On 8/15/2012 5:58 PM, Rob Day wrote:
> Madison May wrote:
The list nlist inside of function xx is not the same as the variable
u outside of the function: nlist and u refer to two separate list
objects. When you modify nlist, you are not modifying u.
<http://mail.python.org/mailman/listinfo/python-list>
This is confused and wrong. The parameter *name* 'nlist' of function xx
is not the same as the *name* 'u' outside the function. The call xx(u)
binds nlist to the same object that u is bound to. At that point, the
two name *are* bound to the same list object. The statement
"nlist+=[999]" dodifying nlist *does* modify u. The subsequent
assignment statement "nlist=nlist[:-1]" rebinds 'nlist' to a *new* list
object. That new object gets deleted when the function returns. So the
rebinding is completely useless.
This sequence, modifying the input argument and then rebinding to a new
object, is bad code.
Well - that's not quite true. Before calling the function, u is [1, 2,
3, 4] - but after calling the function, u is [1, 2, 3, 4, 999]. This is
a result of using 'nlist += [999]' - the same thing doesn't happen if
you use 'nlist = nlist+[999]' instead.
I'm not completely aware of what's going on behind the scenes here, but
you got it right.
I think the problem is that 'nlist' is actually a reference to a list
object - it points to the same place as u.
Calling a python function binds parameter names to argument objects or
(for *args and **kwds parameters) a collection based on argument objects.
When you assign to it within
the function, then it becomes separate from u - which is why nlist =
nlist+[999] and nlist = nlist[:-1] don't modify u - but if you modify
nlist in place before doing that, such as by using +=, then it's still
pointing to u, and so u gets modified as well.
--
Terry Jan Reedy
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