From: Jakub Kicinski <kuba@kernel.org>
To: "Paul E. McKenney" <paulmck@kernel.org>
Cc: Kirill Tkhai <tkhai@ya.ru>,
Linux Kernel Network Developers <netdev@vger.kernel.org>,
davem@davemloft.net, edumazet@google.com, pabeni@redhat.com,
kuniyu@amazon.com, gorcunov@gmail.com
Subject: Re: [PATCH net-next] unix: Guarantee sk_state relevance in case of it was assigned by a task on other cpu
Date: Thu, 26 Jan 2023 13:33:22 -0800 [thread overview]
Message-ID: <20230126133322.3bfab5e0@kernel.org> (raw)
In-Reply-To: <20230126202511.GL2948950@paulmck-ThinkPad-P17-Gen-1>
On Thu, 26 Jan 2023 12:25:11 -0800 Paul E. McKenney wrote:
> > Me trying to prove that memory ordering is transitive would be 100%
> > speculation. Let's ask Paul instead - is the above valid? Or the fact
> > that CPU1 observes state from CPU0 and is strongly ordered with CPU2
> > implies that CPU2 will also observe CPU0's state?
>
> Hmmm... What is listen() doing? There seem to be a lot of them
> in the kernel.
>
> But proceeding on first principles...
>
> Sometimes. Memory ordering is transitive only when the ordering is
> sufficiently strong.
>
> In this case, I do not see any ordering between CPU 0 and anything else.
> If the listen() function were to acquire the same mutex as CPU1 and CPU2
> did, and if it acquired it first, then CPU2 would be guaranteed to see
> anything CPU0 did while holding that mutex.
The fuller picture would be:
[CPU0] [CPU1] [CPU2]
WRITE_ONCE(sk->sk_state,
TCP_LISTEN);
val = READ_ONCE(sk->sk_state)
mutex_lock()
shared_mem_var = val
mutex_unlock()
mutex_lock()
if (shared_mem_var == TCP_LISTEN)
BUG_ON(READ_ONCE(sk->sk_state)
!= TCP_LISTEN)
mutex_unlock()
> Alternatively, if CPU0 wrote to some memory, and CPU1 read that value
> before releasing the mutex (including possibly before acquiring that
> mutex), then CPU2 would be guaranteed to see that value (or the value
> written by some later write to that same memory) after acquiring that
> mutex.
Which I believe is exactly what happens in the example.
> So here are some things you can count on transitively:
>
> 1. After acquiring a given lock (or mutex or whatever), you will
> see any values written or read prior to any earlier conflicting
> release of that same lock.
>
> 2. After an access with acquire semantics (for example,
> smp_load_acquire()) you will see any values written or read
> prior to any earlier access with release semantics (for example,
> smp_store_release()).
>
> Or in all cases, you might see later values, in case someone else also
> did a write to the location in question.
>
> Does that help, or am I missing a turn in there somewhere?
Very much so, thank you!
next prev parent reply other threads:[~2023-01-26 21:33 UTC|newest]
Thread overview: 11+ messages / expand[flat|nested] mbox.gz Atom feed top
2023-01-22 22:21 [PATCH net-next] unix: Guarantee sk_state relevance in case of it was assigned by a task on other cpu Kirill Tkhai
2023-01-24 17:57 ` Kuniyuki Iwashima
2023-01-24 21:05 ` Kirill Tkhai
2023-01-24 22:32 ` Cyrill Gorcunov
2023-01-25 1:35 ` Jakub Kicinski
2023-01-25 21:09 ` Kirill Tkhai
2023-01-26 6:10 ` Jakub Kicinski
2023-01-26 20:25 ` Paul E. McKenney
2023-01-26 21:33 ` Jakub Kicinski [this message]
2023-01-26 21:47 ` Paul E. McKenney
2023-02-02 19:42 ` Kirill Tkhai
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