"Sean Parent" <sparent@adobe.com> skrev i meddelandet news:BC803577-3DC2-4D20-BB64-84E72A501CEB@adobe.com...
Christoph Ludwig posted the correct reason why std::complex should not have an operator < () defined - sorry, my follow up using addition was poorly constructed.
What I am proposing is that std::less should be defined for std::complex.
Disagree. That could potentially break other template programming using std::less, to support std::unary_negate etc. Suddenly this would "work" for std::complex, when it shouldn't.
Why would we have a separate default relation for set/map then std::less? std::less is intended to map to operator <() on the type _or_ to a total ordering for the type if operator <() is not defined.
No, that's just for pointers. 20.5.7/8 says: "For templates greater, less, greater_equal, and less_equal, the specializations for any pointer type yield a total order, even if the built-in operators <, >, <=, >= do not." It doesn't mention std::complex or any other composite types.
For pair<shared_ptr<>, int> - what makes shared_ptr special?
The fact that it is a pointer? I don't care much if it doesn't have an ordering.
std::less is supposed to be operator <, unless operator < is not defined in which case std::less provides a total ordering. It isn't ambiguous.
Where does the standard say that? If you want to have an ordered collection of otherwise unordered data, the relation must be an application specific attribute. So define an appropriate functor, and use that with std::map, etc. Bo Persson