[Tutor] confusion with dictionaries

[Karl Pflästerer]

On 19 Dec 2003, Thomi Richards <- thomi at imail.net.nz wrote:

> I was playing around with dictionaries, and came across something which I 
> found rather... unexpected. here's a demonstration:

>>>> def foo(l = {}):
> ...     l[random.choice(string.lowercase)] = random.random()
> ...     return l
[...]
> I thought that if the dictionary 'l' was not stored in a variable outside the 
> function it would be erased whenever the function foo() was terminated.

No.  Some days ago we had a thread here with exactly the same problem.
Don't you read the list?

I cite from the Python tutorial:

,----[ Python tutorial ]
| *Important warning:*  The default value is evaluated only once.  This
| makes a difference when the default is a mutable object such as a list,
| dictionary, or instances of most classes.  For example, the following
| function accumulates the arguments passed to it on subsequent calls:
| 
|      def f(a, L=[]):
|          L.append(a)
|          return L
|      
|      print f(1)
|      print f(2)
|      print f(3)
| 
| This will print
| 
|      [1]
|      [1, 2]
|      [1, 2, 3]
| 
| If you don't want the default to be shared between subsequent calls,
| you can write the function like this instead:
| 
|      def f(a, L=None):
|          if L is None:
|              L = []
|          L.append(a)
|          return L
`----

So what happens here is that your default value gets only once
evaluated: the time you create the function.  All further calls to the
function which don't give the default value will use exactly the same
dictionary: the one which was created the moment the function definition
was evaluated.

> This is very wierd... can anyone explain this? Furthermore, how can I ensure 

see above.

> that a dictionary is clean when passing it as an optional argument to a 
> function like this?

When you pass a dictionary the default one won't get used.  So there's
no problem.  Or use the technique shown in the tutorial.


   Karl
-- 
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