Re: [boost] [function] how to create function objects from bind expr automatically

From: Tobias Schwinger (tschwinger_at_[hidden])
Date: 2007-05-30 08:28:13

Hello Markus,

Markus Werle wrote:
> Hi!
> 
> What I want to achieve: automatically build
> a boost::function from a bind expression
> without having to specifiy the signature.

the function objects returned from boost::bind have a templatized 
operator(), so they don't really have a signature.

When you assign such a thing to Boost.Function it instantiates 
operator() for a given signature and removes the type.

You can't turn a template into code without instantiating it and you 
can't store a template - only instances thereof (there is no runtime 
representation - stuff like this would require a (partially) interpreted 
language).

>>From what I understand in the boost::bind code
> it is not possible to extract the signature, arity or 
> return type information from an object returned by boost::bind.

Yes, and for signature it's impossible to implement (in general) because 
what's bound might again be a function object with a templatized 
operator(). Plus the result type might be a function of the argument types.

> So I have no idea, how to write code that allows the 
> registration of arbitrary function pointers as boost::function
> objects depending on the argument list.

So I guess you want something like

     some_class c;
     c.register( & some_function );
     c.register( & another_function );

, right?

> Any hint to possible solutions appreciated.

In either case the recipe is to have the 'register' template generate 
some kind of "invoker" based on the signature, so you can "bury" the 
type in the invoker and forget about it once the invoker (which always 
has the same type) is stored. Here's an example that shows the exact case:

     $(CVS_HEAD)/libs/function_types/example/interpreter*

This one might also be interesting to you:

     $(CVS_HEAD)/fusion/example/cookbook/do_the_bind.cpp

Regards,
Tobias