How to Do Modulo in Scheme
How Would I Do the Following in Sicp/Scheme/Dr. Racket? (Define (Even? N) (= (% N 2) 0)) Currently It Seems Like That's Not a Primitive Symbol: %: Unbound...
How would I do the following in sicp/scheme/dr. racket?
(define (even? n) (= (% n 2) 0))
Currently it seems like that's not a primitive symbol: %: unbound identifier in: %.
This may be the stupidest way in the world to do it, but without a % or bitwise-&1 I am doing (without logs or anything else):
(define (even? n)
(if (< (abs n) 2)
(= n 0)
(even? (- n 2))))
2 Answers
mod is modulo in scheme:
(define (even? n)
(= (modulo n 2) 0))
I think it's a good practice to get comfortable writing your own procedures when it feels like they are "missing". You could implement your own mod as -
(define (mod a b)
(if (< a b)
a
(mod (- a b) b)))
(mod 0 3) ; 0
(mod 1 3) ; 1
(mod 2 3) ; 2
(mod 3 3) ; 0
(mod 4 3) ; 1
(mod 5 3) ; 2
(mod 6 3) ; 0
(mod 7 3) ; 1
(mod 8 3) ; 2
But maybe we make it more robust by supporting negative numbers and preventing caller from divi
(define (mod a b)
(if (= b 0)
(error 'mod "division by zero")
(rem (+ b (rem a b)) b)))
(define (rem a b)
(cond ((= b 0)
(error 'rem "division by zero"))
((< b 0)
(rem a (neg b)))
((< a 0)
(neg (rem (neg a) b)))
((< a b)
a)
(else
(rem (- a b) b))))
Maya is a hardware enthusiast who tests and reviews smart home devices, smartphones, wearables, and audio gear. She focuses on practical consumer value and build quality.
