you wrote a different problem. The tutorial should be like this:
Important warning: The default value is evaluated only once. This makes a difference when the default is a mutable object such as a list, dictionary, or instances of most classes. For example, the following function accumulates the arguments passed to it on subsequent 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]If you don't want the default to be shared between subsequent calls, you can write the function like this instead:
def f(a, L=None):
if L is None:
L = []
L.append(a)
return L
Look the second example use None instead of []
To explain more detail let's create an example :
def f1(a, L=[]):
L.append(a)
return L
def f2(a, L=None):
if L is None:
L = []
L.append(a)
return L
def main():
m = f2(3)
print m # produce [3]
m = f2(4)
print m # produce [4]
# ----------------
m = f1(3)
print m # produce[3]
m = f1(4)
print m # produce[3,4]
pass
in f1 : when we don't put the second argument the L=[] is created only once. The second input in f1 will be accumulated.
in f2: when we don't put the second argument L is given a None value. Inside this function if L is None then L = [] this will always make L = [] that's way no accumulation happened.
Hope this help.
pujo
On 3/29/06, Josh Adams <[EMAIL PROTECTED]> wrote:Hi all,
I was going through the tutorial at http://docs.python.org/tut/node6.html when I
came to the bit about default arguments with this code:
def f(a, L=[]):
L.append(a)
return L
print f(1)
print f(2)
print f(3)
returns:
[1]
[1, 2]
[1, 2, 3]
>From the postings here, I think I understand that this occurs because L is only
initialized when f is first run. However, this code gives some different results:
def f2(a, L=[]):
if L == []:
L = []
L.append(a)
return L
print f2(1)
print f2(2)
print f2(3)
returns:
[1]
[2]
[3]
I'm not too clear on why this doesn't return the same results as the first. Can
someone enlighten me?
Thanks,
Josh
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