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Discussion on: FizzBuzz challenge in as many languages as possible

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Avalander

I did write solutions in Javascript and Racket for a similar challenge a while ago.

Since nobody has done Javascript yet, here's a crazy implementation.

const worder = (predicate, patterner) => (prev, n) =>
  predicate(prev, n)
    ? patterner(prev, n)
    : prev

const isDivisible = d => (_, n) =>
  n % d === 0

const isEmpty = s =>
  s.length === 0

const append = x => (prev) => prev + x

const setNumber = (_, n) => n

const fizzer = worder(isDivisible(3), append('fizz'))

const buzzer = worder(isDivisible(5), append('buzz'))

const numberer = worder(isEmpty, setNumber)

const reducer = (...worders) => n =>
  worders.reduce(
    (prev, w) => w(prev, n),
    ''
  )

const fizzbuzzer = reducer(
  fizzer,
  buzzer,
  numberer
)

for (let i = 0; i <= 100; i++) {
  console.log(fizzbuzzer(i))
}

Consider how easy it is to extend to print 'fazz' for multiples of 7.

const fazzer = worder(isDivisible(7), append('fazz'))

const fizzbuzzfazzer = reducer(
  fizzer,
  buzzer,
  fazzer,
  numberer
)

Here's my implementation in Racket

(define (mult-15? n)
  (and (mult-5? n)
       (mult-3? n)))

(define (mult-5? n)
  (= (modulo n 5) 0))

(define (mult-3? n)
  (= (modulo n 3) 0))

(define (fizzbuzz n)
  (cond
    [(mult-15? n) "FizzBuzz"]
    [(mult-5? n) "Buzz"]
    [(mult-3? n) "Fizz"]
    [else n]))

(define (print-list xs)
  (map displayln xs))

(print-list (map fizzbuzz (range 1 101)))

The print-list function is a bit redundant, since (map fizzbuzz (range 1 101)) will already print the resulting list to the console.