on Fri Mar 23 2012, Arno Schödl <aschoedl-AT-think-cell.com> wrote:
Hello,
- make adaptor expressions return const rvalues. The const rvalues will bind to T& arguments (where T is a template parameter) by deducing T to be const.
- make even the const_iterators for such adaptors mutable (writable) iterators
This accurately reflects the fact that such adaptors don't own their values.
If boost::sort is supposed to be chainable (which seems a good idea), then let's not start with adaptors. We already need a solution for
boost::sort( std::vector() );
Do we? You can't pass a std::vector() rvalue to any other mutating function. std::swap(std::vector<int>(), std::vector<int>()); // ERROR
If the solution is
template<typename Range> boost::sort( Range && );
I think the interesting question is really: what should boost::sort return? For example, it *could* return an adaptor that holds a permutation vector and sorts it lazily. Then you could slice the first ten elements off the sorted result and it would only have to do a partial_sort in the implementation. Or it could return an adaptor that holds a permutation vector and sorts it eagerly. Or, it could return a brand new container. Or, it could sort in-place and return and return a const pair of mutable iterators.
then in
boost::sort( boost::adaptors::reverse( std::vector() ) );
reverse should probably derive mutable iterators from mutable rvalues,
I don't know what you mean by "derive mutable iterators from mutable rvalues." If you mean that you ought to be able to get a mutable iterator from a mutable rvalue... IMO that sort of goes against the grain of the type system. std::vector<int>& v = std::vector<int>(); // Error std::vector<int> const& v = std::vector<int>(); // OK v.begin() -- Dave Abrahams BoostPro Computing http://www.boostpro.com