[Boost-users] Re: Functors and function_traits<>

From: David Abrahams (dave_at_[hidden])
Date: 2004-12-27 13:39:42

Hossein Haeri wrote:
> Dear all,
> 
> Yep, again it's me, and it's agian
> function_traits<>...
> 
> AFAIK, std::iterator_traits<> is one of the
> counterparts of function_traits<>, not?
> 
> A sort of symmetry which iterator_traits<> has got is
> that it can work for Iterators as well as plain
> pointers. 

That's the wrong way to think about it.  It just works for iterators.
Plain pointers just happen to be iterators, by the *standard's
definition* of "iterator."

> This seems to be not the case for function_traits<>.
> That is, you can pass a function type to it, but you
> can't do that for a functor. Is that right?

That's because function object types (what you're calling functors) are
not functions by the standard's definition of "function."

> If it is, a big drawback turns out to happen for
> variations of the following which are not that few
> common:
> 
> template <
>           typename ...,
>           ...,
>           typename Function
>          >
> struct Foo
> {
>   typedef 
>     typename 
>       function_traits<Function>
>         ::result_type result_type;
> 
>   Foo (Function f) : f_(f) {}
> 
>   //...
>   result_type operator () () {/*..*/}
> 
>   Function f_;
> };

It doesn't work because there's no way to code it in general.  First of
all there's no universal standard for a way to get the result type of a
function object.  Accessing a nested result_type sometimes works, but
then, sometimes the operator() is overloaded, or even templated.

You might look at boost::result_of for a more general solution.

-- 
Dave Abrahams
Boost Consulting
http://www.boost-consulting.com