Re: Missing a write memory barrier in tls_init()

[Sabrina Dubroca]

2023-11-07, 18:53:24 -0800, Jakub Kicinski wrote:
> On Tue, 7 Nov 2023 23:45:46 +0100 Sabrina Dubroca wrote:
> > Wouldn't it be enough to just move the rcu_assign_pointer after ctx is
> > fully initialized, ie just before update_sk_prot? also clearer wrt
> > RCU.
> 
> I'm not sure, IIUC rcu_assign_pointer() is equivalent to
> WRITE_ONCE() on any sane architecture, it depends on address
> dependencies to provide ordering.

Not what the doc says:

    /**
     * rcu_assign_pointer() - assign to RCU-protected pointer
     [...]
     * Inserts memory barriers on architectures that require them
     * (which is most of them), and also prevents the compiler from
     * reordering the code that initializes the structure after the pointer
     * assignment.
     [...]
     */

And it uses smp_store_release (unless writing NULL).


rcu_dereference is the one that usually doesn't contain a barrier:

    /**
     * rcu_dereference_check() - rcu_dereference with debug checking
     [...]
     * Inserts memory barriers on architectures that require them
     * (currently only the Alpha), prevents the compiler from refetching
     * (and from merging fetches), and, more importantly, documents exactly
     * which pointers are protected by RCU and checks that the pointer is
     * annotated as __rcu.
     */


> Since here we care about
> ctx->sk_prot being updated, when changes to sk->sk_prot
> are visible there is no super-obvious address dependency.
> 
> There may be one. But to me at least it isn't an obvious
> "RCU used right will handle this" case.

Ok, I think you're right. Looking at smp_store_release used by rcu_assign_pointer:

    #define __smp_store_release(p, v)					\
    do {									\
    	compiletime_assert_atomic_type(*p);				\
    	barrier();							\
    	WRITE_ONCE(*p, v);						\
    } while (0)

it's only going to make sure ctx->sk_proto is set when ctx is visible,
and not guarantee that ctx is visible whenever sk->sk_prot has been
switched over.

-- 
Sabrina