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re: Unconditional Challenge: FizzBuzz without `if` VIEW POST

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re: function map(compare, say, next = v => v) { return function(value) { return [() => say, next][Math.sign(compare(value))](value) ...
 

Slightly shorter without bothering to call out for variable potential conditions and using anons:


let map = (compare, say, next = v => v) => v => [() => say, next][Math.sign(v % compare)](v)
const process = map(15, "FizzBuzz", map(5, "Buzz", map(3, "Fizz")))

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