a function in lisp called xtoy which returns a list from x to y

Question

I am very new to lisp. I am trying to write a function named x2y which takes 2 arguments x and y which are integers and return a list of integers which starts from x and ends at y

(defun xtoy (X Y)
  (cond ((> X Y) (list nil))
        ((= X Y) (list Y)))
        (T (append (list X) x2y(+ 1 X) Y)))))

the code was not working which i pasted along with the question but its solved now — Deepak, Oct 23 '13 at 00:14
@shunya: Please not that elisp and common-lisp are **different** languages which both are in the lisp family. If you are going to ask a question please tag with the language the problem is in. — Baggers, Oct 24 '13 at 07:20
@Baggers sorry i just started learning lisp i was not aware that elisp is a different language than common lisp. i removed the tag of elisp now — Deepak, Nov 13 '13 at 04:23
@shunya: No worries! reading my comment again now it seems a bit stern so sorry about that. — Baggers, Nov 13 '13 at 10:48

score 3 · Answer 1 · answered Oct 22 '13 at 20:57

3

In elisp, you're looking for C-hf number-sequence RET.

answered Oct 22 '13 at 20:57

phils

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score 2 · Answer 2 · answered Oct 22 '13 at 21:06

2

The code that you give in your question is in a very messy state. Emacs can highlight parens for you. That was enough for me to fix your code without any debugging:

(defun xtoy (X Y)
  (cond ((> X Y) (list nil))
        ((= X Y) (list Y))
        (t (append (list X) (xtoy (+ 1 X) Y)))))

This is how you turn on highlighting:

(setq show-paren-mode t)

Do most of your editing in *scratch* or ielm - they make things easier.

answered Oct 22 '13 at 21:06

abo-abo

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No, the OP just had a confusion about the invocation syntax, which is `(xtoy (1+ x))` and not `x2y(1+ x)`. I see this confusion extraordinarily frequently in the questions I see here on SO. – C. K. Young Oct 22 '13 at 21:10
@ChrisJester-Young, there's still two extra closing parens in the question code. – abo-abo Oct 22 '13 at 21:18

score 2 · Accepted Answer · answered Oct 22 '13 at 21:35

Starting with abo-abo's version, you can simplify a lot:

1) get rid of (= X Y) and replace (list nil) by nil in (> X Y)

(defun xtoy (X Y)
  (cond ((> X Y) nil)
        (t (append (list X) (xtoy (+ 1 X) Y)))))

2) simplify cond to an ifstatement

(defun xtoy (X Y)
  (if (<= X Y)
    (append (list X) (xtoy (+ 1 X) Y))
    nil))

3) leave out the final nil, because that's what's implicitly returned when the condition doesn't match

(defun xtoy (X Y)
  (if (<= X Y)
    (append (list X) (xtoy (+ 1 X) Y))))

4) use cons instead of append

(defun xtoy (X Y)
  (if (<= X Y)
    (cons X (xtoy (+ 1 X) Y))))

Ah I see there is an elisp and common lisp tags on this question, blimey — Baggers, Oct 22 '13 at 23:37

score 1 · Answer 4 · answered Oct 22 '13 at 21:20

1

If you need help with your function I suggest you use cons instead of list and make it from end to start with an accumulator.

(defun xtoy (from to &optional acc)
  (if (> from to)
      acc
      (xtoy from (- to 1) (cons to acc))))

answered Oct 22 '13 at 21:20

Sylwester

47,942
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a function in lisp called xtoy which returns a list from x to y

4 Answers4