On Tue, Mar 25, 2008 at 2:43 PM, Alexander Belopolsky <[EMAIL PROTECTED]> wrote: > On Tue, Mar 25, 2008 at 4:26 PM, Adam Olsen <[EMAIL PROTECTED]> wrote: > > It does not even have to be a frozenset. A set works just as well, > > never modified by the produced bytecode. > > With the current implementation, precomputed constants must be > hashable because the compiler uses a dictionary lookup in order to > eliminate duplicates. This is of course just an implementation > detail, but it would actually be hard to work around it.
Only builtin types with literal syntax may be deemed "constant" anyway. I fail to see how that's relevant to my frozenset vs set comment. -- Adam Olsen, aka Rhamphoryncus _______________________________________________ Python-3000 mailing list [email protected] http://mail.python.org/mailman/listinfo/python-3000 Unsubscribe: http://mail.python.org/mailman/options/python-3000/archive%40mail-archive.com
