Logo Questions Linux Laravel Mysql Ubuntu Git Menu
 

Practical meaning of std::strong_ordering and std::weak_ordering

I've been reading a bit about C++20's consistent comparison (i.e. operator<=>) but couldn't understand what's the practical difference between std::strong_ordering and std::weak_ordering (same goes for the _equality version for this manner).
Other than being very descriptive about the substitutability of the type, does it actually affect the generated code? Does it add any constraints for how one could use the type?
Would love to see a real-life example that demonstrates this.

like image 962
Mr. Anderson Avatar asked Jul 03 '18 17:07

Mr. Anderson


2 Answers

Does it add any constraints for how one could use the type?

One very significant constraint (which wasn't intended by the original paper) was the adoption of the significance of strong_ordering by P0732 as an indicator that a class type can be used as a non-type template parameter. weak_ordering isn't sufficient for this case due to how template equivalence has to work. This is no longer the case, as non-type template parameters no longer work this way (see P1907R0 for explanation of issues and P1907R1 for wording of the new rules).

Generally, it's possible that some algorithms simply require weak_ordering but other algorithms require strong_ordering, so being able to annotate that on the type might mean a compile error (insufficiently strong ordering provided) instead of simply failing to meet the algorithm's requirements at runtime and hence just being undefined behavior. But all the algorithms in the standard library and the Ranges TS that I know of simply require weak_ordering. I do not know of one that requires strong_ordering off the top of my head.

Does it actually affect the generated code?

Outside of the cases where strong_ordering is required, or an algorithm explicitly chooses different behavior based on the comparison category, no.

like image 152
Barry Avatar answered Oct 17 '22 06:10

Barry


There really isn't any reason to have std::weak_ordering. It's true that the standard describes operations like sorting in terms of a "strict" weak order, but there's an isomorphism between strict weak orderings and a totally ordered partition of the original set into incomparability equivalence classes. It's rare to encounter generic code that is interested both in the order structure (which considers each equivalence class to be one "value") and in some possibly finer notion of equivalence: note that when the standard library uses < (or <=>) it does not use == (which might be finer).

The usual example for std::weak_ordering is a case-insensitive string, since for instance printing two strings that differ only by case certainly produces different behavior despite their equivalence (under any operator). However, lots of types can have different behavior despite being ==: two std::vector<int> objects, for instance, might have the same contents and different capacities, so that appending to them might invalidate iterators differently.

The simple fact is that the "equality" implied by std::strong_ordering::equivalent but not by std::weak_ordering::equivalent is irrelevant to the very code that stands to benefit from it, because generic code doesn't depend on the implied behavioral changes, and non-generic code doesn't need to distinguish between the ordering types because it knows the rules for the type on which it operates.

The standard attempts to give the distinction meaning by talking about "substitutability", but that is inevitably circular because it can sensibly refer only to the very state examined by the comparisons. This was discussed prior to publishing C++20, but (perhaps for the obvious reasons) not much of the planned further discussion has taken place.

like image 38
Davis Herring Avatar answered Oct 17 '22 06:10

Davis Herring