Red@reddthat.com to linuxmemes@lemmy.worldEnglish · 8 months 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 · 8 months agomessage-square128fedilink
minus-square_thebrain_linkfedilinkarrow-up7·8 months agoNot one person in the comments has attempted to answer any of the questions either.
minus-squarethemusicman@lemmy.worldlinkfedilinkarrow-up15·8 months agoHaha good try. Hope your interview goes well
minus-squarebasdiljhs@lemmy.worldlinkfedilinkarrow-up14·8 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-squaremoog@lemm.eelinkfedilinkarrow-up5·8 months agoThank you holy shit I was beginning to think no one has ever seen a fizz buzz before
minus-squareLostXOR@fedia.iolinkfedilinkarrow-up5arrow-down1·8 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-squareJeena@jemmy.jeena.netlinkfedilinkarrow-up4arrow-down1·8 months agoStrictly speaking this one does not find the odd numbers, it just prints them.
minus-squareGoun@lemmy.mllinkfedilinkarrow-up16arrow-down3·8 months agofor (i%1=0; i+2; int) odd++; cout(3)
minus-squareI Cast Fist@programming.devlinkfedilinkarrow-up1·8 months agoWill you give me the position if I answer the problems? 😀
minus-square_thebrain_linkfedilinkarrow-up1·8 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.
minus-squareBolt@lemmy.worldlinkfedilinkarrow-up1·8 months ago(0..=100).filter(|n| n % 2 == 1).for_each(|n| println!("{n}"))
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)
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.
(0..=100).filter(|n| n % 2 == 1).for_each(|n| println!("{n}"))