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Sat Mar 17 23:07:33 CET 2012 · https://mail.python.org/pipermail/python-dev/2012-March/117770.html

On 3/17/2012 4:47 PM, Michael Foord wrote:
> 
> On 17 Mar 2012, at 08:49, Georg Brandl wrote:
> 
>> On 03/15/2012 01:17 AM, victor.stinner wrote:

>>> +   If available, a monotonic clock is used. By default, if *strict* is False,
>>> +   the function falls back to another clock if the monotonic clock failed or is
>>> +   not available. If *strict* is True, raise an :exc:`OSError` on error or
>>> +   :exc:`NotImplementedError` if no monotonic clock is available.
>>
>> 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.

I have to agree with Georg. Looking at the code, it appears OSError can
be raised with both strict=True and strict=False (since floattime() can
raise OSError). The text needs to make it clear OSError can always be
raised.

I also think "By default, if strict is False" confuses things. If
there's a default behavior with strict=False, what's the non-default
behavior? I suggest dropping "By default".

Eric.

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Showing content from https://mail.python.org/pipermail/python-dev/2012-March/117770.html below:

[Python-Dev] cpython: Issue #10278: Add an optional strict argument to time.steady(), False by default