[Tutor] newbie question about default arguments

[Kent Johnson]

Josh Adams wrote:
> Thanks for your help.  That makes a lot more sense.  
> 
> Not to ask too many stupid questions, but why does the L2 assignment in the
> if-block create a new L variable?  Shouldn't the scope from the function
> definition dominate the inner scope of the if-block?

It doesn't create a new variable, it binds a new value to the name L. if 
statements don't introduce a new scope so you are right about that.

Default values for functions are evaluated just once, when the function 
is defined. Rebinding L to a new list inside the function makes each 
execution get a fresh list.

This might help you understand Python name-binding semantics:
http://effbot.org/zone/python-objects.htm

Kent

> 
> Thanks,
> Josh
> 
> 
>>Josh,
>>
>>If you print the id() of your L inside your f2(), you will notice something..
>>In short, the default value stays the same; what you modified was another
>>copy of []. Hope it helps.
>>
>>def f2(a, L=[]):
>>      print "id(L) = ", id(L)
>>      if L==[]:
>>            L=[]
>>            print "id(L2) =", id(L)
>>      L.append(a)
>>      return L
>>
>>
>>>>>print f2(1)
>>
>>id(L)= 11788336
>>id(L2)= 12047184
>>[1]
>>
>>Kenny
>>
>>
> 
> 
> 
> 
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