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Subject: Re: [boost] [iterator] reverse_iterator<counting_iterator<T>>
From: Stewart, Robert (Robert.Stewart_at_[hidden])
Date: 2010-04-13 11:09:24
Jeffrey Hellrung wrote:
> Stewart, Robert wrote:
> > Arno Schödl wrote:
> >> "Stewart, Robert" <Robert.Stewart_at_[hidden]
> >
> >> boost::prior(...) internally makes a copy of its iterator
> >> argument, decrements it and returns it by value, so far so
> >> good. counting_iterator::operator* then returns a reference
> >> into this (temporary) iterator. Returning from
> >> reverse_iterator::dereference() destroys the temporary
> >> iterator, but still returns the reference pointing into it.
> >
> > boost::prior's T is deduced as a reference type. The copy
> is of a reference. There's no temporary.
>
> You can think of the code as doing the equivalent of the following:
>
> reference dereference() const
> {
> counting_iterator<T> temp_it = this->base(); // holds a T value
> --temp_it;
> return *temp_it; // return a reference to a member object of
> temp_it --> BAD
> }
Taking another look:
reference
dereference() const
{
return *boost::prior(this->base());
}
this->base() is:
Base const &
base() const
{
return m_iterator;
}
That returns Iterator const &, which is passed to boost::prior(), which is:
template <class T>
T
prior(T x)
{
return --x;
}
If T is deduced as Iterator const &, x is a reference, which was my earlier inference, but operator --(), which is non-const, would be applied to that reference to const and wouldn't compile. That's the detail I missed earlier.
Clearly, the compiler deduces T as Iterator, copy constructs a temporary Iterator from the reference returned by this->base(), and then decrements that temporary. That leads to the OP's concern and you see that I'm finally on board.
_____
Rob Stewart robert.stewart_at_[hidden]
Software Engineer, Core Software using std::disclaimer;
Susquehanna International Group, LLP http://www.sig.com
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