To clarify how Python handles two equal objects

[MRAB]

On 2023-01-10 20:41, Jen Kris via Python-list wrote:
> 
> Thanks for your comments.  I'd like to make one small point.  You say:
> 
> "Assignment in Python is a matter of object references. It's not
> "conform them as long as they remain equal". You'll have to think in
> terms of object references the entire way."
> 
> But where they have been set to the same object, an operation on one will affect the other as long as they are equal (in Python).  So I will have to conform them in those cases because Python will reflect any math operation in both the array and the matrix.
> 
It's not a 2D matrix, it's a 1D list containing references to 1D lists, 
each of which contains references to Python ints.

In CPython, references happen to be pointers, but that's just an 
implementation detail.

> 
> 
> Jan 10, 2023, 12:28 by rosuav at gmail.com:
> 
>> On Wed, 11 Jan 2023 at 07:14, Jen Kris via Python-list
>> <python-list at python.org> wrote:
>>
>>>
>>> I am writing a spot speedup in assembly language for a short but computation-intensive Python loop, and I discovered something about Python array handling that I would like to clarify.
>>>
>>> For a simplified example, I created a matrix mx1 and assigned the array arr1 to the third row of the matrix:
>>>
>>> mx1 = [ [ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ] ]
>>> arr1 = mx1[2]
>>>
>>> The pointers to these are now the same:
>>>
>>> ida = id(mx1[2]) - 140260325306880
>>> idb = id(arr1) - 140260325306880
>>>
>>> That’s great because when I encounter this in assembly or C, I can just borrow the pointer to row 3 for the array arr1, on the assumption that they will continue to point to the same object.  Then when I do any math operations in arr1 it will be reflected in both arrays because they are now pointing to the same array:
>>>
>>
>> That's not an optimization; what you've done is set arr1 to be a
>> reference to that object.
>>
>>> But on the next iteration we assign arr1 to something else:
>>>
>>> arr1 = [ 10, 11, 12 ]
>>> idc = id(arr1) – 140260325308160
>>> idd = id(mx1[2]) – 140260325306880
>>>
>>> Now arr1 is no longer equal to mx1[2], and any subsequent operations in arr1 will not affect mx1.
>>>
>>
>> Yep, you have just set arr1 to be a completely different object.
>>
>>> So where I’m rewriting some Python code in a low level language, I can’t assume that the two objects are equal because that equality will not remain if either is reassigned.  So if I do some operation on one array I have to conform the two arrays for as long as they remain equal, I can’t just do it in one operation because I can’t rely on the objects remaining equal.
>>>
>>> Is my understanding of this correct?  Is there anything I’m missing?
>>>
>>
>> Assignment in Python is a matter of object references. It's not
>> "conform them as long as they remain equal". You'll have to think in
>> terms of object references the entire way.
>>
>> ChrisA
>> -- 
>> https://mail.python.org/mailman/listinfo/python-list
>>
>