Messages in this thread | | | Date | Tue, 29 Jun 2010 15:31:31 +0200 | From | Oleg Nesterov <> | Subject | Re: [PATCH] sanitize task->comm to avoid leaking escape codes |
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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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