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