this blew me away, thank you for posting it

Awesome! I think you're really getting at the heart of the prompt by definingtrue andfalse as functions.

Nice solution! For hard mode, I think either do it without.filter or show how.filter could be implemented according to the rules.

And here is my object-oriented approach (deeper article on it to come soon)

constbaseBoolean={setThen:function(then){return{...this,then};},setOtherwise:function(otherwise){return{...this,otherwise};},};constobjectOrientedTrue={...baseBoolean,evaluate:function(){returnthis.then;},};constobjectOrientedFalse={...baseBoolean,evaluate:function(){returnthis.otherwise;},};constobjectOrientedNumber={value:0,isaMultipleCache:[objectOrientedFalse],setValue:function(n){return{...this,value:n,isaMultipleCache:[objectOrientedTrue,...Array(n).fill(objectOrientedFalse)]};},isaMultipleOf:function(dividend){returnthis.isaMultipleCache[dividend.value%this.value];}};constobjectOrientedFizzBuzz={for:function(n){constnumber=objectOrientedNumber.setValue(n);returnthis.three.isaMultipleOf(number).setThen(this.five.isaMultipleOf(number).setThen("FizzBuzz").setOtherwise("Fizz").evaluate()).setOtherwise(this.five.isaMultipleOf(number).setThen("Buzz").setOtherwise(number.value).evaluate()).evaluate();},three:objectOrientedNumber.setValue(3),five:objectOrientedNumber.setValue(5),};

Similarly to the functional approach; the main part of this solution is defining "true" and "false" as objects. But now instead of parameters, the objects have the same interface and return one or the other of the attributes set on the object.

As promised, here is my functional approach to solving this (deeper article on it to come soon)

constfunctionalTrue=(onTrue,onFalse)=>onTrue;constfunctionalFalse=(onTrue,onFalse)=>onFalse;constisDivisible=(dividend,divisor)=>[functionalTrue,...Array(divisor).fill(functionalFalse)][dividend%divisor];constfunctionalFizzBuzz=(n)=>{constdivisible_by_three=isDivisible(n,3);constdivisible_by_five=isDivisible(n,5);returndivisible_by_three(divisible_by_five("FizzBuzz","Fizz"),divisible_by_five("Buzz",n));};

Similar to Heiker's solution above; the main part of this solution is defining "true" and "false" as functions taking the same two parameters but returning one or the other.

You can see both solutions in action on this codepen (planning one final post doing a compare/contrast between the two approaches)

Clever! I'll accept it for the normal challenge :)

If you want hard mode though, I think you'll have to show an implementation of.filter that doesn't useif,?:,||,&&,?., or??. (I'll allow.reduce to be used though, as well as.map)

Well done, thanks for commenting!

Well done! Clever use ofMath.sign to select either the first or second element from an array

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

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

It's a nice answer! It satisfies the normal rules (noif or?:)

Thanks for taking the time to submit a solution!

Nice! I like how it usesfalse andtrue as keys on an object to select the resulting string.

If you can get rid of the|| usage, then this will meet the hard mode requirements...

I'm glad you liked it! I think its good to do a little self reflection on the way we do things every once in a while.

And WOW is that an impressive use of destructuring in JS. I knew about each of those things independently; I never have thought about putting them all together like that!