Re: [boost] C++03 unique_ptr emulation

Subject: Re: [boost] C++03 unique_ptr emulation
From: David Abrahams (dave_at_[hidden])
Date: 2009-01-09 20:39:35

on Fri Jan 09 2009, Steven Watanabe <watanabesj-AT-gmail.com> wrote:

> AMDG
>
> David Abrahams wrote:
>>> David Abrahams wrote:
>>>     
>>>> So you're saying, in other words, that "move(x)" really means "you have
>>>> permission to move x" but the one I proposed would mean "move it, now."
>>>>
>>>> OK, good point.
>>>>
>>>> So what about this horrible little proposal?
>>>>
>>>>   template <class T>
>>>>   struct rv<T> : T
>>>>   {
>>>>    private:
>>>>       rv();
>>>>       ~rv();
>>>>       rv(rv const&);
>>>>       void operator=(rv const&);
>>>>   };
>>>>
>>>>   template <class T>
>>>>   boost::enable_if<is_class<T>, rv<T>&>
>>>>   move(T& x)
>>>>   {
>>>>     return static_cast<rv<T>& >(x);
>>>>   }
>>>>
>>>> Does that solve any problems?
>>>>       
>> Well, I haven't thought this through deeply (nor really tested
>> anything), but the idea is that movable types would have a stealing
>> conversion from rv<T>& but non-movable types should just use the copy
>> ctor that slices off the non-existent derived class.
>>   
>
> What if T has a fully generic constructor?

That's detectable, at least.  For such types, you disable this move
unless some trait is specialized.

-- 
Dave Abrahams
BoostPro Computing
http://www.boostpro.com