Discussion on: Fizz Buzz in Every Language

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 )

Jeremy Grifski • Apr 25 '19

I appreciate the commitment to the obscure. Haha these are great.

Avalander • Apr 25 '19

That's the whole point of the exercise, right? :D

Jeremy Grifski • Apr 25 '19

Oh absolutely! Got any code golf solutions?

Avalander • Apr 25 '19 • Edited

Hmm... the best I can come up with right now is 85 chars. Nothing really clever, just sacrificed readability for space.

let i=0;while(i++<101){console.log(i%15==0?'fizzbuzz':i%5==0?'buzz':i%3==0?'fizz':i)}

Another fun one, albeit longer, is this.

console.log(new Array(101) .fill(1) .map((_, i) => i % 15 == 0 ? 'fizzbuzz' : i % 3 == 0 ? 'fizz' : i % 5 == 0 ? 'buzz' : i ) .slice(1) .join('\n'))