3 ms·
Surely, the lispiest way is to write an interpreter! Write a function fizzbuzz-eval so that the following is a solution (left as an exercise). ;) (fizzbuz
by ericbb 15y ago
Surely, the lispiest way is to write an interpreter! Write a function fizzbuzz-eval so that the following is a solution (left as an exercise). ;)
(fizzbuzz-eval
100
'(((3 5) (print "FizzBuzz"))
((3) (print "Fizz"))
((5) (print "Buzz"))
(() print-number)))
- deleted 15y ago[deleted]
- digitailor 15y agoHere'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?