Re: Userapi, Reflection

[Galen Sampson <galen_sampson@xxxxxxxxx>]

Emiel Kollof wrote:
> * Lewis, Todd (todd.lewis@xxxxxx) wrote:
>
> > Maybe I'm misunderstanding, I don't understand why code needs to be in the
> > kernel.  Why not just, at run-time, link in the appropriate
> > syscall-decoder/rewriter depending on the personality of the binary?  New
> > syscalls for linux would just require updating
> > /lib/libpersonality_linux.so...
>
>
> I like this idea. The more we can push into userspace, the better.

  This is the idea.  When you said you would like to "link in the 
appropriate syscall-decoder/rewriter depending on the personality of the 
binary" there are a couple of things that must happen:
  1) The kernel must know which peronality to map in
  2) The kernel must re-arrange the interrupt trap to jump to user space
     instead of the kernel
  3) This must happen for staticly linked programs and dynamically
     linked programs.

> > Also, Matt, when you're talking about 0x80 being a userspace interrupt,
> > would that require rewriting the binary, valgrind-style, or is there some
> > hardware trick that could be used, or is there some other way?  My
> > understanding of x86 is that 0x80 always context-switches into the kernel,
> > but I am a novice in such matters.
>
>
> I don't think that apps need to be rewritten. int 0x80 can always be
> hooked up to do something else (like figure out which syscall is called
> from which personality, and then send the required dragonfly message).
>
> Of course, IANAAH (asm hacker), so I could be completely wrong about
> this.

   "Installing" a personality into the kernel does not make it "part of 
the kernel".  It lets the kernel know where on disk this peronality is, 
which address to map this code to in the process, and which int 0xXX 
call the personality thinks is a system call.  So for linux "installing" 
would be something like "hey kernel, when you see a binary marked 
'linux' you should: map /usr/libexec/peronality/linux.so to (user space) 
0x0b000000, int 0x80 should jump to (user space) 0x0b000000".
   The reason this is different than usual is rtld is only used if the 
binary is dynamically linked.  This means that the kernel must do the 
mapping, not rtld, because statically linked programs do not have rtld 
invoked at all.  The kernel must also determine which personality to 
map, not rtld.  Another reason the kernel must be involved is because 
mapping int 0xXX to an address is a privileged operation, it can't be 
done by a user process.  This will need clarification from someone who 
knows about these things.

Regards,
Galen