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]
---
(PythonWin 2.3.4 (#53, Oct 18 2004, 20:35:07) [MSC v.1200 32 bit
(Intel)] on win32.)
Tried it, it works as expected.
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 parameter values are evaluated when the function definition
is executed. This means that the expression is evaluated once, when
the function is defined, and that that same ``pre-computed'' value is
used for each call. This is especially important to understand when a
default parameter is a mutable object, such as a list or a dictionary:
if the function modifies the object (e.g. by appending an item to a
list), the default value is in effect modified. "
Surely the variable i is a mutable object?
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 mutable objects and not the other mutable objects?
I'm stymied.
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