On 24 Apr, 19:54, Arnaud Delobelle <arno...@googlemail.com> wrote: > Mark Tarver <dr.mtar...@ukonline.co.uk> writes: > > Ah; so this > > > def cons (x,y): > > return [x] + y > > > is not accurate? > > Depends what you mean by accurate! > > in lisp, if you do: > > (setq a '(1 2)) > (setq b (cons 0 a)) > (rplaca a 3) > > Then > a is now (3 2) > b is now (0 3 2) > > In Python, if you do: > > a = [1, 2] > b = cons(0, a) # with your definition of cons > a[0] = 3 > > Then > a is now [3, 2] > b is now [0, 1, 2] > > So in this respect, it is not accurate. But that's because Python lists > are vectors not conses. > > -- > Arnaud
Thanks for that. OK; I think I get it. RPLACA and RPLACD are part of the id of Common Lisp which I rarely contemplate. However what it seems to be is that the difference is this. Lisp operates a destructive operation like RPLACA in such a way that RPLACA on a global G not only changes G, but all globals that reference G. Python on the other hand has a destructive operation a[0] = .... which acts a bit like RPLACA but whose change is local to the global changed. I take it that this is because Python essentially copies the list, creating a brand new vector rather than playing with pointers. Is that more or less it? Which brings me to my next question. Assuming that we ban the use of destructive operations like a[0] = ... Lisp's RPLACA and all the rest. Assuming the following Python encodings, and ignoring questions of performance, would Python and Lisp lists then be observationally indistinguishable? i.e. would these then be fair encodings? def car (x): return x[0] def cdr (x): return x[1:] def cons (x,y): return [x] + y Mark -- http://mail.python.org/mailman/listinfo/python-list
