>>> This is not clear to me. Why wouldn't it raise OSError on error even with >>> strict=False? Please clarify which exception is raised in which case. >> >> It seems clear to me. It doesn't raise exceptions when strict=False because >> it falls back to a non-monotonic clock. If strict is True and a non-monotonic >> clock is not available it raises OSError or NotImplementedError. > > So errors are ignored when strict is false?
Said differently: time.steady(strict=True) is always monotonic (*), whereas time.steady() may or may not be monotonic, depending on what is avaiable. time.steady() is a best-effort steady clock. (*) time.steady(strict=True) relies on the OS monotonic clock. If the OS provides a "not really monotonic" clock, Python cannot do better. For example, clock_gettime(CLOCK_MONOTONIC) speed can be adjusted by NTP on Linux. Python tries to use clock_gettime(CLOCK_MONOTONIC_RAW) which doesn't have this issue. Victor _______________________________________________ Python-Dev mailing list Python-Dev@python.org http://mail.python.org/mailman/listinfo/python-dev Unsubscribe: http://mail.python.org/mailman/options/python-dev/archive%40mail-archive.com
