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x += ... is not the same than x = x + ... if x is mutable

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  x += ... is not the same than x = x + ... if x is mutable bartolome.sintes@gmail.com - 2013-03-20 07:17 -0700
    Re: x += ... is not the same than x = x + ... if x is mutable Jussi Piitulainen <jpiitula@ling.helsinki.fi> - 2013-03-20 16:26 +0200
    Re: x += ... is not the same than x = x + ... if x is mutable Nobody <nobody@nowhere.com> - 2013-03-21 04:27 +0000
      Re: x += ... is not the same than x = x + ... if x is mutable "Colin J. Williams" <cjw@ncf.ca> - 2013-03-21 08:35 -0400
        Re: x += ... is not the same than x = x + ... if x is mutable Steven D'Aprano <steve+comp.lang.python@pearwood.info> - 2013-03-21 12:39 +0000

#41593 — x += ... is not the same than x = x + ... if x is mutable

Hi,

I thought that x += ... was the same than x = x + ..., but today I have realized it is not true when operating with mutable objects.

In Python 3.3 or 2.7 IDLE (Windows) compare:
>>> a = [3]
>>> b = a
>>> a = a + [1]
>>> b
[3]

and
>>> a = [3]
>>> b = a
>>> a += [1]
>>> b
[3, 1]

Is this behaviour explained in the Python documentation? 

Thanking you in advance,
Bartolomé Sintes

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#41597

bartolome.sintes@gmail.com writes:

> Hi,
> 
> I thought that x += ... was the same than x = x + ..., but today I
> have realized it is not true when operating with mutable objects.
> 
> In Python 3.3 or 2.7 IDLE (Windows) compare:
> >>> a = [3]
> >>> b = a
> >>> a = a + [1]
> >>> b
> [3]
> 
> and
> >>> a = [3]
> >>> b = a
> >>> a += [1]
> >>> b
> [3, 1]
> 
> Is this behaviour explained in the Python documentation? 

Yes, it's documented in the language reference, specifically in the
latter half of the paragraph quoted below.

<http://docs.python.org/3/reference/simple_stmts.html#assignment-statements>

# An augmented assignment expression like x += 1 can be rewritten as x
# = x + 1 to achieve a similar, but not exactly equal effect. In the
# augmented version, x is only evaluated once. Also, when possible,
# the actual operation is performed in-place, meaning that rather than
# creating a new object and assigning that to the target, the old
# object is modified instead.

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#41630

On Wed, 20 Mar 2013 07:17:08 -0700, bartolome.sintes wrote:

> I thought that x += ... was the same than x = x + ..., but today I have
> realized it is not true when operating with mutable objects.

It may or may not be the same. x += y will invoke x.__iadd__(y) if x has
an __iadd__ method, otherwise x + y will be evaluated and the result
assigned to x.

In the first case, x will always continue to refer to the same object
(i.e. id(x) won't change). In the second case, x will typically (but not
always) refer to a different object.

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#41646

On 21/03/2013 12:27 AM, Nobody wrote:
> On Wed, 20 Mar 2013 07:17:08 -0700, bartolome.sintes wrote:
>
>> I thought that x += ... was the same than x = x + ..., but today I have
>> realized it is not true when operating with mutable objects.
>
> It may or may not be the same. x += y will invoke x.__iadd__(y) if x has
> an __iadd__ method, otherwise x + y will be evaluated and the result
> assigned to x.
>
Does this depend on whether Py27 or Py32 is used?

Colin W.
> In the first case, x will always continue to refer to the same object
> (i.e. id(x) won't change). In the second case, x will typically (but not
> always) refer to a different object.
>

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#41647

On Thu, 21 Mar 2013 08:35:26 -0400, Colin J. Williams wrote:

> On 21/03/2013 12:27 AM, Nobody wrote:
>> On Wed, 20 Mar 2013 07:17:08 -0700, bartolome.sintes wrote:
>>
>>> I thought that x += ... was the same than x = x + ..., but today I
>>> have realized it is not true when operating with mutable objects.
>>
>> It may or may not be the same. x += y will invoke x.__iadd__(y) if x
>> has an __iadd__ method, otherwise x + y will be evaluated and the
>> result assigned to x.
>>
> Does this depend on whether Py27 or Py32 is used?

No. It is equally true for all versions of Python.


-- 
Steven

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