Re: [boost] [core] Breaking change to boost::ref in 1.56

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.