Hi Ian,
First of all, thanks a lot for your answer and help about this.
I read what you pointed and, I believe, cannot fully grasp it. First,
> There is a conversion from iterator to const_iterator.
which makes me think that, because of that conversion, there would be
no need of the flexibility given by using two different template
parameters.
Then,
> Specifically to permit mixing iterator and const_iterator, yes.
but, conversely, if two different template type parameters are
allowed, no conversion will happen. I believe that each one will bind
to the declared type of the variable.
To check and see what's going on, I did
g++ -E main.cc
, got the full sources and modified the operator- by adding (with the
proper #include <typeinfo>)
std::cout << typeid(_IteratorL).name() << std::endl;
std::cout << typeid(_IteratorR).name() << std::endl;
Running, I get the following
PKi
Pi
which, I think, stands for int const* and int*.
So it seems that each iterator type stays as declared, because the
output corresponds, respectively, to the types used in the definition
of const_iterator and iterator.
typedef __gnu_cxx::__normal_iterator<pointer, vector_type> iterator;
typedef __gnu_cxx::__normal_iterator<const_pointer, vector_type>
const_iterator;
where
typedef const _Tp* const_pointer;
typedef _Tp* pointer;
Because of this, I understand that there's no conversion, and so (as
the standard is only requiring that "a - b" is legal when their type
is the same) have some doubts about the portability of the expression
(in the sense of being guaranteed to work by the standard).
Cheers, and thanks a lot one more time.
--
Rodolfo Federico Gamarra
On Tue, Aug 24, 2010 at 11:25, Ian Lance Taylor <iant@xxxxxxxxxx> wrote:
> rgamarra <rgamarra@xxxxxxxxx> writes:
>
>> int
>> main()
>> {
>> std::vector<int>::const_iterator i;
>> std::vector<int>::iterator j;
>> i - j;
>> // std::distance(i, j); // not the same iterator types
>> }
>
>> Now, the call to operator- succeeds. Is it legal standard-C++?
>> Particularly, is that mixing of types in the expression required to be
>> possible by the standard? Or is it an gnu-extension? I'm using g++
>> (Ubuntu 4.3.3-5ubuntu4) 4.3.3.
>
> There is a conversion from iterator to const_iterator. So in this case
> j is converted to const_iterator in order to make the types the same.
>
>> Now in the sources, I see that (unlike distance) operator- is defined
>> with two different template type parameters, which, I understand,
>> allow to perform, directly, that mixing.
>
> Specifically to permit mixing iterator and const_iterator, yes.
>
> Ian
>
