Re: [boost] [tr1/result of] and l/r values

From: Tobias Schwinger (tschwinger_at_[hidden])
Date: 2007-01-08 11:32:11

Doug Gregor wrote:
> On Dec 27, 2006, at 7:55 AM, Tobias Schwinger wrote:
>> here's another result_of question:
>>
>> Given a function object
>>
>>     struct f
>>     {
>>         template< typename T >
>>         r<T> operator()(T const &) const;
>>
>>         template< typename T >
>>         r<T&> operator()(T &) const;
>>
>>         template< typename Sig >
>>         struct result;
>>     };
>>
>> ,
>>
>>     result_of< f(some_obj &) >::type      // is r<T&>
>>     result_of< f(some_obj const & >::type // is r<T>
>>
>> but what about:
>>
>>     result_of< f(some_obj) >::type        // is r<T>?
> 
> Yes, that's right.
> 
>> That's what I'd do intuitively. I also found an example (written by  
>> Dave) on the net that seems to do the same:
>>
>>     http://tinyurl.com/yzeenu
>>
>> But wait: the TR1 text is talking about lvalues so I probably  
>> should let F::result return r<T&>?!
> 
> TR1 says...
> 
> 	Given an rvalue f of type F and values t1, t2, ..., tN of types T1,  
> T2, ..., TN, respectively, the type member
> 	is the result type of the expression f(t1, t2, ...,tN). The values  
> ti are lvalues when the corresponding type Ti is
> 	a reference type, and rvalues otherwise.

Sorry - it seems I missed that last part of the sentence, somehow (maybe 
by looking at the wrong version?).

> 
> So with result_of< F(some_obj)>, we treat some_obj as an rvalue, and  
> get the same result as for result_of< f(some_obj const & >.

OK, got it. Thank you!

Regards,
Tobias