Red to linuxmemesEnglish • 8 months agoCode interviews for a PHP developer rolesreddthat.comimagemessage-square128fedilinkarrow-up1350arrow-down122
arrow-up1328arrow-down1imageCode interviews for a PHP developer rolesreddthat.comRed to linuxmemesEnglish • 8 months agomessage-square128fedilink
minus-square@[email protected]linkfedilink7•8 months agoNot one person in the comments has attempted to answer any of the questions either.
minus-square@basdiljhslink14•8 months agofor(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
minus-square@[email protected]linkfedilink5•8 months agoThank you holy shit I was beginning to think no one has ever seen a fizz buzz before
minus-squareLostXORlinkfedilink4•8 months agoSlightly 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); }
minus-squareJeenalinkfedilink3•8 months agoStrictly speaking this one does not find the odd numbers, it just prints them.
minus-squareI Cast Fistlinkfedilink1•8 months agoWill you give me the position if I answer the problems? 😀
minus-square@[email protected]linkfedilink1•8 months agoSure! I’ll hire you without even answering the questions. Of course I’m not the op, I dont work in the it field (any more) and none of my open positions involve programming… But you have a job with my company whenever you need one.
Not one person in the comments has attempted to answer any of the questions either.
Haha good try. Hope your interview goes well
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
Thank you holy shit I was beginning to think no one has ever seen a fizz buzz before
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); }Strictly speaking this one does not find the odd numbers, it just prints them.
for (i%1=0; i+2; int) odd++; cout(3)
(0..=100).filter(|n| n % 2 == 1).for_each(|n| println!("{n}"))Will you give me the position if I answer the problems? 😀
Sure! I’ll hire you without even answering the questions. Of course I’m not the op, I dont work in the it field (any more) and none of my open positions involve programming… But you have a job with my company whenever you need one.