On Tue, Feb 23, 2010 at 1:13 PM, Giorgio <anothernetfel...@gmail.com> wrote:
> I have an update:
> I can easily undertand why this example doesn't work:
> def nochange(x):
> x = 0
> y = 1
> nochange(y)
> print y # Prints out 1
> X is a local variable, and only gets modified in the function, that doesn't
> return any value.
> But it's very difficult for me to understand WHY this works:
> def change(some_list):
> some_list[1] = 4
> x = [1,2,3]
> change(x)
> print x # Prints out [1,4,3]
> some_list is a "local" list, isn't it? Maybe i can't have lists that are
> only existing in a function?
Here is what happens, as I understand it:
When you enter the function, a new name (x, in your case) is created
in the local scope. That name points to the object you supplied when
you called the function (an integer object, with a value of 1). the x
= 0 statement creates a new object, and has the name x now pointing to
this new object. The old integer object still exists, and y still
points to it. This is why the global y name is not affected by the
change in x
Now, in the second example, the same thing basically happens. A new
name is created and pointed at the object you supplied. However, the
statement some_list[1] = 4 is different from the assignment, in that
it doesn't create a new object; It modifies the existing one. Since
the global and local names both point to the same object, the change
you make is reflected in both.
You can of course create a new list object, so that the original is
not affected:
def nochange(some_list):
# create shallow copy of list. NOTE: Shallow copies may still bite
you if you change the list members.
some_list = some_list[:]
some_list[1] = 2
>>> x = [1, 2, 3]
>>> nochange(x)
>>> x
[1, 2, 3]
HTH,
Hugo
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