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 >> >> > > > > > _______________________________________________ > Tutor maillist - Tutor@python.org > http://mail.python.org/mailman/listinfo/tutor > > _______________________________________________ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
