Question

In code:

//I know that to get this effect (being able to use it with std algorithms) I can inherit like I did in line below:

    class Iterator //: public std::iterator<std::bidirectional_iterator_tag,T>

    {
    private:
        T** itData_;
    public:
        //BUT I WOULD LIKE TO BE ABLE TO DO IT BY HAND AS WELL
        typedef std::bidirectional_iterator_tag iterator_category;
        typedef T* value_type;//SHOULD IT BE T AS value_type or T*?
        typedef std::ptrdiff_t difference_type;
        typedef T** pointer;//SHOULD IT BE T* AS pointer or T**?
        typedef T*& reference;//SHOULD IT BE T& AS reference or T*&?
};

Basically what I’m asking is if I have my variable of type T** in iterator class is it right assumption that value type for this iterator will be T* and so on as I described in comments in code, right next to relevant lines.
Thank you.

Answer 1

The definition in the standard (section 24.3.2) is:

template<class Category, class T, class Distance = ptrdiff_t,
         class Pointer = T*, class Reference = T&>
struct iterator {
    typedef T value_type;
    typedef Distance difference_type;
    typedef Pointer pointer;
    typedef Reference reference;
    typedef Category iterator_category;
};

As you can see, the defaults are:

typedef T value_type;
typedef ptrdiff_t difference_type;
typedef T* pointer;
typedef T& reference;

This is assuming that the iteration is over a container of elements of type T. If your container holds elements of type T* instead, the typedefs in your question would be correct.