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submitted 5 months ago by reddthat@reddthat.com to c/linuxmemes@lemmy.world

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[–] _thebrain_@sh.itjust.works 7 points 5 months ago (9 children)

Not one person in the comments has attempted to answer any of the questions either.

[–] basdiljhs@lemmy.world 14 points 5 months ago (2 children)

for(var i=0;i<=100;i++){ if((i%2)==1) console.log(i); }

btw % is the modulo operator, x%y returns the remainder of division of x by y

[–] moog@lemm.ee 5 points 5 months ago

Thank you holy shit I was beginning to think no one has ever seen a fizz buzz before

[–] LostXOR@fedia.io 4 points 5 months ago (1 children)

Slightly simpler, start at 1 and increment by 2 so you don't have to check whether i is odd.

for (var i = 1; i < 100; i += 2) {
  console.log(i);
}

[–] jeena@jemmy.jeena.net 3 points 5 months ago

Strictly speaking this one does not find the odd numbers, it just prints them.

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