Pascal J. Bourguignon wrote:
count from 1 to 100
if the number is divizible by three print fizz
if by 5 print bazz
if by both 3 and 5, print fizzbazz
if none of those, print the number
post your lisp solution!
Just put parentheses around your own solution!
(defun divisiblep (n d) (zerop (mod n d)))
(loop for number from 1 to 100
if (divisiblep number 3) do (princ "fizz")
if (divisiblep number 5) do (princ "bazz")
if (not (or (divisiblep number 3) (divisiblep number 5))) do (princ number)
do (terpri))
Gauche Scheme and Racket:
(use srfi-42) ;; do-ec for Gauche
or
(require srfi/42) ;; do-ec for Racket
(require srfi/1) ;; any for Racket
(define (maybe-print n d text)
(and (zero? (modulo n d)) (display text)))
(define (fizz)
(define stuff '((3 fizz) (5 buzz) (7 hum)))
(do-ec (:range i 1 36)
(begin
(or (any values (map (lambda (xs) (apply maybe-print i xs)) stuff))
(display i))
(newline))))
(fizz)
===>
1
2
fizz
4
buzz
fizz
hum
8
fizz
buzz
11
fizz
13
hum
fizzbuzz
16
17
fizz
19
buzz
fizzhum
22
23
fizz
buzz
26
fizz
hum
29
fizzbuzz
31
32
fizz
34
buzzhum
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