Subject: Re: [boost] [core] Breaking change to boost::ref in 1.56
From: Andrey Semashev (andrey.semashev_at_[hidden])
Date: 2014-07-14 07:03:45
On Monday 14 July 2014 13:43:03 Peter Dimov wrote:
> Andrey Semashev wrote:
> > template< typename T >
> > void foo1(T const& t)
> > {
> >
> > foo(boost::ref(t));
> >
> > }
>
> That's getting a bit artificial now. What is the purpose of foo1 except to
> break the code?
Well, you might want to do something on t before invoking the common logic in
foo.
I admit I invented this example just now, but I can imagine something like
this in real life.
> > The point was that (a) unwrap_reference doesn't remove all nested levels
> > of reference_wrappers, just the top level, so class_type does not have bar
> > when foo1 is called and (b) boost::bind saves a reference to a reference,
> > so get_pointer does not work as intended, not to mention that a dangling
> > reference may be left if the function object call is delayed.
>
> If you delay the function call in
>
> > template< typename T >
> > void foo(T const& t)
> > {
> >
> > typedef typename boost::unwrap_reference<T>::type class_type;
> > boost::bind(&class_type::bar, boost::ref(t))();
> >
> > }
>
> it will result in a dangling reference when called with an rvalue of any
>
> type. This line, for example:
> > foo(my_class_const());
>
> will store a dangling reference to the my_class_const temporary.
Right. That comment about the dangling reference was more related to the call
of foo1.