3 ms·
Here's a macro. There's some issues :) The biggest one is you have to make sure to put the multiple divisors list first. The call is (fizzbuzz (100 (((3 5)
by digitailor 15y ago
Here's a macro. There's some issues :) The biggest one is you have to make sure to put the multiple divisors list first.
The call is
(fizzbuzz (100 (((3 5) "fizzbuzz")(3 "fizz")(5 "buzz"))))
(defun make-cond (item)
(let ((divisors (first item))
(exclamation (second item)))
(if (listp divisors)
(let ((div1 (first divisors))
(div2 (second divisors)))
`((and (= 0 (mod n ,div1)) (= 0 (mod n ,div2)))
(format t "~a: ~a~%" n ,exclamation)))
`((= 0 (mod n ,divisors)) ;else
(format t "~a: ~a~%" n ,exclamation)))))
(defmacro fizzbuzz (params-list)
(destructuring-bind (num-to req-list) params-list
(let ((cond-list '(((= 0 n)))))
(loop for item in req-list do
(push (make-cond item) cond-list))
`(dotimes (n ,num-to)
(cond ,@(reverse
(push '(t (format t "~a~%" n))
cond-list)))))))
- ericbb 15y agoHehe. That's quite a monster. I'll try to be a bit more helpful. 1. Use loop instead of dotimes so you can make the index range over 1-100 instead of 0-99. 2. Factor out the string calculation so you only need one print statement. 3. Maybe use zerop. 4. Don't write ")(". Use a space in between. Example: (loop for n from 1 to 100 do (format t "~a~%" (cond ((and (zerop (mod n 3)) (zerop (mod n 5))) "FizzBuzz") ((zerop (mod n 3)) "Fizz") ((zerop (mod n 5)) "Buzz") (t (format nil "~a" n)))))
- digitailor 15y agoThanks! With the macro you can use arbitrary divisors though ;) Any desire to refactor that beast?