Re: [boost] C++03 unique_ptr emulation

Subject: Re: [boost] C++03 unique_ptr emulation
From: Anthony Williams (anthony.ajw_at_[hidden])
Date: 2009-01-09 11:11:01

David Abrahams <dave_at_[hidden]> writes:

> on Fri Jan 09 2009, Ion Gaztañaga <igaztanaga-AT-gmail.com> wrote:
>
>> David Abrahams wrote:
>>> on Thu Jan 08 2009, Ion Gaztañaga <igaztanaga-AT-gmail.com> wrote:
>>>
>>>> The same happens when implementing forwarding functions, so I'm starting to think
>> that
>>>> putting T() in functions taking movable-only types by value is not a bad idea ;-)
>>>
>>> Sorry, but what do you mean by "putting T() in functions?"
>>>
>>> putting-milk-in-his-T'ly y'rs,
>>
>> void function (movable m);
>>
>> int main()
>> {
>>    movable m;
>>    function(movable(boost::move(m)));
>>    function(movable());
>>    return 0;
>> }
>>
>> I mean that to pass movable-only objects per-value you need to put movable() in the
>> argument. 
>
> do you mean, generically, "movable(...)"?  I couldn't tell whether you
> think the default-ctor-ness of that has significance.
>
>> This is needed if boost::move returns boost::detail::rv<T> instead of
>> T.
>
> But boost::move should return T.

Sorry for jumping in here, but I'm not sure that it should. Consider

movable m;
boost::move(m);

In C++0x, move(m) is equivalent to static_cast<movable&&>(m), which
just obtains a reference. If boost::move<T> returns a T then the code
above will create a temporary which moves the data out of m, and then
destroy that temporary, leaving m a hollow shell.

Anthony
-- 
Anthony Williams
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