Why doesn't my custom iterator work with the STL copy?

Question

I wrote an OutputIterator for an answer to another question. Here it is:

#include 

using namespace std;

template< typename T, typename U &g         


        

Answer 1

#include <queue>
#include <algorithm>
#include <iterator>
#include <iostream>

using namespace std;

template< typename T, typename U >
class queue_inserter
{
    queue<T, U> &qu;

public:
    // for iterator_traits to refer
    typedef output_iterator_tag iterator_category;
    typedef T value_type;
    typedef ptrdiff_t difference_type;
    typedef T* pointer;
    typedef T& reference;

    queue_inserter(queue<T,U> &q) : qu(q) { }
    queue_inserter<T,U>& operator ++ () { return *this; }
    queue_inserter<T,U> operator * () { return *this; }
    void operator = (const T &val) { qu.push(val); }
};

template< typename T, typename U >
queue_inserter<T,U> make_queue_inserter(queue<T,U> &q)
{
    return queue_inserter<T,U>(q);
}

int main()
{
    // uses initalizer list (C++0x), pass -std=c++0x to g++
    vector<int> v({1, 2, 3});
    queue<int, deque<int>> q;
    copy(v.cbegin(), v.cend(), make_queue_inserter(q));
    while (!q.empty())
    {
        cout << q.front() << endl;
        q.pop();
    }
}

This should do it with iterator_traits; a helper struct in <iterator> which defines all types an iterator should typically define. Functions in <algorithm>, refer to these types when required like iterator_traits<it>::iterator_category or say iterator_traits<it>::value_type, etc. Just defining them inside one's custom iterator would do the trick. This is the modern way of writing iterators, as opposed to the classical way of inheriting from std::iterator. Having a look at <iterator> reveals that even std::iterator defines these types i.e. iterator_category, difference_type, etc. This is the reason, when inherited from std::iterator, the derived iterator class gets these due to heredity.

Answer 2

Your iterator doesn't meet the requirement for an 'assignable' type which is a requirement for an output iterator because it contains a reference and assignable types need to ensure that after t = u that t is equivalent to u.

You can provide a suitable specialization for iterator_traits for your iterator either by deriving from a specialization of std::iterator or by providing one explicitly.

namespace std
{
    template<> struct iterator_traits<MyIterator>
    {
        typedef std::output_iterator_tag iterator_category;
        typedef void value_type;
        typedef void difference_type;
    };
}

Answer 3

Derive it from std::iterator. If you are interested the Dr. Dobb's has an article about custom containers and iterators.

Answer 4

Your queue_inserter needs to be derived from std::iterator so that all the typedefs such as value_type are properly defined since these are used inside STL algorithms This definition works:

template< typename T, typename U >
class queue_inserter : public std::iterator<std::output_iterator_tag, T>{
    queue<T, U> &qu;  
public:
    queue_inserter(queue<T,U> &q) : qu(q) { }
    queue_inserter<T,U> operator ++ (int) { return *this; }
    queue_inserter<T,U> operator ++ () { return *this; }
    queue_inserter<T,U> operator * () { return *this; }
    void operator = (const T &val) { qu.push(val); }
};