Preparation for std::iterator Being Deprecated - c++

Question

On March 21st the standards committee voted to approve the deprecation of std::iterator proposed in P0174:
The long sequence of void arguments is much less clear to the reader than simply providing the expected typedefs in the class definition itself, which is the approach taken by the current working draft, following the pattern set in c++14
Before c++17 inheritance from std::iterator was encouraged to remove the tedium from iterator boilerplate implementation. But the deprecation will require one of these things:
An iterator boilerplate will now need to include all required typedefs
Algorithms working with iterators will now need to use auto rather than depending upon the iterator to declare types
Loki Astari has suggested that std::iterator_traits may be updated to work without inheriting from std::iterator
Can someone enlighten me on which of these options I should expect, as I design custom iterators with an eye towards c++17 compatibility?

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.