Beginner Question

[Igor Korot]

Steven,

On Thu, Jun 2, 2016 at 1:20 AM, Steven D'Aprano
<steve+comp.lang.python at pearwood.info> wrote:
> On Thursday 02 June 2016 14:21, Igor Korot wrote:
>
>> Hi, guys,
>>
>> On Wed, Jun 1, 2016 at 9:42 PM, boB Stepp <robertvstepp at gmail.com> wrote:
>>> On Wed, Jun 1, 2016 at 7:55 PM, Marcin Rak <mrak at sightlineinnovation.com>
>>> wrote:
>>>> Hi to all
>>>>
>>>> I have a beginner question to which I have not found an answer I was able
>>>> to understand.  Could someone explain why the following program:
>>>>
>>>> def f(a, L=[]):
>>>>     L.append(a)
>>>>     return L
>>>>
>>>> print(f(1))
>>>> print(f(2))
>>>> print(f(3))
>>>>
>>>> gives us the following result:
>>>>
>>>> [1]
>>>> [1,2]
>>>> [1,2,3]
>>>>
>>>> How can this be, if we never catch the returned L when we call it, and we
>>>> never pass it on back to f???
>>
>> I think the OP question here is:
>>
>> Why it is printing the array?
>
> Because he calls the function, then prints the return result.
>
> print(f(1))
>
> calls f(1), which returns [1], then prints [1].
>
> Then he calls:
>
> print(f(2))
>
> which returns [1, 2] (but he expects [2]), then prints it. And so on.
>
>
>> There is no line like:
>>
>> t = f(1)
>> print t
>
> Correct. But there are lines:
>
> print(f(1))
> print(f(2))
> print(f(3))

I think you missed the point.

Compare:

def f(a, L=[]):
     L.append(a)
     return L

print(f(1))
print(f(2))
print(f(3))

vs.

def f(a, L=[]):
     L.append(a)
     return L

t = f(1)
print t
t = f(2)
print t
t = f(3)
print t

For people that comes from C/C++/Java, the first syntax is kind of weird:
you return a value from the function but the caller does not save it anywhere.
Especially since the return is not a basic type and most of them are
not familiar
with scalar vs list context (sorry for the Perl terminology here)

Thank you.


>
>
>
> --
> Steve
>
> --
> https://mail.python.org/mailman/listinfo/python-list