检查模板参数是否是std :: vector <T> ::迭代

Question

How to check if a template argument is a std::vector<T>::iterator ?

For void type, we have std::is_void. Is there something like that for std::vector<T>::iterator ?

Answer 1

You could create a trait for that:

#include <vector>
#include <list>
#include <type_traits>


template <class T, class = void>
struct is_vector_iterator: std::is_same<T, std::vector<bool>::iterator> { };

template <class T>
struct is_vector_iterator<T, decltype(*std::declval<T>(), std::enable_if_t<!std::is_same<T, std::vector<bool>::iterator>::value>())>: std::is_same<T, typename std::vector<std::decay_t<decltype(*std::declval<T>())>>::iterator> { };

int main() {
    static_assert(is_vector_iterator<std::vector<int>::iterator>::value, "Is not a vector iterator");
    static_assert(is_vector_iterator<std::vector<bool>::iterator>::value, "Is not a vector iterator");
    static_assert(!is_vector_iterator<std::list<int>::iterator>::value, "Is a vector iterator");
    static_assert(!is_vector_iterator<std::list<int>::iterator>::value, "Is a vector iterator");
}

[live demo]

Answer 2

Better using std::iterator_traits I think:

#include <list>
#include <vector>
#include <iterator>

template <class It, class = void>
struct is_vector_iterator : std::false_type { };

template <class It>
struct is_vector_iterator<It, std::enable_if_t<
                                  std::is_same<
                                      It,
                                      typename std::vector< 
                                          typename std::iterator_traits<It>::value_type
                                      >::iterator
                                  >::value
                         >> : std::true_type { };

int main() {
    static_assert(is_vector_iterator<std::vector<int>::iterator>::value, "Is not a vector iterator");
    static_assert(is_vector_iterator<std::vector<bool>::iterator>::value, "Is not a vector iterator");
    static_assert(!is_vector_iterator<std::list<int>::iterator>::value, "Is a vector iterator");
}

DEMO

Answer 3

An alternative solution also using std::iterator_traits :

#include <iostream>
#include <vector>
#include <list>

template <typename T>
struct is_vector_iterator 
{
    typedef char yes[1];
    typedef char no[2];

    template <typename C>
    static yes& test(
        typename std::enable_if<
            std::is_same<T, typename std::vector<typename C::value_type>::iterator>::value
        >::type*);

    template <typename>
    static no& test(...);

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

};

int main() {
    std::cout << is_vector_iterator<int>::value << std::endl;
    std::cout << is_vector_iterator<int*>::value << std::endl;
    std::cout << is_vector_iterator<std::list<int>::iterator>::value << std::endl;
    std::cout << is_vector_iterator<std::vector<int>::iterator>::value << std::endl;
    return 0;
}

live demo

Answer 4

You can write a trait for this:

namespace detail
{

template<typename T> constexpr std::false_type is_vector_iterator(T&&, ...)
{
    return {};
}

template<typename T>
constexpr auto is_vector_iterator(T&& t, void* = nullptr) ->
decltype(std::is_same<typename std::vector<std::decay_t<decltype(*t)>>::iterator, std::decay_t<T>>{})
{
    return {};
}

}

template<typename T>
struct is_vector_iterator : decltype(detail::is_vector_iterator(declval<T>(), 0)) {};

Here I'm getting the decayed type of *t to make vector<type>::iterator and check that for equality with T . It works with the exception of vector<bool> which isn't really a vector anyway.

demo