Hi Alex, Op di 31 jul. 2018 om 17:34 schreef Alexander Burger <a...@software-lab.de>:
> > But 'cut' is *not* a destructive function. It does not modify the list in > any > way. Only the variable which points to the list is modified. > OK. That isn't totally obvious, but of course you are right as shown per this example: : (setq X '(1 2 3 4 5 6 7 8)) > -> (1 2 3 4 5 6 7 8) > : (setq S X) > -> (1 2 3 4 5 6 7 8) ; Now S points to same memory location as X > : (cut 3 'S) > -> (1 2 3) > : S > -> (4 5 6 7 8) ; Now S points to the 4th element of that same memory > location > : X > -> (1 2 3 4 5 6 7 8) ; X still points to the same memory location Ergo: Alex is totally right :) > In general, all destructive functions are marked as such in the reference. > Good to get some feedback and find out how it *really* works :) Thx very much! Arie
