This function will return a None if it ends though your failsafe else OR
if it ends through a non-zero i.  It is not ending through the failsafe
branch since we do not see the printing of failsafe.

When the i==0 path returns the deck, it is returning it to the i==1
invocation of shuffle. That invocation does NOT have an explicit return,
so it returns None to the i == 2 invocation and so on.

The fix is:
....
     if i >  0:
         print "if i >  0"
         shuffle(i-1, deck)
         return deck
....

Ole Jensen wrote:

>  WAIDW == (W)hat (A)m (I) (D)oing (W)rong
>   
>  (Question put short:
>  Why does this function return; None?
>  ###
>  def shuffle(i=24, deck=[]):
>      a = [ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, jack, queen, king]
>      deck.append(a[:])
>      print deck
>      if i > 0:
>          print "if i > 0"
>          shuffle(i-1, deck)
>      elif i == 0:
>          print "elif i == 0"
>          print deck
>          return deck
>      else: print "failsafe"
>  ###
>   
>  Read below for more detailed describtion of my wrong doings ;)
>  )
>   
>  I'm trying to create a list representing a stack of cards containing six 
>  card decks, I do not really care much of the about colour so my stack 
>  would come to look like this:
>   
>  ###
>  stack = [[ace, 2, 3,....., king], [ace, 2, 3,....., king],........,[ace, 
>  2, 3,...., king]
>  ###
>   
>  The above should symbolise a list (stack) with 24 lists of different 
>  colours (allthough the colours e.g. clubs or hearts, are not shown as such).
>   
>  Now call me lazy but I would prefer not to type all those those 
>  identical lists manually (both easthatically and labour wise), so I saw 
>  this as a good time to try to play around  with recursive functions 
>  (easily done in a forloop with two lines, but now I'm stubborn).
>  The best bit of code I have come up with untill now is this:
>   
>  ###
>  def shuffle(i=24, deck=[]):
>      a = [ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, jack, queen, king]
>      deck.append(a[:])
>      print deck
>      if i > 0:
>          print "if i > 0"
>          shuffle(i-1, deck)
>      elif i == 0:
>          print "elif i == 0"
>          print deck
>          return deck
>      else: print "failsafe"
>  ###
>  (The print statements are for testing)
>   
>  Which of course doesn't work (Otherwise I wouldn't be writing this mail)!
>  The the output ends up like this:
>  [[11, 2, 3, 4, 5, 6, 7, 8, 9, 10, 10, 10, 10]]
>  if i > 0
>  [[11, 2, 3, 4, 5, 6, 7, 8, 9, 10, 10, 10, 10], [11, 2, 3, 4, 5, 6, 7, 8, 
>  9, 10, 10, 10, 10]]
>  if i > 0
>  etc. etc.
>  These few lines above shows me that everything shaping about nicely, the 
>  list is looking the way it's supposed to.
>   
>  In the end of the output i finally equals 0, and the list is printed a 
>  final time looking complete (len(deck) == 24). When the next command in 
>  the sourcecode reads: return deck.
>   
>  ###
>  elif i == 0:
>      print "elif i == 0"
>      print deck
>      return deck
>  ###
>   
>  My question is this:
>  If the list "deck" is as it is supposed to (and it looks to be), then 
>  why does the function return; None?
>   
>  In advance thanks for any enlightment.
>   
>  
>  
>  ------------------------------------------------------------------------
>  
>  _______________________________________________
>  Tutor maillist  -  Tutor@[...].org
>  http://mail.python.org/mailman/listinfo/tutor

-- 
Lloyd Kvam
Venix Corp.
1 Court Street, Suite 378
Lebanon, NH 03766-1358

voice:	603-443-6155
fax:	801-459-9582


_______________________________________________
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