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Comparing fizzbuzz implementations in Haskell and JavaScript
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fizzes = cycle ["", "", "Fizz"] | |
buzzes = cycle ["", "", "", "", "Buzz"] | |
fbs = zipWith (++) fizzes buzzes | |
nums = map show [1..] | |
fizzbuzz n = take n (zipWith max nums fbs) | |
-- Usage: | |
-- fizzbuzz 3 | |
-- -> ["1", "2", "Fizz"] |
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function* cycle(vals) { | |
let len = vals.length | |
let index = 0 | |
while (true) { | |
yield vals[index % len] | |
index += 1 | |
} | |
} | |
function* zipStreams(...streams) { | |
while (true) { | |
yield streams.map(g => g.next().value) | |
} | |
} | |
function* zipStreamsWith(f, ...streams) { | |
while (true) { | |
yield streams | |
.map(g => g.next().value) | |
.reduce(f) | |
} | |
} | |
function take(n, stream) { | |
let result = [] | |
for (let i = 0; i < n; i++) { | |
result.push(stream.next().value) | |
} | |
return result | |
} | |
function* range(start, stop = Infinity) { | |
let i = start | |
while (i < stop) { | |
yield i | |
i += 1 | |
} | |
} | |
const concat = (a, b) => a + b | |
const longer = (a, b) => String(a).length > String(b).length ? String(a) : String(b) | |
const fizzes = cycle(["", "", "fizz"]) | |
const buzzes = cycle(["", "", "", "", "buzz"]) | |
const fbs = zipStreamsWith(concat, fizzes, buzzes) | |
const nums = range(1) | |
const fizzBuzz = (n) => take(n, zipStreamsWith(longer, nums, fbs)) |
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