Giovanni Piero Deretta wrote:
On 9/25/07, Neal Becker <ndbecker2@gmail.com> 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.
Thank you. I understand that now. But doesn't this conflict with the range documentation? Look at the last statement: size(x): ... std::distance(p.first,p.second) if p is of type std::pair<T> sz if a is an array of size sz end(s) - s if s is a string literal or a Char* boost_range_size(x) if that expression would invoke a function found by ADL t.size() otherwise