Re: plain iterators and reverse iterators on vector

From:

Michael Doubez <michael.doubez@free.fr>

Newsgroups:

comp.lang.c++

Date:

Thu, 6 Aug 2009 06:02:59 -0700 (PDT)

Message-ID:

<4cef8010-354f-45c8-8331-63dda7f46691@k6g2000yqn.googlegroups.com>

On 6 ao=FBt, 14:35, "subramanian10...@yahoo.com, India"
<subramanian10...@yahoo.com> wrote:

Consider the following program x.cpp:

#include <cstdlib>
#include <iostream>
#include <vector>

using namespace std;

int main()
{
        vector<int> c;

        for (int i = 0; i != 10; ++i)
                c.push_back(i);

        cout << c.end() - c.begin() << endl;
        cout << c.rend() - c.rbegin() << endl;

        return EXIT_SUCCESS;

}

I compiled with g++3.4.3 as
g++ -std=c++98 -pedantic -Wall -Wextra x.cpp

When I ran it, it produced the output
10
10

The first 10 in the output is fine. It corresponds to c.end() - c.begin
().
Shouldn't the second 10 in the output be -10 because I am using
reverse iterators in calculating c.rend() - c.rbegin() ie isn't this
expression equivalent to c.begin() - c.end() in which case -10 would
be printed?

It is not

Is my understanding wrong ?

It is.

operator-() computes the distance between two iterators not the
difference between pointers.
Example:
distance = it2 - it1
means that
it2 == it1 + distance

In the case of reverse iterator:
distance = rend() - rbegin()
rend() == rbegin() + size()

Is the difference operator-() between the RandomAccessIterators
defined in <iterator> ?

RandomAccessIterators is not a type, it is a concept, each iterator
type which are RandomAccessIterators will define their own operator-
().

If so, how does the program compile even when
I do not #include <iterator> ?

<vector> may define it, you are guaranteed that it works without
additional header (vector may include <iterator> and I guess it does
for reverse_iterator).

--
Michael