On 17Aug2017 01:03, Mok-Kong Shen <mok-kong.s...@t-online.de> wrote:
Am 17.08.2017 um 00:39 schrieb Chris Angelico:
On Thu, Aug 17, 2017 at 8:29 AM, Mok-Kong Shen
<mok-kong.s...@t-online.de> wrote:
Chris wrote:
objects exist independently of names, and names refer to
objects. If you do "x = y", you're saying "figure out which object 'y'
means, and make the name 'x' refer to it". If you do "x[1] = y",
you're saying "figure out which object 'y' means, and tell the object
that 'x' means that it should make [1] refer to that object". So if
you have multiple names referring to the same object, any change you
ask that object to do will be seen by every other name that also
refers to it - because it's all about the object.
I may have misunderstood you. But I don't think what you wrote
above would explain why the program below produces the output:
[1, 2, 3]
[3, 6, 9]
M. K. Shen
-----------------------------------------------------
def test2(alist):
alist[0],alist[1],alist[2]=3,6,9
alist=[30,60,90]
return
This is because "alist" in the function test2 is a _local_ variable. It is a
reference to the same list whose reference was passed to the function.
So:
alist[0],alist[1],alist[2]=3,6,9
This modifies the references within that list.
alist=[30,60,90]
This makes the local name "alist" refer to a shiny new list [30,60,90], totally
unrelated to the list whose reference was first passed in. Importantly, in the
main program "ss" still refers to the original list, which now has references
to the values 3, 6 and 9 in it.
def test3(alist):
alist=[30,60,90]
alist[0],alist[1],alist[2]=3,6,9
return
This code first points "alist" to a new, unrelated, list. Then modifies the
references inside that list. Put different values in this function (by using
the same values as in test2 you can see whether changes came from one or the
other) eg use 5,6,7 or something.
ss=[1,2,3]
test3(ss)
print(ss)
test2(ss)
print(ss)
So test3 first discards its reference to the [1,2,3] list, then modifies an
unrelate new list. So "ss" is unchanged, still holding [1,2,3].
Then test modifies the original list (affecting what "ss" holds) and then
points its local "alist" at something else (another shiny new list). But that
doesn't change what "ss" refers to, so it now has [3,6,9].
Cheers,
Cameron Simpson <c...@cskk.id.au> (formerly c...@zip.com.au)
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