Bypassing hiding Linux processes?

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1















Hiding Linux Processes



Recently, I've followed this tutorial, which detailed how to hide processes from users without root access, by means of setting hidepid=n on /proc/.



After following this tutorial step-by-step, I found that it is mostly effective at preventing a normal user from viewing other running processes(e.g. htop,top,ps).



Bypass



From messing around, I have also found that I can bypass these restrictions on /proc/ by using loginctl provided by systemd.



For example, if I were to run:



loginctl user-status root


I would be able to see all processes that the root user is running(that is if they are logged in)?



Questions



So, my questions remain:



  • Why is this allowed?

  • How does this work(via direct daemon)?

  • Is this a vulnerability









share|improve this question



















  • 1





    hidepid doesn't hide other evidence. systemd track its own process cgroup status. I don't think this is a vulnerability because it doesn't allow you access important property of root processes

    – 炸鱼薯条德里克
    Feb 7 at 21:33











  • @炸鱼薯条德里克 Why does systemd do this though?

    – NerdOfCode
    Feb 7 at 21:35











  • Because it is designed to manage services and sessions using cgroup, which happens to let you see these processes. It's just a harmless side-effect

    – 炸鱼薯条德里克
    Feb 7 at 21:37












  • Then what's the point of even setting hidepid?

    – NerdOfCode
    Feb 7 at 21:56











  • "it doesn't allow you access important property of root processes " Didn't I I just provide the answer?

    – 炸鱼薯条德里克
    Feb 7 at 21:58















1















Hiding Linux Processes



Recently, I've followed this tutorial, which detailed how to hide processes from users without root access, by means of setting hidepid=n on /proc/.



After following this tutorial step-by-step, I found that it is mostly effective at preventing a normal user from viewing other running processes(e.g. htop,top,ps).



Bypass



From messing around, I have also found that I can bypass these restrictions on /proc/ by using loginctl provided by systemd.



For example, if I were to run:



loginctl user-status root


I would be able to see all processes that the root user is running(that is if they are logged in)?



Questions



So, my questions remain:



  • Why is this allowed?

  • How does this work(via direct daemon)?

  • Is this a vulnerability









share|improve this question



















  • 1





    hidepid doesn't hide other evidence. systemd track its own process cgroup status. I don't think this is a vulnerability because it doesn't allow you access important property of root processes

    – 炸鱼薯条德里克
    Feb 7 at 21:33











  • @炸鱼薯条德里克 Why does systemd do this though?

    – NerdOfCode
    Feb 7 at 21:35











  • Because it is designed to manage services and sessions using cgroup, which happens to let you see these processes. It's just a harmless side-effect

    – 炸鱼薯条德里克
    Feb 7 at 21:37












  • Then what's the point of even setting hidepid?

    – NerdOfCode
    Feb 7 at 21:56











  • "it doesn't allow you access important property of root processes " Didn't I I just provide the answer?

    – 炸鱼薯条德里克
    Feb 7 at 21:58













1












1








1


1






Hiding Linux Processes



Recently, I've followed this tutorial, which detailed how to hide processes from users without root access, by means of setting hidepid=n on /proc/.



After following this tutorial step-by-step, I found that it is mostly effective at preventing a normal user from viewing other running processes(e.g. htop,top,ps).



Bypass



From messing around, I have also found that I can bypass these restrictions on /proc/ by using loginctl provided by systemd.



For example, if I were to run:



loginctl user-status root


I would be able to see all processes that the root user is running(that is if they are logged in)?



Questions



So, my questions remain:



  • Why is this allowed?

  • How does this work(via direct daemon)?

  • Is this a vulnerability









share|improve this question
















Hiding Linux Processes



Recently, I've followed this tutorial, which detailed how to hide processes from users without root access, by means of setting hidepid=n on /proc/.



After following this tutorial step-by-step, I found that it is mostly effective at preventing a normal user from viewing other running processes(e.g. htop,top,ps).



Bypass



From messing around, I have also found that I can bypass these restrictions on /proc/ by using loginctl provided by systemd.



For example, if I were to run:



loginctl user-status root


I would be able to see all processes that the root user is running(that is if they are logged in)?



Questions



So, my questions remain:



  • Why is this allowed?

  • How does this work(via direct daemon)?

  • Is this a vulnerability






permissions vulnerability






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Feb 10 at 15:09









ctrl-alt-delor

11.8k42160




11.8k42160










asked Feb 7 at 21:13









NerdOfCodeNerdOfCode

12315




12315







  • 1





    hidepid doesn't hide other evidence. systemd track its own process cgroup status. I don't think this is a vulnerability because it doesn't allow you access important property of root processes

    – 炸鱼薯条德里克
    Feb 7 at 21:33











  • @炸鱼薯条德里克 Why does systemd do this though?

    – NerdOfCode
    Feb 7 at 21:35











  • Because it is designed to manage services and sessions using cgroup, which happens to let you see these processes. It's just a harmless side-effect

    – 炸鱼薯条德里克
    Feb 7 at 21:37












  • Then what's the point of even setting hidepid?

    – NerdOfCode
    Feb 7 at 21:56











  • "it doesn't allow you access important property of root processes " Didn't I I just provide the answer?

    – 炸鱼薯条德里克
    Feb 7 at 21:58












  • 1





    hidepid doesn't hide other evidence. systemd track its own process cgroup status. I don't think this is a vulnerability because it doesn't allow you access important property of root processes

    – 炸鱼薯条德里克
    Feb 7 at 21:33











  • @炸鱼薯条德里克 Why does systemd do this though?

    – NerdOfCode
    Feb 7 at 21:35











  • Because it is designed to manage services and sessions using cgroup, which happens to let you see these processes. It's just a harmless side-effect

    – 炸鱼薯条德里克
    Feb 7 at 21:37












  • Then what's the point of even setting hidepid?

    – NerdOfCode
    Feb 7 at 21:56











  • "it doesn't allow you access important property of root processes " Didn't I I just provide the answer?

    – 炸鱼薯条德里克
    Feb 7 at 21:58







1




1





hidepid doesn't hide other evidence. systemd track its own process cgroup status. I don't think this is a vulnerability because it doesn't allow you access important property of root processes

– 炸鱼薯条德里克
Feb 7 at 21:33





hidepid doesn't hide other evidence. systemd track its own process cgroup status. I don't think this is a vulnerability because it doesn't allow you access important property of root processes

– 炸鱼薯条德里克
Feb 7 at 21:33













@炸鱼薯条德里克 Why does systemd do this though?

– NerdOfCode
Feb 7 at 21:35





@炸鱼薯条德里克 Why does systemd do this though?

– NerdOfCode
Feb 7 at 21:35













Because it is designed to manage services and sessions using cgroup, which happens to let you see these processes. It's just a harmless side-effect

– 炸鱼薯条德里克
Feb 7 at 21:37






Because it is designed to manage services and sessions using cgroup, which happens to let you see these processes. It's just a harmless side-effect

– 炸鱼薯条德里克
Feb 7 at 21:37














Then what's the point of even setting hidepid?

– NerdOfCode
Feb 7 at 21:56





Then what's the point of even setting hidepid?

– NerdOfCode
Feb 7 at 21:56













"it doesn't allow you access important property of root processes " Didn't I I just provide the answer?

– 炸鱼薯条德里克
Feb 7 at 21:58





"it doesn't allow you access important property of root processes " Didn't I I just provide the answer?

– 炸鱼薯条德里克
Feb 7 at 21:58










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