Re: [Boost-users] lambda and shared_ptr?

From: Peter Dimov (pdimov_at_[hidden])
Date: 2005-11-21 21:38:21

Mark Santaniello (AMD) wrote:
>>> template< typename T >
>>> boost::function<T(T)> foo( T n_ )
>>> {
>>>   //Works, but leaks n
>>>   //T *n = new T( n_ );
>>>
>>>   //Doesn't work
>>>   boost::shared_ptr<T> n( new T(n_) );
>>>
>>>   return ( *n += _1 );
>>
>> I think that you need to wrap n in boost::lambda::constant:
>>
>>     return *constant(n) += _1;
>>
>> Without it, *n is evaluated immediately.
>>
>>> }
>>
>
> Peter,
>
> Thanks for the reply.  I must be doing something dumb, because now I'm
> getting a build error.
>
> I think this *might* be the important part:
> /usr/include/boost/detail/is_incrementable.hpp:59: error: no match for
> 'operator++' in
> '++boost::detail::is_incrementable_::impl<boost::shared_ptr<int> >::x'

Apparently, boost::detail::is_incrementable doesn't work on your compiler; 
this in turn breaks boost::indirect_reference and the return type deduction 
for lambda's operator*. But I should also note that *constant(n) += _1 
doesn't seem to work even when the compiler supports is_incrementable, 
because for some reason, Lambda thinks that dereferencing a const pointer 
yields a const reference.

   return ret<int&>( *constant(pn) ) += _1;

compiles for me, though.