How to implement std::distance() for custom templated iterator?

Question

I have a templated bidirectional iterator. I do not want to make it random access because the it += n operation would not be constant time. However, the it2 - it1 operation is constant time. I wanted to specialize std::distance() for this iterator so that algorithms that use it (such as std::vector::assign()) can make use of the efficient difference operation. How can I do this if the iterator is a template?

Here's a toy example:

#include <iterator>
#include <iostream>

// template bidirectional iterator
template<typename T>
class iter : public std::iterator<std::bidirectional_iterator_tag, T> {
    T *ptr;
public:
    iter(T *ptr) : ptr(ptr) { }

    iter() = default;
    iter(const iter &) = default;
    iter &operator = (const iter &) = default;

    T *operator * () { return ptr; }

    bool operator == (const iter &it) { return ptr == it.ptr; }
    bool operator != (const iter &it) { return ptr != it.ptr; }

    iter &operator ++ () { ++ptr; return *this; }
    iter operator ++ (int) { iter tmp(*this); operator++(); return tmp; }

    iter &operator -- () { --ptr; return *this; }
    iter operator -- (int) { iter tmp(*this); operator--(); return tmp; }

    // Would not be used for a bidirectional iterator.
    // Implemented only so we can use it in std::distance() below.
    ptrdiff_t operator - (const iter &it) { return ptr - it.ptr; }
};

namespace std {
    // We could specialize std::distance() for iter<int> like this:
    template<>
    iter<int>::difference_type distance(iter<int> first, iter<int> last) {
        std::cout << "my distance called\n";
        return last - first;
    }

    // QUESTION: Can we do it in general, for iter<T> ?
}

// Just to test that everything works as intended.
int main() {
    int arr[5];
    iter<int> it1(&arr[0]);
    iter<int> it2(&arr[5]);

    std::cout << std::distance(it1, it2) << std::endl;

    return 0;
}

---

This is a follow-up of Is it reasonable to overload std functions such as std::distance?

We could in principle do something like this:

namespace std {
    template<class T>
    typename iter<T>::difference_type distance(iter<T> first, iter<T> last) {
        std::cout << "my distance called\n";
        return last - first;
    }
}

But that would be an overload of std::distance(), which is not allowed for std namespace functions according to the standard.

Answer 1

The correct way to do it is to define your distance method in the same namespace as your iter-template, (in this case global namespace).

....
    typename iter::difference_type operator -(const iter &it)
    {
        return ptr - it.ptr;
    }
}; // close template<typename T> class iter

template<typename T>
typename iter<T>::difference_type distance( iter<T> first,  iter<T> last)
{
    std::cout << "my distance called\n";
    return last - first;
}

And later using ADL as seen in this example:

int main()
{
    int arr[5];
    iter<int> it1(&arr[0]);
    iter<int> it2(&arr[5]);

    using std::distance;
    using std::begin;
    using std::end;

    std::cout << distance(it1, it2) << '\n';
    std::cout << "using std::distance\n";
    std::cout << distance(begin(arr), end(arr)) << '\n';

    return 0;
}

will output:

my distance called
5
using std::distance
5

A good explanation of this problem of partial specialization of templates of methods from std is given by Scott Meyers in his book "Effective C++", third edition, item 25.