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;}
};