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.
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