On Tue, 01 Dec 2015 16:18:49 -0500, Terry Reedy wrote:

> On 12/1/2015 3:32 PM, Denis McMahon wrote:
>> On Tue, 01 Dec 2015 03:32:31 +0000, MRAB wrote:
>>
>>> In the case of:
>>>
>>>       tup[1] += [6, 7]
>>>
>>> what it's trying to do is:
>>>
>>>       tup[1] = tup[1].__iadd__([6, 7])
>>>
>>> tup[1] refers to a list, and the __iadd__ method _does_ mutate it, but
>>> then Python tries to put the result that the method returns into
>>> tup[1].
>>> That fails because tup itself is a tuple, which is immutable.
>>
>> I think I might have found a bug:
> 
> What you found is an specific example of what MRAB said in general
> above.
> 
>> $ python Python 2.7.3 (default, Jun 22 2015, 19:33:41)
>> [GCC 4.6.3] on linux2 Type "help", "copyright", "credits" or "license"
>> for more information.
>>>>> tup = [1,2,3],[4,5,6]
>>>>> tup
>> ([1, 2, 3], [4, 5, 6])
>>>>> tup[1]
>> [4, 5, 6]
>>>>> tup[1] += [7,8,9]
>> Traceback (most recent call last):
>>    File "<stdin>", line 1, in <module>
>> TypeError: 'tuple' object does not support item assignment
> 
> The bug is trying to replace a member of a tuple.  The correct code, to
> avoid the exception while extending the list, is
> 
> tup[1].extend([7,8,9])
> 
>>>>> tup[1]
>> [4, 5, 6, 7, 8, 9]

You snipped the important bit of my original post, which was the state of 
tup after the TypeError occurred.

After the error, 

>>> tup[1]
[4, 5, 6, 7, 8, 9]
>>> tup
([1, 2, 3], [4, 5, 6, 7, 8, 9])

The "bug" I refer to is that despite giving the TypeError, the tuple 
allowed the assignment of the mutated list to replace the original list.

-- 
Denis McMahon, denismfmcma...@gmail.com
-- 
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