Steven, Thanks! That is quite helpful to know the nuts and bolts of how that works.
--Bill On Tue, Jan 25, 2011 at 07:45, Steven D'Aprano <st...@pearwood.info> wrote: > Bill Allen wrote: > >> Ok, I have definately verified this to myself. The following works >> perfectly and is a little easier to understand. In this version, I am >> plainly modifying my parts_list iterator thus producing the effect of an >> iterator that is growing over the course of the operation of the code. >> So, >> I am convinced that I had previously assigned part_list to out_list by >> reference, not value as I mistaken thought when I first wrote the code, >> > > I believe you're working under a misunderstanding there. Python is neither > by reference nor value, if I've understood what you mean by that. (Normally > people talk about "by reference" when discussing calling functions, not > assignment, so I may have misunderstood you.) > > After executing the line `out_list = part_list`, *both* of these statements > are *incorrect*: > > "out_list is a copy of part_list" +++ WRONG +++ > "out_list and part_list are the same variable (out_list is an alias for > part_list)" +++ ALSO WRONG +++ > > In a nutshell, Python uses the *name binding* assignment model. When you do > this: > > part_list = [1, 2] > > Python creates a *name* "part_list" and binds it to the *object* [1, 2]. > Then when you do this: > > out_list = part_list > > the name "out_list" is created and bound to the same object as part_list. > This means that out_list and part_list are two names for the same object, > but this is not a permanent state of affairs -- they happen to be bound to > the same object now, but you can re-bind one without changing the other. > > A month or so ago, somebody raised a similar question about pass by > reference and pass by value. If you will excuse me blowing my own trumpet, I > think my response then may be useful for you now: > > http://www.mail-archive.com/tutor%40python.org/msg46612.html > > To use a copy of part_list, you need to explicitly copy it. For lists, the > easiest way is with a slice: > > out_list = part_list[:] > > You can also use the copy module, and functions copy.copy and > copy.deepcopy, to copy other objects. > > > > which explains it. It was a silly mistake born from still being new in >> Python and thinking in terms of another language I know that typically >> assigns by value instead. It had no occurred to me initially that it was >> possible to modify an iterator in this way. I do not think most languages >> would allow this. >> > > My understanding is that this behaviour -- multiple names for one object -- > is standard semantics for most modern languages, including Perl, Ruby, Java, > and many others. > > > > > -- > Steven > > _______________________________________________ > Tutor maillist - Tutor@python.org > To unsubscribe or change subscription options: > http://mail.python.org/mailman/listinfo/tutor >
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