Converting a dotted pair to a two-element list in LISP
common-lisp, lisp
Solution
A cons cell is a structure that has two parts, called its `car` and its `cdr`. The pair `(1 . 2)` is a cons cell whose `car` is `1` and whose `cdr` is `2`. Lists in Lisps are built up from cons cells and `nil`. How this works is described in lots of places, including the answer to Recursive range in Lisp adds a period? A list is either the empty list `()` (also called `nil`), or a cons whose `car` is the first element of the list and whose `cdr` is another list which is the rest of the list. That means that a list
(1 2)
is built of cons cells and `nil` as
(cons 1 (cons 2 nil))
If you've already got `(1 . 2)`, then you can get `1` and `2` with `car` and `cdr`. You'd put them back together as just described. That is,
(let ((x '(1 . 2)))
(cons (car x) (cons (cdr x) nil)))
Alternatively, you could just use `list`:
(let ((x '(1 . 2)))
(list (car x) (cdr x)))
If you want to reuse the same cons cell, you could replace the `cdr` of the cell with `(cons 2 nil)`. For instance (and note that we're not quoting the pair anymore, because modifying literal data is undefined behavior):
(let ((x (cons 1 2)))
(setf (cdr x) (cons (cdr x) nil))
x)
That could also be
(let ((x (cons 1 2)))
(setf (cdr x) (list (cdr x)))
x)
You could also use `rplacd`:
(let ((x (cons 1 2)))
(rplacd x (list (cdr x)))
x)
Problem
I am new to lisp and working on a homework problem to flatten a nested list. I have my funciton working except it needs to 'remove' dotted pairs. So given (1 (2 3) (4 . 5) ((6 7) (89))) my function should output (1 2 3 4 5 6 7 8 9). So.. my actual question.. Given a dotted pair e.g `(1 . 2)`, how can I get the list `'(1 2)`?