On 09/18, Peter Zijlstra wrote:
>
> On Thu, Sep 17, 2015 at 08:09:19PM +0200, Oleg Nesterov wrote:
>
> > I need to recheck, but afaics this is not possible. This optimization
> > is fine, but probably needs a comment.
>
> For sure, this code doesn't make any sense to me.
So yes, after a sleep I am starting to agree that in theory this fast-path
check is wrong. I'll write another email..
> As an alternative patch, could we not do:
>
> void put_pid(struct pid *pid)
> {
> struct pid_namespace *ns;
>
> if (!pid)
> return;
>
> ns = pid->numbers[pid->level].ns;
> if ((atomic_read(&pid->count) == 1) ||
> atomic_dec_and_test(&pid->count)) {
>
> + smp_read_barrier_depends(); /* ctrl-dep */
Not sure... Firstly it is not clear what this barrier pairs with. And I
have to admit that I can not understand if _CTRL() logic applies here.
The same for atomic_read_ctrl().
OK, please forget about put_pid() for the moment. Suppose we have
X = 1;
synchronize_sched();
Y = 1;
Or
X = 1;
call_rcu_sched( func => { Y = 1; } );
Now. In theory this this code is wrong:
if (Y) {
BUG_ON(X == 0);
}
But this is correct:
if (Y) {
rcu_read_lock_sched();
rcu_read_unlock_sched();
BUG_ON(X == 0);
}
So perhaps something like this
/*
* Comment to explain it is eq to read_lock + read_unlock,
* in a sense that this guarantees a full barrier wrt to
* the previous synchronize_sched().
*/
#define rcu_read_barrier_sched() barrier()
make sense?
And again, I simply can't understand if this code
if (READ_ONCE_CTRL(Y))
BUG_ON(X == 0);
to me it does _not_ look correct in theory.
Oleg.
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