Red@reddthat.com to linuxmemes@lemmy.worldEnglish · 7 months agoCode interviews for a PHP developer rolesreddthat.comimagemessage-square123fedilinkarrow-up1336arrow-down120
arrow-up1316arrow-down1imageCode interviews for a PHP developer rolesreddthat.comRed@reddthat.com to linuxmemes@lemmy.worldEnglish · 7 months agomessage-square123fedilink
minus-square_thebrain_@sh.itjust.workslinkfedilinkarrow-up7·7 months agoNot one person in the comments has attempted to answer any of the questions either.
minus-squarethemusicman@lemmy.worldlinkfedilinkarrow-up14·7 months agoHaha good try. Hope your interview goes well
minus-squarebasdiljhs@lemmy.worldlinkfedilinkarrow-up13·7 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-squareLostXOR@fedia.iolinkfedilinkarrow-up5arrow-down1·7 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-squareJeena@jemmy.jeena.netlinkfedilinkarrow-up4arrow-down1·7 months agoStrictly speaking this one does not find the odd numbers, it just prints them.
minus-squaremoog@lemm.eelinkfedilinkarrow-up4·7 months agoThank you holy shit I was beginning to think no one has ever seen a fizz buzz before
minus-squareBolt@lemmy.worldlinkfedilinkarrow-up1·7 months ago(0..=100).filter(|n| n % 2 == 1).for_each(|n| println!("{n}"))
minus-squareI Cast Fist@programming.devlinkfedilinkarrow-up1·7 months agoWill you give me the position if I answer the problems? 😀
minus-square_thebrain_@sh.itjust.workslinkfedilinkarrow-up1·7 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
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.
Thank you holy shit I was beginning to think no one has ever seen a fizz buzz before
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.