如何打印通用 std::list 迭代器？

Question

I want to be able to print any std::list iterator by printing its value. My initial code looked like this:

template<typename T>
std::ostream& operator<<(std::ostream& os, const typename std::list<T>::const_iterator& x)
{
   return os << "&" << *x;
}

Which did not work, since the compiler can not determine the parameter T . I then tried making it generic over the iterator type itself and using iterator_traits to limit it to iterators.

template<
    typename It,
    typename = typename std::iterator_traits<It>::value_type
>
std::ostream &operator<<(std::ostream &os, const It &x)
{
    return os << "&" << *x;
}

But then, of course, I get two conflicting implementations for std::ostream << *const char , since pointers are also iterators. How can I limit the implementation to std::list iterators so I don't get a conflict?

Answer 1

You can constrain the type as iterator or const_iterator of std::list . Eg

template<typename It>
std::enable_if_t<std::is_same_v<It, typename std::list<typename std::iterator_traits<It>::value_type>::iterator> ||
                 std::is_same_v<It, typename std::list<typename std::iterator_traits<It>::value_type>::const_iterator>
                 , std::ostream &> 
operator<<(std::ostream &os, const It &x) {
    return os << "&" << *x;
}

Answer 2

You can SFINAE the const char* out from the operator<< overload.

#include <type_traits> // std::enable_if_t, std::is_same_v, std::remove_reference_t

template<
    typename It,
    typename = typename std::iterator_traits<It>::value_type
>
auto operator<<(std::ostream &os, const It &x)
-> std::enable_if_t< !std::is_same_v<std::remove_reference_t<It>, const char*>, std::ostream&>
{
    return os << "&" << *x;
}

( See a Demo )

Note that, the above is not only restricted for std::list::iterator , meaning the iterators from the other containers, can also consider this overload. This might not be the behaviour you want.

---

Since we could not get the container type from the iterator , I would suggest the same as @super mentioned in the comments. Provide a operator<< overload for the Legacy Bidirectional Iterator which is what the std::list has.

Following is an example code, which will work for your expected cases as well as all the containers, which meet the requirements of a bidirectional iterator.

#include <list>
#include <iostream>
#include <iterator>    // std::iterator_traits, std::bidirectional_iterator_tag
#include <type_traits> // std::is_same_v, std::enable_if_t

// SFINAE helper  type for bidirectional_iterator_t
template<typename Iterator, typename ReType = void>
using enable_for_bidirectional_iterator_t
= std::enable_if_t<
   std::is_same_v<std::bidirectional_iterator_tag, typename std::iterator_traits<Iterator>::iterator_category>
   , ReType
>;

template<typename Iterator>
auto operator<<(std::ostream& os, const Iterator x) noexcept
-> enable_for_bidirectional_iterator_t<Iterator, std::ostream&>
{
   return os << "&" << *x;
}

( See a Demo )

---

However, usually, you provide an operator<< overload for the container, not for the iterators. You might want to rethink about the design.