how use loop when list element only one

[avinash00002002]

I have some list in which has only one element, how to run loop

(("Hole2" . 18)) 

if more than one we can use

(repeat (setq i (1- (length lst1)))

but for one element it gives error.

Thanks,

Avinash

[ВeekeeCZ · Accepted solution]

Because you need to loop it "length-times", not len-1 times

If length of your lst1 is 1, then (1- (length lst1)) is 0, so it runs 0-times

(defun c:test ()
  (setq lst1 '(("Hole2" . 18)))
  (repeat (setq i (length lst1)) ; length of entire list
    (print
      (nth (setq i (1- i)) lst1) ; last index is length-1
      ))  
  (princ)
  )

[calderg1000 · Accepted solution]

Regards @avinash00002002 

When you have a single element in your list, it is not necessary to apply a loop, since in advance you can know that the position it occupies within the list is i = 0
Here I show you some examples to get the only item on your list.

(setq lst'(("Hole2" . 18)))

1. Option
(setq elist(nth 0 lst))

2. Option
(setq elist(car lst))

3. Option
(setq elist(last lst))

4. Option
(repeat (setq i (length lst))
(setq elist (nth (setq i(1- i)) lst)))