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

Terry Reedy tjreedy at udel.edu
Thu Nov 22 04:41:01 CET 2012


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:] == ''

>> 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.

-- 
Terry Jan Reedy



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