On Feb 27, 5:38 pm, [EMAIL PROTECTED] wrote:
> On Feb 27, 4:16 pm, "Gabriel Genellina" <[EMAIL PROTECTED]>
> > For a), you use something like obj.a.somemethod(). "obj.a" still refers to
> > the same object, even if it changed internally; if obj.a and foo.bar both
> > were refering to the same objec
On Feb 27, 4:16 pm, "Gabriel Genellina" <[EMAIL PROTECTED]>
wrote:
> En Tue, 26 Feb 2008 05:58:52 -0200, <[EMAIL PROTECTED]> escribió:
>
> > It works to keep a reference to the object, then access the member
> > again. If your only reference to the object is the member itself,
> > obj.a= {} breaks
En Tue, 26 Feb 2008 05:58:52 -0200, <[EMAIL PROTECTED]> escribió:
> It works to keep a reference to the object, then access the member
> again. If your only reference to the object is the member itself,
> obj.a= {} breaks d, but obj.a.clear() keeps it alive.
>
> In the case of immutables, there i
On Feb 26, 1:11 pm, Dennis Lee Bieber <[EMAIL PROTECTED]> wrote:
> On Mon, 25 Feb 2008 23:58:52 -0800 (PST), [EMAIL PROTECTED] declaimed
> the following in comp.lang.python:
>
> > The generic solution involves a second level of indirection: tuple* d=
> > &obj.a. I can't land a clean solution to it
On Feb 25, 11:30 pm, Dennis Lee Bieber <[EMAIL PROTECTED]> wrote:
> On Mon, 25 Feb 2008 17:55:18 -0800 (PST), [EMAIL PROTECTED] declaimed
> the following in comp.lang.python:
>
> > I'd like to do this:
>
> > a= list( range( 5 ) )
> > assert a== [ 0, 1, 2, 3, 4 ]
> > for i in ref( a ):
> > i.ref
I'd like to do this:
a= list( range( 5 ) )
assert a== [ 0, 1, 2, 3, 4 ]
for i in ref( a ):
i.ref*= 2
a= deref( a )
assert a== [ 0, 2, 4, 6, 8 ]
In the for loop, i objects maintain their identities, while still
being reassigned. The first way I think of is this:
class Ref:
def __init__(
