A style question
in [Lisp]
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From: justinhj on 28 Feb 2007 10:51 On Feb 27, 11:31 pm, job-271842...(a)craigslist.org wrote: > justinhj wrote: > > (defun multiple(x n) > > (= 0 (mod x n))) > > > (defun output-multiple(x n str) > > (if (and (multiple x n) (princ str)) > > 1 > > 0)) > > > (defun fizzbuzz(n) > > (loop for x from 1 to n do > > (if (> (+ (output-multiple x 3 "fizz") (output-multiple x 5 > > "buzz")) 0) > > (format t "~%")))) > > Hmm.. this doesn't seem to work. > > [1]> (defun multiple(x n) > (= 0 (mod x n))) > MULTIPLE > [2]> > (defun output-multiple(x n str) > (if (and (multiple x n) (princ str)) > 1 > 0)) > OUTPUT-MULTIPLE > [3]> > (defun fizzbuzz(n) > (loop for x from 1 to n do > (if (> (+ (output-multiple x 3 "fizz") (output-multiple x 5 > "buzz")) 0) > (format t "~%")))) > FIZZBUZZ > [4]> (fizzbuzz 12) > fizz > buzz > fizz > fizz > buzz > fizz > NIL > [5]> Unless I'm misunderstanding something what doesn't work? The output is correct. Justin
From: Barry Fishman on 28 Feb 2007 11:05 "Tim Bradshaw" <tfb+google(a)tfeb.org> writes: > Incidentally, I'm deeply disappointed in the quality of answers in > this thread. In the elder days there would have been at least a few > followups showing how to do this in the proper "FORMAT string > indistinguishable from line noise" way. No true CL programmer ever > uses any other construct when the problem can be solved with a > combination of FORMAT, LOOP & GO (FORMAT being always preferable, > obviously). There may yet be those reading cll who know this, though > I suspect they have all gone into the west now. I'm not a good practitioner of FORMAT, but there needs to be at least one FORMAT solution presented. (defun fizz-buzz (n) (loop for i from 1 to n do (format t "~[~d~;Fizz~;Buzz~;FizzBuzz~]~%" (+ (if (zerop (mod i 3)) 1 0) (if (zerop (mod i 5)) 2 0)) i))) -- Barry Fishman
From: justinhj on 28 Feb 2007 11:31 On Feb 28, 10:51 am, "justinhj" <justi...(a)gmail.com> wrote: > On Feb 27, 11:31 pm, job-271842...(a)craigslist.org wrote: > > > > > justinhj wrote: > > > (defun multiple(x n) > > > (= 0 (mod x n))) > > > > (defun output-multiple(x n str) > > > (if (and (multiple x n) (princ str)) > > > 1 > > > 0)) > > > > (defun fizzbuzz(n) > > > (loop for x from 1 to n do > > > (if (> (+ (output-multiple x 3 "fizz") (output-multiple x 5 > > > "buzz")) 0) > > > (format t "~%")))) > > > Hmm.. this doesn't seem to work. > > > [1]> (defun multiple(x n) > > (= 0 (mod x n))) > > MULTIPLE > > [2]> > > (defun output-multiple(x n str) > > (if (and (multiple x n) (princ str)) > > 1 > > 0)) > > OUTPUT-MULTIPLE > > [3]> > > (defun fizzbuzz(n) > > (loop for x from 1 to n do > > (if (> (+ (output-multiple x 3 "fizz") (output-multiple x 5 > > "buzz")) 0) > > (format t "~%")))) > > FIZZBUZZ > > [4]> (fizzbuzz 12) > > fizz > > buzz > > fizz > > fizz > > buzz > > fizz > > NIL > > [5]> > > Unless I'm misunderstanding something what doesn't work? The output is > correct. > > Justin I'm not having much luck trying to make it elegant and concise but I quite like this recursive version... (defun fizzbuzz2(n) (labels ((fizzbuzz-rec (current last fizz buzz) (let ((output nil)) (if (= 0 fizz) (push "fizz" output)) (if (= 0 buzz) (push "buzz" output)) (if output (format t "~{~a~}~%" output)) (unless (= current last) (fizzbuzz-rec (1+ current) last (mod (1- fizz) 3) (mod (1- buzz) 5)))))) (fizzbuzz-rec 1 n (1- 3) (1- 5)))) CL-USER> (fizzbuzz2 15) fizz buzz fizz fizz buzz fizz buzzfizz Justin
From: Raffael Cavallaro on 28 Feb 2007 11:37 On 2007-02-28 02:33:08 -0500, job-271842874(a)craigslist.org said: > Looks like read macros are involved, but I don't want to get ahead of myself... To clarify, rather than use mod, Rob's version has loop iterate on two circular lists, one representing (mod i 3) the other representing (mod i 5). The #integer= and #integer# notations allow the lisp reader to construct wholly or partially self-referential list structure. Heed Rob's warning about setting *print-circle* to t unless you want a stack overflow.
From: Tim Bradshaw on 28 Feb 2007 11:53
On Feb 28, 3:40 pm, Richard M Kreuter <kreu...(a)progn.net> wrote: > (dotimes (i 100) > (format t "~[~[FizzBuzz~:;Fizz~]~:;~[Buzz~:;~D~]~]~%" (mod i 3) (mod i 5) i)) > > Best I could do on one cup of coffee. Good, but you'd get extra points for avoiding the explicit FizzBuzz in there. Off the top of my head I think: (format t "~&~[Fizz~;~]~[Buzz~;~D~]~%" (mod i 3) (mod i 5) i). But I'd have to check, and it's not really squiggly enough. --tim |