How to check if a template parameter is an iterator type or not?

Question

template<class T>
struct is_iterator
{
    static const bool value = ??? // What to write ???
};

int main()
{
    assert(false == is_iterator<int>::value);
    assert(true == is_iterator<vector<int>::iterator>::value);
    assert(true == is_iterator<list<int>::iterator>::value);
    assert(true == is_iterator<string::iterator>::value);
    assert(true == is_iterator<char*>::value); // a raw pointer is also an iterator
}

The question is: How to make the five assert statements pass?

Answer 1

Coming in here a few years later, where C++11 and C++14 make it a lot easier to do such things. An iterator is, at its core, something that is dereferencable, incrementable. If it's an input iterator, then also comparable. Let's go with the latter - since that looks like what you want.

The simplest version would be to use void_t:

template <typename... >
using void_t = void;

Base case:

template <typename T, typename = void>
struct is_input_iterator : std::false_type { };

Valid case specialization:

template <typename T>
struct is_input_iterator<T,
    void_t<decltype(++std::declval<T&>()),                       // incrementable,
           decltype(*std::declval<T&>()),                        // dereferencable,
           decltype(std::declval<T&>() == std::declval<T&>())>>  // comparable
    : std::true_type { };

Alias:

template <typename T>
using is_input_iterator_t = typename is_input_iterator<T>::type;

No need to rely on iterator_category or using the tedious C++03 style of check things using overload resolution. Expression SFINAE is where it's at.

---

As Mr. Wakely points out in the comments, [iterator.traits] requires that:

it is required that if Iterator is the type of an iterator, the types

iterator_traits<Iterator>::difference_type
iterator_traits<Iterator>::value_type
iterator_traits<Iterator>::iterator_category

be defined as the iterator’s difference type, value type and iterator category, respectively.

So we can define our iterator trait to simply check for that:

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

template <class T>
struct is_iterator<T, void_t<
    typename std::iterator_traits<T>::iterator_category
>> : std::true_type { };

If iterator_traits<T>::iterator_category is ill-formed, then T is not an iterator.