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

28 Nov 2003

      Matt S Trentini wrote:
...
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 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.
They aren't required to. You should use
std::iterator_traits<It>::value_type.

[...]
...
Am I overlooking something here?
1. Iterators are typically passed by value.
2. You don't need to use boost::function in a template. It is typically only
used when you want a fixed compile-time signature.

template
void Extract(InIt first, InIt last, OutIt out, F f)
{
    for (; first != last; ++first, ++out)
    {
        *out = f(*first);
    }
}

Now look at std::transform. :-)

(You'll no longer be able to pass &X::f directly, though, use mem_fn
explicitly.)