[TUHS] Reading UNIX V6: Does eash process has its own user

[jigsaw]

hi Brantley,

Now I start to understand what's going on.

But do you mean 0744 by 0743?
0743    mov    (sp), r0
0744    mov    $_u, r0

And 2230 should be 2229, which is:
2229    sureg();

Thanks  &
Regards,

Qinglai

On 10/3/06, Brantley Coile <brantley at coraid.com> wrote:
> Rp->p_addr is the address of the swappable image in core.  The process
> image begins with the user segment for that process.  Line 0743 maps
> the upage into the current address space (KISA6) and _retu loads
> previously saved sp and r5 from there.  Notice that on line 2230
> Ken reloads the other memory mapping registers.
>
> Read the section `Memory Management' starting on page 2-4 for background
> on this.
>
>
> U.u_rsave is just a constant location in memory.  Notice that rp->p_addr
> isn't a byte address but a core click address in units of 64 bytes.
>
> Hope that helps.
>
>  Brantley
>
> > The final question is about how savu/retu work.
> >
> > savu:
> > line 0729 and line 0730: r5 and sp are saved to (r0) and (r0)+, which
> > are the address of u.u_rsav.
> >
> > retu:
> > 0746 and 0747: sp and r5 are read from (r0) and (r0)+, which is
> > "rp->p_addr" (see line 2228). It looks weird to me. (Okay...I have to
> > confess I look stupid here...) When making call to retu, why bother
> > "retu(rp->p_addr)"? Why not calling with "retu(u.u_rsav)"? Does it
> > mean that rp->p_addr == u.u_rsav?
>
>