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

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(a)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? > >