retain values between fun calls

[Kent Johnson]

George Sakkis wrote:
> Gary Wessle wrote:
>> Hi
>>
>> the second argument in the functions below suppose to retain its value
>> between function calls, the first does, the second does not and I
>> would like to know why it doesn't? and how to make it so it does?
>>
>> thanks
>>
>> # it does
>> def f(a, L=[]):
>>     L.append(a)
>>     return L
>> print f('a')
>> print f('b')
>>
>>
>> # it does not
>> def f(a, b=1):
>>     b = a + b
>>     return b
>> print f(1)
>> print f(2)
> 
> It's a FAQ:
> http://www.python.org/doc/faq/general/#why-are-default-values-shared-between-objects.
> 
> Whenever you want to control one or more objects beyond the lifetime of
> a single function call, your first thought should be to use a class to
> couple behaviour with state:
> 
> class SomeFancyClassName(object):
>     def __init__(self, b=1):
>         self.b = b
>     def f(self, a):
>         self.b += a
>         return self.b
> 
> x = SomeFancyClassName()
> print x.f(1)
> print x.f(2)

If you make the class callable you can match the original syntax:
In [40]: class F(object):
    ....:     b=1
    ....:     def __call__(self, a):
    ....:         F.b += a
    ....:         return F.b
    ....:
    ....:

In [41]: f=F()

In [42]: f(1)
Out[42]: 2

In [43]: f(2)
Out[43]: 4

Alternately you can use an attribute of the function to save the state:

In [35]: def f(a):
    ....:     f.b += a
    ....:     return f.b
    ....:

In [36]: f.b=1

In [37]: f(1)
Out[37]: 2

In [38]: f(2)
Out[38]: 4

Kent