On 06/29, Artem Bityutskiy wrote:
>
> On Wed, 2010-06-23 at 21:41 +0200, Oleg Nesterov wrote:
> > On 06/23, Kees Cook wrote:
> > >
> > > @@ -956,7 +957,15 @@ void set_task_comm(struct task_struct *tsk, char *buf)
> > > */
> > > memset(tsk->comm, 0, TASK_COMM_LEN);
> > > wmb();
> >
> > Off-topic. I'd wish I could understand this barrier. Since the lockless
> > reader doesn't do rmb() I don't see how this can help.
>
> This wmb() looks wrong to me as well. To achieve what the comment in
> this function says, it should be smp_wmb() and we should have smp_rmb()
> in the reading side, AFAIU.
>
> > OTOH, I don't
> > understand why it is needed, we never change ->comm[TASK_COMM_LEN-1] == '0'.
>
> I think the idea was that readers can see incomplete names, but not
> messed up names, consisting of old and new ones.
OK, agreed, comm[TASK_COMM_LEN-1] == '0' can't help to avoid the
messed names.
But nether can help smp_rmb() in the reading (lockless) side?
Say,
printk("comm=%s\n", current->comm);
if we add rmb() before printk, it can't make any difference. The lockless
code should do something like
get_comm_lockless(char *to, char *comm)
{
while (*comm++ = *to++)
rmb();
}
to ensure it sees the result of
memset(tsk->comm, 0, TASK_COMM_LEN);
wmb();
strcpy(tsk->comm, buf);
in the right order.
otherwise printk("comm=%s\n", current->comm) can read, say, comm[1]
before set_task_comm->memset(), and comm[0] after set_task_comm->strcpy().
So, afaics, set_task_comm()->wmb() buys nothing and should be removed.
The last zero char in task_struct->comm[] is always here, at least this
guarantees that strcpy(char *dest, tsk->comm) is always safe.
(I cc'ed the expert, Paul can correct me)
Oleg.
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