Re: [RFC nf-next v3 1/2] netfilter: bpf: support prog update

On 12/23/23 6:23 AM, Alexei Starovoitov wrote:

On Thu, Dec 21, 2023 at 11:06 PM D. Wythe [off-list ref] wrote:

quoted

On 12/21/23 5:11 AM, Alexei Starovoitov wrote:

quoted

On Wed, Dec 20, 2023 at 6:09 AM D. Wythe [off-list ref] wrote:

quoted

From: "D. Wythe" <alibuda@linux.alibaba.com>

To support the prog update, we need to ensure that the prog seen
within the hook is always valid. Considering that hooks are always
protected by rcu_read_lock(), which provide us the ability to
access the prog under rcu.

Signed-off-by: D. Wythe <alibuda@linux.alibaba.com>
---
   net/netfilter/nf_bpf_link.c | 63 ++++++++++++++++++++++++++++++++++-----------
   1 file changed, 48 insertions(+), 15 deletions(-)

diff --git a/net/netfilter/nf_bpf_link.c b/net/netfilter/nf_bpf_link.c
index e502ec0..9bc91d1 100644
--- a/net/netfilter/nf_bpf_link.c
+++ b/net/netfilter/nf_bpf_link.c

@@ -8,17 +8,8 @@
   #include <net/netfilter/nf_bpf_link.h>
   #include <uapi/linux/netfilter_ipv4.h>

-static unsigned int nf_hook_run_bpf(void *bpf_prog, struct sk_buff *skb,
-                                   const struct nf_hook_state *s)
-{
-       const struct bpf_prog *prog = bpf_prog;
-       struct bpf_nf_ctx ctx = {
-               .state = s,
-               .skb = skb,
-       };
-
-       return bpf_prog_run(prog, &ctx);
-}
+/* protect link update in parallel */
+static DEFINE_MUTEX(bpf_nf_mutex);

   struct bpf_nf_link {
          struct bpf_link link;

@@ -26,8 +17,20 @@ struct bpf_nf_link {
          struct net *net;
          u32 dead;
          const struct nf_defrag_hook *defrag_hook;
+       struct rcu_head head;

I have to point out the same issues as before, but
will ask them differently...

Why do you think above rcu_head is necessary?

quoted

   };

+static unsigned int nf_hook_run_bpf(void *bpf_link, struct sk_buff *skb,
+                                   const struct nf_hook_state *s)
+{
+       const struct bpf_nf_link *nf_link = bpf_link;
+       struct bpf_nf_ctx ctx = {
+               .state = s,
+               .skb = skb,
+       };
+       return bpf_prog_run(rcu_dereference_raw(nf_link->link.prog), &ctx);
+}
+
   #if IS_ENABLED(CONFIG_NF_DEFRAG_IPV4) || IS_ENABLED(CONFIG_NF_DEFRAG_IPV6)
   static const struct nf_defrag_hook *
   get_proto_defrag_hook(struct bpf_nf_link *link,

@@ -126,8 +129,7 @@ static void bpf_nf_link_release(struct bpf_link *link)
   static void bpf_nf_link_dealloc(struct bpf_link *link)
   {
          struct bpf_nf_link *nf_link = container_of(link, struct bpf_nf_link, link);
-
-       kfree(nf_link);
+       kfree_rcu(nf_link, head);

Why is this needed ?
Have you looked at tcx_link_lops ?

Introducing rcu_head/kfree_rcu is to address the situation where the
netfilter hooks might
still access the link after bpf_nf_link_dealloc.

Why do you think so?

Hi Alexei,


IMMO, nf_unregister_net_hook does not wait for the completion of the execution of the hook that is being removed,
instead, it allocates a new array without the very hook to replace the old arrayvia rcu_assign_pointer() (in __nf_hook_entries_try_shrink),
then it use call_rcu() to release the old one.

You can find more details in commit 8c873e2199700c2de7dbd5eedb9d90d5f109462b.

In other words, when nf_unregister_net_hook returns, there may still be contexts executing hooks on the
old array, which means that the `link` may still be accessed after nf_unregister_net_hook returns.

And that's the reason why we use kfree_rcu() to release the `link`.

quoted

                                                       nf_hook_run_bpf
                                                       const struct
bpf_nf_link *nf_link = bpf_link;

bpf_nf_link_release
      nf_unregister_net_hook(nf_link->net, &nf_link->hook_ops);

bpf_nf_link_dealloc
      free(link)
bpf_prog_run(link->prog);


I had checked the tcx_link_lops ,it's seems it use the synchronize_rcu()
to solve the

Where do you see such code in tcx_link_lops ?

I'm not certain if the reason that it choose to use synchronize_rcu()is the same as mine,
but I did see it here:


tcx_link_release() -> tcx_entry_sync()


static inline void tcx_entry_sync(void)
{
    /* bpf_mprog_entry got a/b swapped, therefore ensure that
     * there are no inflight users on the old one anymore.
     */
    synchronize_rcu();
}

quoted

same problem, which is also the way we used in the first version.

https://lore.kernel.org/bpf/1702467945-38866-1-git-send-email-alibuda@linux.alibaba.com/ (local)

However, we have received some opposing views, believing that this is a
bit overkill,
so we decided to use kfree_rcu.

https://lore.kernel.org/bpf/20231213222415.GA13818@breakpoint.cc/ (local)

quoted

quoted

   }

   static int bpf_nf_link_detach(struct bpf_link *link)

@@ -162,7 +164,34 @@ static int bpf_nf_link_fill_link_info(const struct bpf_link *link,
   static int bpf_nf_link_update(struct bpf_link *link, struct bpf_prog *new_prog,
                                struct bpf_prog *old_prog)
   {
-       return -EOPNOTSUPP;
+       struct bpf_nf_link *nf_link = container_of(link, struct bpf_nf_link, link);
+       int err = 0;
+
+       mutex_lock(&bpf_nf_mutex);

Why do you need this mutex?
What race does it solve?

To avoid user update a link with differ prog at the same time. I noticed
that sys_bpf()
doesn't seem to prevent being invoked by user at the same time. Have I
missed something?

You're correct that sys_bpf() doesn't lock anything.
But what are you serializing in this bpf_nf_link_update() ?
What will happen if multiple bpf_nf_link_update()
without mutex run on different CPUs in parallel ?

I must admit that it is indeed feasible if we eliminate the mutex and use cmpxchg to swap the prog (we need to ensure that there is only one bpf_prog_put() on the old prog).
However, when cmpxchg fails, it means that this context has not outcompeted the other one, and we have to return a failure. Maybe something like this:

if (!cmpxchg(&link->prog, old_prog, new_prog)) {
    /* already replaced by another link_update */
    return -xxx;
}

As a comparison, The version with the mutex wouldn't encounter this error, every update would succeed. I think that it's too harsh for the user to receive a failure
in that case since they haven't done anything wrong.

Best wishes,
D. Wythe