Re: typename iterator_traits::pointer
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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