On 2022-03-12 at 21:45:56 +0100, Marco Sulla <marco.sulla.pyt...@gmail.com> wrote:
[ ... ] > So if I do not cache if the object is unhashable, I save a little > memory per object (1 int) and I get a better error message every time. > On the other hand, if I leave the things as they are, testing the > unhashability of the object multiple times is faster. The code: > > try: > hash(o) > except TypeError: > pass > > execute in nanoseconds, if called more than 1 time, even if o is not > hashable. Not sure if this is a big advantage. Once hashing an object fails, why would an application try again? I can see an application using a hashable value in a hashable situation again and again and again (i.e., taking advantage of the cache), but what's the use case for *repeatedly* trying to use an unhashable value again and again and again (i.e., taking advantage of a cached failure)? So I think that caching the failure is a lot of extra work for no benefit. -- https://mail.python.org/mailman/listinfo/python-list
