Re: typename iterator_traits::pointer

From:

"Victor Bazarov" <v.Abazarov@comAcast.net>

Newsgroups:

comp.lang.c++

Date:

Thu, 24 Jan 2008 08:34:15 -0500

Message-ID:

<fna44n$d0p$1@news.datemas.de>

Ioannis Vranos wrote:

I am reading TC++PL3 and on page 552 it is mentioned:

"The related types of an iterator are described by a small set of
declarations in an iterator_traits template class:

template <class Iter> struct iterator_traits {
typedef typename Iter::iterator_category iterator_category; //19.2.3
typedef typename Iter::value_type value_type; // type of element
typedef typename Iter::difference_type difference_type;
typedef typename Iter::pointer pointer; //return type of
operator->() typedef typename Iter::reference reference; //return
type of operator*() };"

How can we use Iter.operator->() to get an iterator_traits<Iter>::
pointer?

You can't. The code will immediately use the pointer that operator->
returns to access the member expression following the -> token. IOW
the overloaded operator-> can never be used to initialise a pointer,
for example (unless you have a member that returns 'this', then you
need to call that member).

Let's assume vector<int> as an example and we want to get a
pointer to the first element of the sequence using operator->():

#include <vector>
#include <iostream>
#include <iterator>

int main()
{
  using namespace std;

  vector<int> vec(10);

  iterator_traits<vector<int>::iterator>::pointer p= ???
}

There is no syntax in C++ to do what you seem to want/need.

V
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