Re: [boost] boost::result_of and C++03

From: Peter Dimov (pdimov_at_[hidden])
Date: 2008-04-07 13:51:12

Anthony Williams:
> "Peter Dimov" <pdimov_at_[hidden]> writes:
>
>> Anthony Williams:
>>> "Peter Dimov" <pdimov_at_[hidden]> writes:
>>>
>>>> Eric Niebler:
>>>>
>>>>> Can you write what the identity function object would look like in
>>>>> C++0x?
>>>>
>>>> struct identity
>>>> {
>>>>     template<class A> A operator()( A && a ) const
>>>>     {
>>>>         return std::forward<A>( a );
>>>>     }
>>>> };
>>>>
>>>> I guess.
>>>
>>> No. That will convert lvalues to rvalues.
>>
>> Try it.
>
> I have, and it doesn't work.

With what compiler does it not work? Can you post the example program?

> Besides, you can see in the signature: it returns "A" by value.

There's a special deduction rule for A&&, where A is a template parameter: 
when passed an lvalue of type T, A is deduced as T& and not as T. So in this 
case operator() returns a T& "by value".