[Boost-users] Re: boost::function - why does this work??

From: Matt S Trentini (matt_s_trentini_at_[hidden])
Date: 2003-11-27 19:01:46

Peter Dimov wrote:

> When you construct a boost::function from &HandleClass::handle, what
> actually gets stored is boost::mem_fn(&HandleClass::handle). See
> 
>     http://www.boost.org/libs/mem_fn/
> 
> Even
> 
>      std::vector< boost::shared_ptr<HandleClass> > listOfHcs;
> 
> will work.

Thanks Peter, makes sense after reading that documentation now.

I've actually got another question - I'd really prefer to replace 
ReturnContainer with an output iterator type.  So Extract would look 
like this:

     template
     <
         class InIterator,
        class OutIterator,
     >
     void Extract(const InIterator & itBegin,
                  const InIterator & itEnd,
                  OutIterator      & itOut,
                  boost::function1<OutIterator::value_type, 
FwdIterator::value_type> f)
     {
         for (FwdIterator it = itBegin ;it != itEnd; ++it)
         {
             *itOut++ = f(*it);
         }
     };

And then you'd typically call Extract with an inserter:

     Extract(listOfHcs.begin(), listOfHcs.end(), 
back_inserter(needToFill), &HandleClass::handle);

But this doesn't compile because the inserters (at least on my platform) 
don't define value_type.  I'd prefer to take an iterator because it 
gives greater flexibility (can use inserter, front_inserter or 
back_inserter) and it more closely models the standard algorithms. 
However, I need to know the return type of the function and I can't see 
any way to get it without specifying it as a template parameter.

Actually, I think I could use the typedefs that are defined in the 
inserters to retrieve the container, and then the value type, but this 
then forces the client to use inserters.

Am I overlooking something here?

Cheers,
Matt