[9fans] kproc() bug

[9fans] kproc() bug

[erik quanstrom]

from the last few lines of kproc (port/proc.c:/^kproc)

		ready(p);
		/*
		 *  since the bss/data segments are now shareable,
		 *  any mmu info about this process is now stale
		 *  and has to be discarded.
		 */
		p->newtlb = 1;
		flushmmu();
	}

this looks obviously wrong to me.  you can't ready the proc
before messing with it's tlb bits.  it could (just for one example)
have exited and been cleaned up before p is dereferenced to
set newtlb = 1.

wouldn't this be much safer as

		/*
		 *  since the bss/data segments are now shareable,
		 *  any mmu info about this process is now stale
		 *  and has to be discarded.
		 */
		p->newtlb = 1;
		flushmmu();

		ready(p);
	}

since this looks too obvious, is there something that i'm missing?

- erik

Re: [9fans] kproc() bug

[Charles Forsyth]

>it could (just for one example) have exited and been cleaned up
>before p is dereferenced to set newtlb = 1.

it wouldn't matter since processes aren't deallocated, and an extra newtlb
is at worst inefficient. the kproc doesn't, however, share memory with
the current proc, and i don't think it needs to do anything at all.
(even if memory were shared, a new process starts with an empty mmu state.)
the comment is copied from another site, where it does matter,
but only for up (and flushmmu sets up->newtlb), and indeed there isn't a p->newtlb mention there.
i don't think either statement is needed, but neither will they actually cause harm.

Re: [9fans] kproc() bug

[erik quanstrom]

On Thu May 12 14:22:16 EDT 2011, forsyth@terzarima.net wrote:
> is at worst inefficient. the kproc doesn't, however, share memory with
> the current proc, and i don't think it needs to do anything at all.
> (even if memory were shared, a new process starts with an empty mmu state.)
> the comment is copied from another site, where it does matter,
> but only for up (and flushmmu sets up->newtlb), and indeed there isn't a p->newtlb mention there.
> i don't think either statement is needed, but neither will they actually cause harm.

makes sense.  although i'm paranoid.  i just delete the comment and
both lines.  i'd rather not do random things to a random process, even
if i think i can convince myself that it is at worse inefficient.  :-)

- erik