Bulba! wrote:
Thanks everyone (esp. Fredrik, http://www.effbot.org/zone/index.htm
contains lotsa goodies by him I intend to study), I'm going offline to
reboot my brain. ;-)
Bulba, with reference to your "don't laugh" comment, I'd just like to
say that for a newbie your questions do have a habi
Thanks everyone (esp. Fredrik, http://www.effbot.org/zone/index.htm
contains lotsa goodies by him I intend to study), I'm going offline to
reboot my brain. ;-)
--
It's a man's life in a Python Programming Association.
--
http://mail.python.org/mailman/listinfo/python-list
"Bulba!" wrote:
> Surely the variable i is a mutable object?
nope.
> OK, again:
>
def t(a,i=[0]):
> ... i[0]=i[0]+1
> ... return (a, i)
> ...
t(1)
> (1, [1])
t(3)
> (3, [2])
t(5)
> (5, [3])
>
> Yes, it works with the list. But why sharing value between calls
> pertains some m
On Wed, 22 Dec 2004 22:29:44 GMT, "Matt Gerrans" <[EMAIL PROTECTED]>
wrote:
>Actually i was not mutable. Try this:
>i = 1
>id(i)
>i += 1
>id(i)
Looks like I will have to read the Language Reference anyway. :-(
>Because you are assigning the local reference variable L to a new list,
>instead
Actually i was not mutable. Try this:
i = 1
id(i)
i += 1
id(i)
Change both of your sample functions to also print the id of the default
variable and that might shed a little more light on the matter.
"i = i + 1" is only changing the local reference called i to refer to an
instance of a diffe
Bulba! wrote:
> [...]
> But why the following happens?
>
>
def j(a,i=0):
>
> ... i=i+1
> ... return (a, i)
> ...
>
j(2)
>
> (2, 1)
>
j(3)
>
> (3, 1)
>
j(5)
>
> (5, 1)
>
>
> From Language Reference:
>
> http://www.python.org/doc/2.3.4/ref/function.html
>
> "Default
OK. Don't laugh.
There's this example from tutorial, showing how default
value is computed once when defining function and shared
between function calls:
---
def f(a, L=[]):
L.append(a)
return L
print f(1)
print f(2)
print f(3)
This will print
[1]
[1, 2]
[1, 2, 3]
---
(Pyth
