Re: [Boost-users] dereference an iterator

From: Christian Holmquist (c.holmquist_at_[hidden])
Date: 2007-04-16 07:35:44

I want to be able to express the operator *() as a function (or function
object), like in the following trivial example.

template<class Iterator>
typename std::iterator_traits<Iterator>::reference deref(Iterator i)
{
    return *i;
}

std::vector<int *> vp;
std::vector<int> v;
std::transform(vp.begin(), vp.end(), std::back_inserter(v), deref<int*>);

This is, I think, somewhat related to

template<class T, class X> inline
T implicit_cast(X x)
{
     return x;
}

std::vector<boost::reference_wrapper<int> > vp;
std::vector<int> v;
std::transform(vp.begin(), vp.end(), std::back_inserter(v),
implicit_cast<int, int&>);
(well, this would compile with std::copy instead of std::transform..but this
is just for illustration).

I'm looking for something like 'Given any 'dereferencable object T' return
the object it has a reference to'.

On 16/04/07, Paul Giaccone <paulg_at_[hidden]> wrote:
>
> Christian Holmquist wrote:
>
> >
> > Sorry for beeing off-topic, but is there a utility in boost for
> > dereferencing an iterator, such as the below (but that also works for
> > boost::reference_wrapper)
> >
> > template<class Iterator>
> > typename std::iterator_traits<Iterator>::reference deref(Iterator i)
> > {
> >     return *i;
> > }
> >
> > Thanks,
> > Christian
>
> Apart from it looking a little tidier and so perhaps more readable, why
> would you need this when it works in exactly the same way as unary
> operator* (and so means more typing)? It might be useful if it had some
> checks for zero or bad pointers before it did the dereferencing, of
> course, but can that happen with Boost iterators (I'm not familiar
> enough with them to know)?
>
> _______________________________________________
> Boost-users mailing list
> Boost-users_at_[hidden]
> http://lists.boost.org/mailman/listinfo.cgi/boost-users
>