Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
173 views
in Technique[技术] by (71.8m points)

Cannot do sum in lisp with do loop

(defun suma (L)
  (setq var 0)
  (do 
      ((i 0 (+ i 1)))
      ((= i (length L)))
    (+ var (nth i L)))
  var)

Why does it always returns 0?

Shouldn't it return sum of list L?

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

+ does not modify its arguments, so, since you never modify var, its initial value of 0 is returned.

You need to replace (+ var (nth i L)) with (incf var (nth i L)), of, equivalently, (setq var (+ var (nth i L))).

See incf.

Note that you should bind var with let instead of making it global with setq.

Most importantly, note that your algorithm is quadratic in the length of the list argument (because nth scans your list every time from the start).

Here are some better implementations:

(defun sum-1 (l)
  (reduce #'+ l))

(defun sum-2 (l)
  (loop for x in l sum x))

(defun sum-3 (l)
  (let ((sum 0))
    (dolist (x l sum)
      (incf sum x))))

Here is a bad implementation:

(defun sum-4 (l)
  (apply #'+ l))

The problem with sum-4 is that it will fail if the length of the supplied list is larger than call-arguments-limit.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

1.4m articles

1.4m replys

5 comments

57.0k users

...