_traits<T> will refer to T::value_type, T::pointer, T::iterator_category etc. and that happens outside the "immediate context" where SFINAE doesn't apply.

C++14 will fix that was supposed to fix that (it happened post-C++14 with DR 2408), but for C++11 the safe way to define is_iterator is to write a trait that checks for all the required operations an iterator must define. The only operations that all iterators are required to support are operator* and pre- and post-increment. Unfortunately, there can be types that define those operations which are not valid iterators, so writing a correct trait is quite hard.

回答2:

Your check fails if std::iterator_traits<T>::pointer is a type that is not a pointer, for instance if T = std::ostream_iterator<U>.

I think a better test might be whether std::iterator_traits<T>::iterator_category is either std::input_iterator_tag or a derived type, or std::output_iterator_tag.

template <class, class Enable = void> struct is_iterator : std::false_type {};
template <typename T> 
struct is_iterator
<T, 
 typename std::enable_if<
    std::is_base_of<std::input_iterator_tag, typename std::iterator_traits<T>::iterator_category>::value ||
    std::is_same<std::output_iterator_tag, typename std::iterator_traits<T>::iterator_category>::value 
 >::type> 
 : std::true_type {};

回答3:

I think there's no need to check for any particular property of iterator_traits's nested typedefs. It should be enough to check for mere presence of iterator_traits<T>::value_type (or any other nested typedef, for that matter), because every iterator has one.

#include <type_traits>
#include <iostream>
#include <iterator>

template<typename>
struct void_ {
  typedef void type;
};
// remove typename spam below:
template<typename Discard>
using void_t=typename void_<Discard>::type;
template<typename T>
using decay_t=typename std::decay<T>::type;    

// stick helper types into details, so the interface
// for is_iterator is cleaner:
namespace details {
  template<typename T, typename Enable=void>
  sturct is_iterator : is_iterator2<T, Enable> {};

  // special case: void* is not an iterator
  // but T* specialization would pick it up
  // if there weren't the following:

  template<typename V>
  struct is_iterator<V*, decay_t<V>> : std::false_type {};

  // phase 2: SFINAE pass to std::iterator_traits test
  // valid in C++14, and in many C++11 compilers, except
  // for above void issue:
  template<typename, typename Enable = void>
  struct is_iterator2 : std::false_type {};

  template<typename T>
  struct is_iterator2<T, 
    void_t< typename std::iterator_traits<T>::value_type>
  > : std::true_type {};
}
template<typename T>
struct is_iterator : details::is_iterator<T> {};

int main()
{
    std::cout
        << is_iterator<int*>::value
        << is_iterator<double>::value;
}

Unfortunately this isn't guaranteed to work in C++11, but C++14 will fix it (thanks to Jonathan Wakely for pointing it out).