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 -- https://mail.python.org/mailman/listinfo/python-list
