Inconsistent behaviour os str.find/str.index when providing optional parameters

[MRAB]

On 2012-11-22 03:41, Terry Reedy wrote:
> On 11/21/2012 8:32 AM, MRAB wrote:
>> On 2012-11-21 12:43, Giacomo Alzetta wrote:
>>> I just came across this:
>
>   >>> 'spam'.find('')
> 0
>   >>> 'spam'.find('', 1)
> 1
>   >>> 'spam'.find('', 4)
> 4
>
>>>>>> 'spam'.find('', 5)
>>> -1
>>>
>>>
>>> Now, reading find's documentation:
>>>
>>>>>> print(str.find.__doc__)
>>> S.find(sub [,start [,end]]) -> int
>>>
>>> Return the lowest index in S where substring sub is found,
>>> such that sub is contained within S[start:end].  Optional
>>> arguments start and end are interpreted as in slice notation.
>
> This seems not to be true, as 'spam'[4:] == 'spam'[5:] == ''
>
It can't return 5 because 5 isn't an index in 'spam'.

It can't return 4 because 4 is below the start index.

>>> Return -1 on failure.
>>>
>>> Now, the empty string is a substring of every string so how can find
>>> fail?
>>> find, from the doc, should be generally be equivalent to
>>> S[start:end].find(substring) + start, except if the substring is not
>>> found but since the empty string is a substring of the empty string it
>>> should never fail.
>>>
>> [snip]
>> I think that returning -1 is correct (as far as returning -1 instead of
>> raising an exception like .index could be considered correct!) because
>> otherwise it whould be returning a non-existent index. For the string
>> "spam", the range is 0..4.
>
> I tend to agree, but perhaps the doc should be changed. In edge cases
> like this, there sometimes is no 'right' answer. I suspect that the
> current behavior is intentional. You might find a discussion on the tracker.
>

It's a special case, but the Zen has something to say about that! :-)

(The empty string is also the only substring which can start at len(S).)