Re: [PATCH bpf-next] seg6: Ensure that seg6_bpf_srh_states can only be accessed from input_action_end_bpf()

On 2024-07-08 17:03:58 [-0700], Martin KaFai Lau wrote:

quoted

diff --git a/net/core/filter.c b/net/core/filter.c
index 403d23faf22e1..ea5bc4a4a6a23 100644
--- a/net/core/filter.c
+++ b/net/core/filter.c

@@ -6459,6 +6459,8 @@ BPF_CALL_4(bpf_lwt_seg6_store_bytes, struct sk_buff *, skb, u32, offset,
  	void *srh_tlvs, *srh_end, *ptr;
  	int srhoff = 0;
+	if (!bpf_net_ctx_seg6_state_avail())
+		return -EINVAL;

The syzbot stack shows that the seg6local bpf_prog can be run by test_run
like: bpf_prog_test_run_skb() => bpf_test_run(). "return -EINVAL;" will
reject and break the existing bpf prog doing test with test_run.

But wouldn't this be the case anyway because seg6_bpf_srh_states::srh
isn't assigned?

bpf_test_run() has already done the local_bh_disable() and
bpf_net_ctx_set(). How about doing the
local_[un]lock_nested_bh(&seg6_bpf_srh_states.bh_lock) in bpf_test_run()
when the prog->type == BPF_PROG_TYPE_LWT_SEG6LOCAL?

Okay. Sure. And I assume it is limited that only those two call paths
can invoke this type of BPF program.

Sebastian