Add lists to a list lisp

Question

I'm new to Lisp and I'm having trouble figuring out how I add a list to another list. I start with an empty list, and I have to add new lists, each containing three elements. For example,

(add '(1 2 3) '())

would return ((1 2 3)) [let's call it new-list], and adding a new list to this new one, for example

(add '(4 5 6) new-list)

would return ((1 2 3) (4 5 6)) or ((4 5 6) (1 2 3))

I've tried a few different ways, but so far the closest I've come up was ((((1 2 3)) (4 5 6)) (7 8 9))

I was using something like this:

(defun add (lst new-let) (if (null lst) '() (setf new-lst (cons new-lst (cons lst '()))))

I don't quite understand why the first invocation would return ((1 2 3)) instead of ((1 2 3) ()). An explicitly wanted special case? — JB.
– JB., Commented Nov 10, 2013 at 22:24
@JB.: No, I guess it's no big difference. My main doubt is how to get rid of those kind of nested lists... — Rita Pereira
– Rita Pereira, Commented Nov 10, 2013 at 22:32

3uc1id · Accepted Answer · 2013-11-10 22:36:01Z

2

Have you tried :

(defun add (thing lst) (append lst (list thing)))

I haven't tried this with Common Lisp as I am more of a Scheme kind of guy, bu I think it would work.

answered Nov 10, 2013 at 22:36

3uc1id

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JB. · Accepted Answer · 2013-11-10 22:38:33Z

1

The way I read it, your requirement is exactly cons (non destructive) or push (destructive).

answered Nov 10, 2013 at 22:38

JB.

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2 Comments

Joshua Taylor Over a year ago

Except that you can't (push '(1 2 3) '()), and in (defun add (x y) (push x y)) only modifies y, not the list that it contains. I.e., (let ((a (list 1 2 3)) (b (list 4 5 6))) (add a b) b) returns (4 5 6), not ((1 2 3) (4 5 6)) or ((1 2 3) 4 5 6).

JB. Over a year ago

Hello semantic pedantry. Obviously the destructive version is to be used on a place, and defuning add on top of it is redundant. The intent is clear enough. There's a reason I linked to the documentation.

Sylwester · Accepted Answer · 2013-11-10 23:20:36Z

You being new to LISP I'd like to give you an alternative to the accepted answer.

Lists are singly linked list chains. Such structure allows for adding and removing in front to be a constant time operation while adding or removing something from the end would take as many turns as there are elements in the list in both time and space (it will have to recreate the list structure on it's way).

(defun add (element list)
  (cons element list))

This looks very familiar.. It's actually just a wrapper to cons. So lets imagine you have a practical application where you'd like to use add, but needed the elements in the order in your question. One way to do it would then be to finish up first (add whats to add), then do one reverse or nreverse (if every element was made in your function and mutation won't matter for the outside)..

(defun fetch-all (num-elements thunk &optional acc)
  (if (zerop num-elements)
      (nreverse acc)
      (fetch-all (- num-elements 1) thunk (add (funcall thunk) acc)))); add == cons

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