Re: [boost] [fusion] value_of tranform_view

From: Stjepan Rajko (stipe_at_[hidden])
Date: 2007-07-12 16:10:20

On 7/12/07, Peter Dimov <pdimov_at_[hidden]> wrote:
> Stjepan Rajko wrote:
>
> > So, if I have, say, a function object with operator()(int &), I should
> > specify result<F(int &)>, (but maybe if I have operator()(const int
> > &), or operator()(int), I can specify result<F(int)>?)
>
> If you have operator()( int& ), you need result<F(int&)>. If you have
> operator()( int ), you need both result<F(int)> and result<F(int&)> because
> your operator() can take either an lvalue or an rvalue.
>
> In this simple case you can just provide result_type, of course.
>
> You are allowed to provide result<F(int)> in addition to result<F(int&)>
> even in the first case; failing invalid queries is not required.
>
> What this means in practice is that in 99% of the cases you just need to
> strip the reference from the argument types when providing result or a
> result_of specialization. Only odd function objects such as
>
> struct F
> {
>     int operator()( int& );
>     void operator()( long );
> };
>
> need special handling.
>

Thank you for the detailed explanation!  It is much clearer now.

Stjepan