Re: [Boost-users] [iterator] error while accessing member of result of brackets-operator

Subject: Re: [Boost-users] [iterator] error while accessing member of result of brackets-operator
From: Krzysztof Czainski (1czajnik_at_[hidden])
Date: 2014-01-09 11:43:46

2014/1/9 Sebastian Pfützner <s.pfuetzner_at_[hidden]>

> Hello,
>
> the following code does not compile with the last line in main(). The
> error message is
>
> main.cpp: In function 'int main()':
> main.cpp:37:8: error: 'boost::detail::operator_brackets_result<It, S,
> S&>::type' has no member named 'i'
>   it[3].i = 0;
>         ^
>
> What could be the problem?
>
>
> #include <boost/iterator/iterator_facade.hpp>
>
> struct S
> {
>         int i;
>         float f;
> };
>
> class It :
>         public boost::iterator_facade<It, S, boost::random_access_
> traversal_tag>
> {
> private:
>         friend class boost::iterator_core_access;
>
>         static S s;
>
>         reference dereference() const
>         { return s; }
>
>         bool equal(It const&) const
>         { return true; }
>
>         void increment() {}
>         void decrement() {}
>         void advance(difference_type n) {}
>
>         difference_type distance_to(It const&) const
>         { return 0; }
> };
>
> int main()
> {
>         It it;
>         (*(it + 3)).i = 0;
>         S& s = it[3];
>         s.i = 0;
>         it[3].i = 0;
> }

Hello,

In your case, it[3] returns a boost::detail::operator_brackets_proxy<It>
instead of S& you expect:

    // A proxy return type for operator[], needed to deal with
    // iterators that may invalidate referents upon destruction.
    // Consider the temporary iterator in *(a + n)
    template <class Iterator>
    class operator_brackets_proxy;

As the comment sais, it[3] won't return a reference directly "to deal with
    // iterators that may invalidate referents upon destruction.
    // Consider the temporary iterator in *(a + n)". I don't know in what
situations that can happen.

I don't see a clean way to force it[3]  to return a real reference. Can
anyone comment on this?

That said, here's a hack for you: right after the definition of struct S
insert this code:

class It;

namespace boost { namespace detail {
template <> struct operator_brackets_result<It,S,S&>
{ typedef S& type; };
}}

(Your code won't link as it is, buth that's irrelevent here)
HTH
Kris