Preparation for std::iterator Being Deprecated

Question

On March 21^st the standards committee voted to approve the deprecation of std::iterator proposed in P0174:

The long sequence of void argume

Answer 1

The discussed alternatives are clear but I feel that a code example is needed.

Given that there will not be a language substitute and without relying on boost or on your own version of iterator base class, the following code that uses std::iterator will be fixed to the code underneath.

With std::iterator

template<long FROM, long TO>
class Range {
public:
    // member typedefs provided through inheriting from std::iterator
    class iterator: public std::iterator<
                        std::forward_iterator_tag, // iterator_category
                        long,                      // value_type
                        long,                      // difference_type
                        const long*,               // pointer
                        const long&                // reference
                                      >{
        long num = FROM;
    public:
        iterator(long _num = 0) : num(_num) {}
        iterator& operator++() {num = TO >= FROM ? num + 1: num - 1; return *this;}
        iterator operator++(int) {iterator retval = *this; ++(*this); return retval;}
        bool operator==(iterator other) const {return num == other.num;}
        bool operator!=(iterator other) const {return !(*this == other);}
        long operator*() {return num;}
    };
    iterator begin() {return FROM;}
    iterator end() {return TO >= FROM? TO+1 : TO-1;}
};

(Code from http://en.cppreference.com/w/cpp/iterator/iterator with original author's permission).

Without std::iterator

template<long FROM, long TO>
class Range {
public:
    class iterator {
        long num = FROM;
    public:
        iterator(long _num = 0) : num(_num) {}
        iterator& operator++() {num = TO >= FROM ? num + 1: num - 1; return *this;}
        iterator operator++(int) {iterator retval = *this; ++(*this); return retval;}
        bool operator==(iterator other) const {return num == other.num;}
        bool operator!=(iterator other) const {return !(*this == other);}
        long operator*() {return num;}
        // iterator traits
        using difference_type = long;
        using value_type = long;
        using pointer = const long*;
        using reference = const long&;
        using iterator_category = std::forward_iterator_tag;
    };
    iterator begin() {return FROM;}
    iterator end() {return TO >= FROM? TO+1 : TO-1;}
};

Answer 2

Option 3 is a strictly more-typing version of Option 1, since you have to write all the same typedefs but additionally wrap iterator_traits<X>.

Option 2 is unviable as a solution. You can deduce some types (e.g. reference is just decltype(*it)), but you cannot deduce iterator_category. You cannot differentiate between input_iterator_tag and forward_iterator_tag simply by presence of operations since you cannot reflexively check if the iterator satisfies the multipass guarantee. Additionally, you cannot really distinguish between those and output_iterator_tag if the iterator yields a mutable reference. They will have to be explicitly provided somewhere.

That leaves Option 1. Guess we should just get used to writing all the boilerplate. I, for one, welcome our new carpal-tunnel overlords.