I'm making my own vector container and I'm trying to implement an iterator that works like the real one, using the C 98 standard.
This is homework so I don't want the answer just a hint as to where I should look and what I should learn to be able to tackle this problem.
So basically I'm trying to make this code work:
ft::vector<int> v (100, 100);
ft::vector<int>::iterator it = v.begin();
ft::vector<int>::const_iterator cit = it;
std::cout << (cit == it) << std::endl; //comparison 1 /// works
std::cout << (it == cit) << std::endl; //comparison 2 /// doesn't compile
std::cout << (cit 1 == it 1) << std::endl; //doesn't work
std::cout << (it 1 == cit 1) << std::endl; //doesn't work
iterator
and const_iterator
are typedefs like this:
typedef typename ft::iterator_vector<value_type> iterator;
typedef typename ft::iterator_vector<const value_type> const_iterator;
And value type being the type passed to the vector template.
The first comparison didn't work until I added a user-defined conversion operator in my iterator template to convert (the other way around actually as pointed out by @TedLyngmo)
which is iterator<const foo>
into iterator<foo>
operator iterator_vector<const value_type>() const { return _p; }
but the compiler says that I need now to be able to convert a iterator<const foo>
into a const iterator<foo>
and I have no idea how to proceed.
This is the implementation of my iterator:
template <class T>
class iterator_vector : public ft::iterator<std::random_access_iterator_tag, T> {
public:
typedef ft::iterator<std::random_access_iterator_tag, T> iterator;
typedef ft::iterator_traits<iterator> iterator_traits;
typedef typename iterator_traits::difference_type difference_type;
typedef typename iterator_traits::value_type value_type;
typedef typename iterator_traits::pointer pointer;
typedef typename iterator_traits::reference reference;
typedef typename iterator_traits::iterator_category iterator_category;
/*
** Member functions
*/
iterator_vector(pointer p = 0) : _p(p) {}
~iterator_vector(void) {}
operator iterator_vector<const value_type>() const { return _p; }
iterator_vector& operator () { _p; return *this; }
iterator_vector operator (int)
{
iterator_vector r = *this;
_p;
return r;
}
iterator_vector& operator--() { --_p; return *this; }
iterator_vector operator--(int)
{
iterator_vector r = *this;
--_p;
return r;
}
iterator_vector operator (size_t n) const { return iterator_vector(_p n); }
iterator_vector operator-(size_t n) const { return iterator_vector(_p - n); }
iterator_vector& operator =(size_t n) { _p = n; return *this; }
iterator_vector& operator-=(size_t n) { _p -= n; return *this; }
difference_type operator (iterator_vector rhs) const { return _p rhs._p; }
difference_type operator-(iterator_vector rhs) const { return _p - rhs._p; }
reference operator*(void) const { return *_p; }
pointer operator->() const { return _p; }
reference operator[](size_t n) const { return _p[n]; }
bool operator==(const iterator_vector& rhs) const { return _p == rhs._p; }
bool operator!=(const iterator_vector& rhs) const { return _p != rhs._p; }
bool operator<(const iterator_vector& rhs) const { return _p > rhs._p; }
bool operator>(const iterator_vector& rhs) const { return _p < rhs._p; }
bool operator<=(const iterator_vector& rhs) const { return _p <= rhs._p; }
bool operator>=(const iterator_vector& rhs) const { return _p >= rhs._p; }
/*
** Non-member functions
*/
friend iterator_vector operator (size_t n, const iterator_vector& rhs) { return iterator_vector(rhs._p n); }
friend iterator_vector operator-(size_t n, const iterator_vector& rhs) { return iterator_vector(rhs._p - n); }
private:
pointer _p;
};
CodePudding user response:
It seems from your comments in your code about what is failing that you are missing a comparison function for when iterator
is to the left and the const_iterator
is to the right. You could add this free function:
template<typename T>
bool operator==(const iterator_vector<T>& lhs, const iterator_vector<const T>& rhs) {
// just swap the order here and the implicit conversion from `iterator`
// to `const_iterator` from `lhs` solves the rest:
return rhs == lhs;
}
I added a user-defined conversion operator in my iterator template to convert
iterator<const foo>
intoiterator<foo>
No, you added an implicit conversion the other way around. That is, from iterator
to const_iterator
- which is good!
An alternative could be to make the two iterators friend
s to not have to implement similar swapping lhs with rhs functions for all your operators. The operators in your template would then become:
// define these before your class template (borrowed from C 11):
template< class T > struct remove_const { typedef T type; };
template< class T > struct remove_const<const T> { typedef T type; };
//... in your class template:
friend class iterator_vector<typename remove_const<T>::type>;
friend class iterator_vector<const T>;
template<typename U>
bool operator==(const iterator_vector<U>& rhs) const { return _p == rhs._p; }
template<typename U>
bool operator!=(const iterator_vector<U>& rhs) const { return _p != rhs._p; }
template<typename U>
bool operator<(const iterator_vector<U>& rhs) const { return _p > rhs._p; }
template<typename U>
bool operator>(const iterator_vector<U>& rhs) const { return _p < rhs._p; }
template<typename U>
bool operator<=(const iterator_vector<U>& rhs) const { return _p <= rhs._p; }
template<typename U>
bool operator>=(const iterator_vector<U>& rhs) const { return _p >= rhs._p; }