On Thu, May 15, 2008 at 11:37 PM, Daniel Walker <daniel.j.walker@gmail.com> wrote:
Yes, that's basically all there is to it. Really, it's just ....
function<int(int,int)> f0 = *add_floats.target<plus>();
... with error checking.
functional_cast is almost as simple as lexical_cast. They're both just syntactic sugar in a way. They communicate the notion that you're converting between two types that represent the same thing. Check out the implementation I sent yesterday. functional_cast's body is just four lines. It's nothing fancy; it's just more expressive and possibly more convenient than directly manipulating pointers to boost::function's innards.
I've checked out now. I didn't wanted to look before because I wanted to try to understand myself, it helps in my understanding when reading the actual code ;-)...that I have to say is very nice. I've learnt a lot of tricks from that few lines. One thing. Regarding the actual core // Cast between two boost::functions with different call signatures // wrapping the same polymorphic function object. template<class Signature, class CallSignature> function<typename call_signature<signature<Signature> >::type> functional_cast(function<CallSignature> f) { typedef typename callable_object<Signature>::type functor_type; functor_type const* p = f.template target<functor_type>(); if(!p) throw std::bad_cast(); return function<typename call_signature<signature<Signature> >::type>(*p); } I would say that for it to be equivalent to function<int(int,int)> f0 = *add_floats.target<plus>(); It may be return function<typename call_signature<signature<Signature>
::type>(boost::ref(*p));
instead of return function<typename call_signature<signature<Signature> >::type>(*p); because boost::function makes a copy as default, so you have one copy of *p in the boost::function c'tor in the return type and another, outside functional_cast in the assignment function<int(int,int)> f0 = functional_cast<...>(..); Marco