Red@reddthat.com to linuxmemes@lemmy.worldEnglish · 1 year agoCode interviews for a PHP developer rolesreddthat.comimagemessage-square128fedilinkarrow-up1350arrow-down122
arrow-up1328arrow-down1imageCode interviews for a PHP developer rolesreddthat.comRed@reddthat.com to linuxmemes@lemmy.worldEnglish · 1 year agomessage-square128fedilink
minus-square_thebrain_@sh.itjust.workslinkfedilinkarrow-up7·1 year agoNot one person in the comments has attempted to answer any of the questions either.
minus-squarethemusicman@lemmy.worldlinkfedilinkarrow-up15·1 year agoHaha good try. Hope your interview goes well
minus-squarebasdiljhs@lemmy.worldlinkfedilinkarrow-up14·1 year 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-squaremoog@lemm.eelinkfedilinkarrow-up5·1 year agoThank you holy shit I was beginning to think no one has ever seen a fizz buzz before
minus-squareLostXOR@fedia.iolinkfedilinkarrow-up5arrow-down1·1 year 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-down1·1 year agoStrictly speaking this one does not find the odd numbers, it just prints them.
minus-squareGoun@lemmy.mllinkfedilinkarrow-up16arrow-down3·1 year agofor (i%1=0; i+2; int) odd++; cout(3)
minus-squareBolt@lemmy.worldlinkfedilinkarrow-up1·1 year ago(0..=100).filter(|n| n % 2 == 1).for_each(|n| println!("{n}"))
minus-squareI Cast Fist@programming.devcakelinkfedilinkarrow-up1·1 year agoWill you give me the position if I answer the problems? 😀
minus-square_thebrain_@sh.itjust.workslinkfedilinkarrow-up1·1 year 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.