Newbie References Question -> Exemple

[Guy Rabiller]

Sorry I should have said:

Thanks to Gustavo, not sismex :-)

--
guy rabiller
3d animator / td
grabiller at 3dvf.net
http://grabiller.3dvf.net

Guy Rabiller wrote:
> Ok I solved it, thanks to sismex.
>
> If I want to change a coordinate, I have yo write:
>
> p[0][:] = [5,5]
>
> and not
>
> p[0] = [5,5]
>
> I understand that in the first case, the reference pointed ( via
> dictionary ) by p[0] is lost, a new value is created and rebinded to
> p[0], so quad and p[0] are not synchonised anymore.
>
> In the second case, the reference pointed by p[0] is updated with a
> new value. p[0] and quad are still synchronised.
>
> Is my understanding correct ?
>
> Thanks a lot to all.
>
>
> Guy Rabiller wrote:
>> Ok,
>>
>> thanks for your answers, I understand now that a variable name is
>> 'just' a name in a dictionary, and not a container.
>>
>> What I want to do is simple, and I suppose that this is my 'way of
>> thinking' regarding Python that is wrong.
>>
>> I have a list of points, and a list of polygons that should contain
>> 'references' to points.
>>
>> So for me:
>> -> points list:
>> p = []
>> p.append([0,0])
>> p.append([0,1])
>> p.append([1,1])
>> p.append([1,0])
>> -> polygon:
>> quad = []
>> quad.append(p[0])
>> quad.append(p[1])
>> quad.append(p[2])
>> quad.append(p[3])
>>
>> But now, if I change the coordinate of one point:
>> p[0] = [5,5]
>> I will have:
>>>>> p
>> [[5, 5], [0, 1], [1, 1], [1, 0]]
>> but:
>>>>> quad
>> [[0, 0], [0, 1], [1, 1], [1, 0]]
>>
>> For me the quad list of point is not 'updated' with the change of one
>> of its point.
>>
>> What I want to do is that, quad keeps a list of 'references' to
>> points coordinates.
>> Then if I change one point coordinates, the change is reflected in
>> the quad list of points.
>>
>> I have no clue how to do this.
>>
>>
>> Guy Rabiller wrote:
>>> Hi,
>>>
>>> let say I have:
>>> i1 = 1
>>> i2 = 2
>>> and
>>> p = [i1,i2]
>>>
>>> How can I have:
>>> p=[*i1,*i2]
>>> rather than
>>> p=[**i1,**i2] as it is curently ?
>>>
>>> ( Sorry for this nasty hybrid syntax )
>>>
>>> Textualy, how if I want that p[0] refere to the 'i1' container, and
>>> not to the refererence it contains ?
>>>
>>> What I want is that if now I set:
>>> i1 = 4
>>> that automaticaly:
>>> p -> [4,2]
>>> and not keeping [1,2]
>>>
>>> Thanks in advance.