nikhil <nik....@gmail.com> wrote:
Hi,
I am learning Python and came across this example in the Python Online
Tutorial (
http://docs.python.org/tutorial/controlflow.html#default-argument-values)
for Default Argument Values.
ex:
def func(a, L=[ ]):
L.append(a)
return L
print func(1)
print func(2)
print func(3)
*O/P*
[1]
[1,2]
[1,2,3]
Now my doubt is,
1) After the first function call, when ' L' (Reference to List object) goes
out of scope, why isn't it destroyed ?
2) It seems that this behavior is similar to static variable logic in C
language. Is it something similar ? OR
Is it related to scope rules in Python ?
The L parameter is a default parameter. That's a formal parameter with
an initial value (L=[]). The reference for that value is established
when the definition is executed, not when the function is called. The
net effect when the function is at top-level is similar to a static
variable in C. But it's not really L that is retained, but the object
its bound to. Perhaps it's easiest to think that the function object
has a reference to the same object, and copies it to L each time the
function is called without that particular formal parameter.
See the online help (this one's for 2.6.2, but other recent versions
are the same):
http://www.python.org/doc/2.6.2/reference/compound_stmts.html#function-definitions
Usually, you want to make such an initial value a non-mutable value,
such as None, 42, or "". But if it's mutable, it will indeed retain its
value from one call to the next of that function. As you observed.
There are times when this behavior is useful (eg. nested definitions),
and I think it's permanently established in the lanaguage. But it is
perhaps one of the most common beginner confusions. And it's part of
the solution to some of the other common problems.
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