On Wed, 09 Feb 2005 14:12:30 -0700, Bob Gailer <[EMAIL PROTECTED]> wrote: > At 01:14 PM 2/9/2005, Jeffrey Lim wrote: > > > >Example 1 > > >>> def f(a,L=[]): > >... if L==[5]: > >... print 'L==[5] caught' > >... print L > >... print 'resetting L...' > >... L=[] > >... L.append(a) > >... return L > >... > > >>> f(5) > >[5] > > >>> f(5) > >L==[5] caught > >[5] > >resetting L... > >[5] > > >>> f(2) > >L==[5] caught > >[5] > >resetting L... > >[2] > > That works as expected. I assume you are happy with the results.
did you even read what i gave as an example properly? So tell me then - why are you happy with the results when 'f(2)' is called? I am not happy with the results. Calling function 'f' with argument 2 should not (unless u are satisfied with such a result) get caught in the 'if L==[5]' > >So when the default value for 'L' is an empty list, the test > >condition, _once triggered_, *always* catches? > > No. What leads you to think that is happening? This code empties the list > each time you invoke it. > pls read http://www.python.org/doc/2.4/tut/node6.html#SECTION006710000000000000000 for the whole context to my questions. Read also my quotes from the doc in my original post. I assume that you can cope with the level of the presentation of my questions. > > > >How is this possible? > > > >Or even consider the example given - > >def f(a, L=None): > > if L is None: > > L = [] > > L.append(a) > > return L > > > >How is 'L == None' even possible all the time, given that for > >def f(a, L=[]): > > L.append(a) > > return L > >, L isn't even [] except for the first call to 'f'? > > From the 1st reference you cited above: 'The default values are evaluated > at the point of function definition" > perhaps you might like to explain that further, seeing as how you seem to be adopting a "I am smarter than you" attitude. -jf _______________________________________________ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
