On Thu, Nov 27, 2014 at 9:53 AM, Nick Coghlan <ncogh...@gmail.com> wrote: > The implicit stop decorator would then check the flags on the code object > attached to the passed in function. If GENERATOR wasn't set, that would be > an immediate ValueError, while if EXPLICIT_STOP wasn't set, the generator > function would be passed through unmodified. However, if EXPLICIT_STOP *was* > set, the generator function would be replaced by a *new* generator function > with a *new* code object, where the only change was to clear the > EXPLICIT_STOP flag.
Is it possible to replace the code object without replacing the function? Imagine if you have multiple decorators, one of which retains a reference to the function and then this one which replaces it - the order of decoration would be critical. OTOH, I don't know that anyone would retain references to __code__. ChrisA _______________________________________________ Python-Dev mailing list Python-Dev@python.org https://mail.python.org/mailman/listinfo/python-dev Unsubscribe: https://mail.python.org/mailman/options/python-dev/archive%40mail-archive.com
