Path: utzoo!utgpu!water!watmath!clyde!att!rutgers!cs.utexas.edu!milano!perseus!rcp
From: rcp@perseus. (Rob Pettengill)
Newsgroups: comp.lang.lisp
Subject: Re: Summing a list
Message-ID: <1457@perseus.>
Date: 26 Oct 88 19:34:15 GMT
References: <1131@amelia.nas.nasa.gov> <5671@eagle.ukc.ac.uk> <10794@srcsip.UUCP>
Reply-To: rcp@perseus.sw.mcc.com.UUCP (Rob Pettengill)
Organization: MCC Software Technology Program
Lines: 54

In article <10794@srcsip.UUCP> rogers@eagle.UUCP (Brynn Rogers) writes:
;
; I Have a silly question.   I need to take the sum of a list 
; of numbers.
; (SETQ NUMLIST '(1 2 3 6 7))
; This is quite simple, really, but what is the BEST way to do it?
; I like: 
; 
; (EVAL (CONS '+ NUMLIST))
;
; But someone else says this would be faster::
;  
; (LET ((SUM 0))
;    (DOLIST (NUM NUMLIST)
;       (SETQ SUM (+ SUM NUM))))  ;; or maybe (INCF SUM NUM)
;
;
;of course, there is always:
;
;(DEFUN SUM (NUMLIST)
;   (COND ((NULL NUMLIST) 0)
;         (T (+ (FIRST NUMLIST) (SUM (CDR NUMLIST))))))
;
;    I think it's obvious that the last would be slow.
;    Also it's Obvious that there are Many ways to do this.
;What is the BEST (for speed and/or elegance) way to sum a list??
;
;(I Run Gold Hills Common 3.0 on a Compaq 386/20 with a 387)
;
;Thanks,    Brynn Rogers   rogers@src.honeywell.com

Why make life difficult?

<cl> (setq numlist '(1 2 3 6 7))

(1 2 3 6 7) 
<cl> (apply #'+ numlist)

19 

this was too easy since #'+ takes an arbitrary number of arguments.
Say it took only two arguments... A general way to handle this in
common lisp is with reduce:

<cl> (reduce #'+ numlist)

19 

;rob

  Robert C. Pettengill, MCC Software Technology Program
  P. O. Box 200195, Austin, Texas  78720
  ARPA:  rcp@mcc.com            PHONE:  (512) 338-3533
  UUCP:  {ihnp4,seismo,harvard,gatech,pyramid}!ut-sally!im4u!milano!rcp