Re: [boost] boost::range needs help

From: Giovanni Piero Deretta (gpderetta_at_[hidden])
Date: 2007-09-25 11:14:59

On 9/25/07, Neal Becker <ndbecker2_at_[hidden]> wrote:
> Thorsten Ottosen wrote:
>
> > Neal Becker skrev:
> >> boost::size of a std::list gives:
> >>
> >> /usr/local/src/boost.hg/boost/range/size.hpp:28: error: no match
> >> for 'operator-' in 'boost::end [with T = std::list<double,
> >> std::allocator<double> >](((const std::list<double,
> >> std::allocator<double>
> >>> &)((const std::list<double, std::allocator<double> >*)r))) -
> >>> boost::begin
> >> [with T = std::list<double, std::allocator<double> >](((const
> >> std::list<double, std::allocator<double> >&)((const std::list<double,
> >> std::allocator<double> >*)r)))'
> >>
> >> I would hope that boost::size (type with member .size()) would use the
> >> member .size().
> >
> > Use boost::distance( my_list );
> >
> > size() was changed to reflect O(1) complexity.
> >
> > -Thorsten
>
> Sorry, I don't think that was my question.  boost::size(T t) has a default
> implementation using t.end()-t.begin() (in boost/range/size.hpp).  I can
> supply my own implementation of size(std:list<T>).  I was wishing that
> boost::size would implement boost::size using the member .size() if it
> exists.
>

I think that what Thorsten was saying ist that the whole point of
using operator- is to guarantee that boost::size is *always* O(1).

With some list implementations size is O(N), so forwarding to .size()
would violate the guarantee. If you really need to know the size but
do not care of the complexity guarantee, you can use boost::distance.

gpd