Le mardi 11 juillet 2006 à 15:34 +0300, Peter Dimov a écrit :
Any set of computer data can be ordered. I don't think it means that they should all have an operator<.
Yes of course. So which types should have an operator< and which should not?
In my humble opinion, only types which model mathematical types for which mathematicians have defined a "natural" order. For example, std::complex types model the complex numbers which do not have a natural ordering. std::vector types model the word monoids which have a natural ordering (the lexicographic one) as long as the base type is totally ordered. But this point of view only gives a small intuition. Sometimes it may be difficult to take a decision. For example, if you consider that std::set types model typed sets, then they should not have an operator<. But as a matter of fact they don't (in the STL): they model ordered typed sets, so it is not clear anymore that operator< should not be defined for them. Back to the topic of operator< for smart_ptr, I think it was a mistake for the STL to use std::less as the default parameter for ordered containers. It would have been better to define some kind of total_order template class (that would have defaulted to being std::less). For smart_ptr, the situation would then be clear: no operator< but a specialization of total_order. With respect to the current STL, I don't have a strong opinion on whether smart pointers should define or not operator<. Best regards, Guillaume