possible python bug here

[manuel]

In the sample below, the foo function modify the b list,
but I think it should modify only c, not b! It work
correctly if if b is one dimension list instead two.

def foo(aList):
print "use foo..."
aList[2][0] += .35
aList[2][1] += .35
aList[2][2] += .35

p = [2.5,2.5,2.5]
b = [p,p,p,p]

c=b[:] #c is a totally new list cloned from b, it's not an alias!

print "b = ",b[2]

foo(c) #The argument is c, not b! Why b is modified too?

print "b = ", b[2]# after foo(c), the list b now is different!

[Peter Kleiweg]

Re: possible python bug here

manuel schreef:
[color=blue]
> In the sample below, the foo function modify the b list,
> but I think it should modify only c, not b! It work
> correctly if if b is one dimension list instead two.
>
> def foo(aList):
> print "use foo..."
> aList[2][0] += .35
> aList[2][1] += .35
> aList[2][2] += .35
>
> p = [2.5,2.5,2.5]
> b = [p,p,p,p]
>
> c=b[:] #c is a totally new list cloned from b, it's not an alias![/color]

No it isn't. You didn't make a deep copy.



--
Peter Kleiweg L:NL,af,da,de,e n,ia,nds,no,sv, (fr,it) S:NL,de,en,(da, ia)
info: http://www.let.rug.nl/~kleiweg/ls.html

The Halloween Documents: http://www.opensource.org/halloween/

[manuel]

Re: possible python bug here

Peter Kleiweg wrote:[color=blue]
> manuel schreef:[color=green]
>>c=b[:] #c is a totally new list cloned from b, it's not an alias![/color]
>
>
> No it isn't. You didn't make a deep copy.[/color]

Oh..thank you very much!

[Tuure Laurinolli]

Re: possible python bug here

manuel wrote:
[color=blue]
> In the sample below, the foo function modify the b list,
> but I think it should modify only c, not b! It work
> correctly if if b is one dimension list instead two.
>
> def foo(aList):
> print "use foo..."
> aList[2][0] += .35
> aList[2][1] += .35
> aList[2][2] += .35
>
> p = [2.5,2.5,2.5][/color]
Here you bind the name p to a list.[color=blue]
> b = [p,p,p,p][/color]
Here you bind the name b to a list, which has four references to list p[color=blue]
>
> c=b[:] #c is a totally new list cloned from b, it's not an alias![/color]
Here you bind the name c to a list, which has four copies of the
references to list p[color=blue]
>
> print "b = ",b[2][/color]
Here you print the third reference to p of list b[color=blue]
>
> foo(c) #The argument is c, not b! Why b is modified too?[/color]
Here you modify the third reference to p of list c[color=blue]
>
> print "b = ", b[2]# after foo(c), the list b now is different![/color]
Here you print the third reference to p of list b, it's still just
reference to list (originally bound to) p.

Note that also the other references to the changed list (b[0:4], c[0:4],
p) refer to the same changed list.