Something in the function tutorial confused me.

[Alex Popescu]

Stargaming <stargaming at gmail.com> wrote in news:46b6df49$0$26165
$9b622d9e at news.freenet.de:

> On Sun, 05 Aug 2007 23:50:24 -0700, Lee Fleming wrote:
> 
>> Hello,
>> I have a simple question. Say you have the following function:
>> 
>> def f(x, y = []):
>>     y.append(x)
>>     return y
>> 
>> print f(23)  # prints [23]
>> print f(42)  # prints [23, 42]
>> 
>> As far as I understand, the default value y, an empty list, is 
created
>> when the def statement evaluates. With this thought in mind, the 
above
>> calls
>> to f make sense.
>> 
>> But this, the code that "fixes" the list accumulation confounds me: 
def 
>> f(x, y=None):
>>     if y is None: y = []
>>     y.append(x)
>>     return y
>> 
>> print f(23)  # prints [23]
>> print f(42)  # prints [42]
>> 
>> Why didn't the second call to f, f(42) return [23, 42]? As I 
understand
>> it, y is only None at the beginning of f(23). Then y changes from 
None
>> to 23. When f ends, doesn't y still have 23 in it,
>> just as it did in the first function I discussed? And if y has 23 in 
it,
>> won't the second call to f not execute what's in the if statement?
>> 
>> In other words, what's going on here? How is it that y accumulates
>> argument values between function calls in the first function, but
>> doesn't in the second one?
> 
> You're just unluckily shadowing the name `y` in the local scope of 
your 
> function. Your code snippet could be rewritten as::
> 
>   def f(x, y=None):
>     if y is None: my_y = []
>     else: my_y = y
>     my_y.append(x)
>     return my_y
> 
> HTH,
> Stargaming

For the given example this will continue to print:

> print f(23)  # prints [23]
> print f(42)  # prints [42]

so this doesn't solve/explain OP's initial question. A previous post has 
already clarified the reasons for seeing this normal behavior.

bests,
./alex
--
.w( the_mindstorm )p.