The order of iterable de-referencing in assignment?

[Chris Angelico]

On Wed, Aug 24, 2016 at 9:03 PM, Joaquin Alzola
<Joaquin.Alzola at lebara.com> wrote:
>>> One might guess a.extend(a) would turn into an infinite loop.  It turns out here Python first gets all the items of `a' and then append them to `a', so the infinite loop is avoided.
>>>> a = [1,2]
>>>> for x in a: a.append(x)
>>...
>>^CTraceback (most recent call last):
>  > File "<stdin>", line 1, in <module>
>>KeyboardInterrupt
>>>>> len(a)
>>6370805
>
>>That right there, folks, is an infinite loop.
>
> If I am correct python points out an infinite loop with the "...", just pointing to more information.

That's true of self-referential objects and circular references:

>>> a = [1, 2]
>>> a.append(a)
>>> a
[1, 2, [...]]

In this case, it's not an infinite loop or anything; it's simply an
object that references itself:

>>> id(a)
139945904550344
>>> [id(x) for x in a]
[9241344, 9241376, 139945904550344]

The list has three elements, one of which is the list itself. Same
applies if the list has a reference to something else which has a
reference to the original list, or anything along those lines.

But "a.extend(a)" isn't quite like that. In its purest form, it means
the same as the 'for' loop that I showed above, but... well, this
comment from the CPython sources says exactly what I'm thinking of:

Objects/listobject.c:795

    /* Special cases:
       1) lists and tuples which can use PySequence_Fast ops
       2) extending self to self requires making a copy first
    */

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ChrisA