5
$\begingroup$

I´m not exactly sure what kind of impact these two things have on kernel design. Are servers more flexible because of running in user space, for instance, and does adding a module require a new build of a kernel?

|cite|improve this question
$\endgroup$
  • 1
    $\begingroup$ Welcome! I am unsure as to the scope of your question. Are you sure it is a CS question, that is refers to operating system (theory) in general? Or would you rather discuss implementation details of specific OSs? $\endgroup$ – Raphael♦ Sep 30 '12 at 16:51
  • 1
    $\begingroup$ It might be more suitable for superuser.com. $\endgroup$ – Dave Clarke Sep 30 '12 at 17:15
  • 3
    $\begingroup$ @DaveClarke I'm pretty sure that Super User would close this question as too theoretical, they're rather focused on “how do I do X”-type questions. Unix & Linux is more open to “how does this work under the hood”, but even there I think the question would be considered borderline. I think this question is ok here as a general question on OS design. $\endgroup$ – Gilles Sep 30 '12 at 21:51
  • $\begingroup$ @Raphael I'm just looking into OS theory, not a specific OS. I'd say it falls within the field of computer science, since it's a general question Gilles stated. $\endgroup$ – 1000 Gibibit Oct 1 '12 at 7:07
  • $\begingroup$ I'm certainly happy for it to stay here. $\endgroup$ – Dave Clarke Oct 1 '12 at 7:20
4
$\begingroup$

Basically, a module runs at the same memory space than the kernel and a server runs in a different one.

Although a module can surely be added/removed on-the-fly (linux kernel does that), any misbehavior caused by it may affect the entire system, whether in the micro-kernel architecture, only the server gets compromised.

|cite|improve this answer
$\endgroup$
  • 1
    $\begingroup$ Rather than "memory space", one should probably say "protection domain". In addition, the protection is not absolute; a server's misbehavior can become a denial of service attack or exploit bugs in other code that would not manifest under normal provision of the server's functionality. $\endgroup$ – Paul A. Clayton Jan 30 at 14:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.